NCERT Solutions for Class 12 Chemistry Chapter 2 Solutions

Access premium articles, webinars, resources to make the best decisions for career, course, exams, scholarships, study abroad and much more with

Plan, Prepare & Make the Best Career Choices

# NCERT Solutions for Class 12 Chemistry Chapter 2 Solutions

Edited By Irshad Anwar | Updated on Mar 06, 2023 02:16 PM IST | #CBSE Class 12th

NCERT Solutions for Class 12 Chemistry Chapter 2 Solutions- In NCERT Class 12 Chemistry chapter 2 Solutions, there are direct answers to the 41 questions which are there in the chapter exercise. The students will be able to find step-by-step NCERT solutions for Class 12 Chemistry Chapter 2 Solutions which will eventually help you to write good answers and get good marks in the CBSE exam. The NCERT solutions which are provided here are free of cost and are easily accessible. By referring to the NCERT solutions for class 12, students can understand all the important concepts and practice questions well.

In our daily life, we come across various mixtures like soft drinks, syrups, and air. All of them are mixtures of two or more pure substances like air is a mixture of mainly nitrogen and oxygen etc. Also, you know about various types of mixtures or solutions like gaseous solutions, liquid solutions, and solid solutions. These topics of NCERT solutions for Class 12 Chemistry Chapter 2 Solutions are not only important for the CBSE 12th exam but also for the various competitive exams like JEE Mains, NEET, BITSAT, VITEEE, etc.

NCERT Exemplar Solutions Class 12 Chemistry Chapter 2

## NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions

### NCERT chemistry class 12 intext questions solutions chapter 2: Exercise 2.1 to 2.12

We know that solute and solvent forms solution.

So mass percentage of benzene (solute) :-

$=\frac{22}{22+122}\times100 = \frac{22}{144}\times100 = 15.28\%$

Similarly mass percentage of CCl 4 :-

$=\frac{122}{22+122}\times100 = \frac{122}{144}\times100 = 84.72\%$

For calculating mole fraction, we need moles of both the compounds.

It is given that benzene is $30\%$ in the solution by mass.

So if we consider 100g of solution then 30g is benzene and 70g is CCl 4 .

$Moles of CCl_{4}= \frac{Given\ mass}{Molar\ mass} = \frac{70}{154} = 0.4545\ mol$

$\left ( Molar\ mass\ of\ CCl_4 = 12 + 4(35.5) \right )$

Similarly moles of benzene :

$= \frac{30}{78} = 0.3846$ $\left ( Molar\ mass\ of\ benzene = 6(12) + 6(1) \right )$

So mole fraction of benzene is given :

$= 0.458$

$\inline 30\; g$ of $\inline Co(NO_{3})_{2}.\; 6H_{2}O$ in $\inline 4.3\: L$ of solution

For finding molarity we need the moles of solute and volume of solution.

So moles of solute :

$= \frac{Given\ mass}{Molar\ mass} = \frac{30}{291} = 0.103$

$= 291\ g\ mol^{-1}$

Now, $Molarity = \frac{No.\ of\ moles\ of\ solute}{Volume\ of\ solution\ (l)}$

$= \frac{0.103}{4.3} = 0.023\ M$

$\inline 30\; mL$ of $\inline 0.5\; M$ $\inline H_{2}SO_{4}$ diluted to $\inline 500 \; mL$ .

By conservation of moles we can write :

M 1 V 1 = M 2 V 2

Given that M 1 = 0.5 M and V 1 = 30 ml ; V 2 = 500 ml

$M_2 = \frac{M_1V_1}{V_2} = 0.03\ M$

Let us assume that the mass of urea required be x g.

So moles of urea will be :

$Moles = \frac{Given\ mass}{Molar\ mass} = \frac{x}{60}\ moles$

$Molality = \frac{Moles}{Mass\ of\ solvent\ in\ Kg} = \frac{\frac{x}{60}}{2.5-0.001x} = 0.25$

we get x = 37

Thus mass of urea required = 37 g.

Question 2.5 Calculate

(a) molality

(b) molarity and

(c) mole fraction

of KI if the density of $20\%$ (mass/mass) aqueous KI is $\inline 1.202\; g\; mL^{-1}$ .

If we assume our solution is 100 g. Then according to question, 20 g KI is present and 80 g is water.

So moles of KI :

$=\frac{20}{166}$ $\left ( Molar\ mass = 39+127 = 166\ g\ mol^{-1} \right )$

(a) Molality :-

$Molality = \frac{Moles}{Mass\ of\ solvent\ in\ Kg} = \frac{\frac{20}{166}}{0.08}= 1.506\ m.$

(b) Molarity :-

$Density = \frac{Mass}{Volume}$

$Volume = \frac{Mass}{Density} = \frac{100}{1.202} = 83.19 mL$

$Molarity = \frac{Moles}{Volume(l)} = \frac{\frac{20}{166}}{83.19\times10^{-3}} = 1.45\ M$

(c) Mol fraction :- Moles of water :-

$= \frac{80}{18} = 4.44$

So, mol fraction of KI :-

$= \frac{0.12}{0.12+4.44} = 0.0263$

For finding Henry's constant we need to know about the mole fraction of H 2 S.

Solubility of H 2 S in water is given to be 0.195 m .

i.e., 0.195 moles in 1 Kg of water.

$Moles\: of\: water :=\frac{1000}{18} = 55.55\ moles$

So $x_{H_2S} = Mole\ fraction\ of\ H_2S$

$= \frac{0.195}{0.195+55.55} = 0.0035$

At STP conditions, pressure = 1 atm or 0.987 bar

Equation is : $p_{H_2S} = K_h\times x_{H_2S}$

So we get :

$K_h =\frac{0.987}{0.0035} = 282\ bar$

We know that ,

$p = k_h\times x$

Pressure of CO 2 = 2.5 atm

We know that : $1\ atm = 1.01\times10^5\ Pa$

So, Pressure of CO 2 = $2.53\times10^5$ Pa

By Henry Law we get,

$x = \frac{p}{k_h} = \frac{2.53\times10^5}{1.67\times10^8} = 1.52\times10^{-3}$

Taking density of soda water = 1 g/ml

We get mass of water = 500 g.

So, Moles of water :

$= \frac{500}{18} = 27.78$

Also, $x_{H_2O} = \frac{n_{CO_2}}{n_{H_2O}+n_{CO_2}} \approx \frac{n_{CO_2}}{n_{H_2O}}$

So, moles of CO 2 = 0.042 mol

Using relation of mole and given mass, we get

Mass of CO 2 = 1.848 g.

Let the composition of liquid A (mole fraction) be x A .

So mole fraction of B will be x B = 1 - x A .

Given that, $P^{\circ}_A = 450\ mm\ of\ Hg\ ;\ P^{\circ}_B = 700\ mm\ of\ Hg$

Using Raoult’s law ,

$p_{total} = p^{\circ}_A\ x_A\ +\ p^{\circ} _B\ (1-x_A)$

Putting values of p total and vapour pressure of pure liquids in the above equation, we get :

600 = 450.x A + 700.(1 - x A )

or 600 - 700 = 450x A - 700x A

or x A = 0.4

and x B = 0.6

Now pressure in vapour phase :

$P_A = p^{\circ}_A\ x_A$

= 450(0.4) = 180 mm of Hg

$P_B = p^{\circ}_B\ x_B$

= 700(0.6) = 420 mm of Hg

$Mole\ fraction\ of\ liquid\ A = \frac{P_A}{P_A\ + P_B }$

$= \frac{180}{180\ + 420 } = 0.30$

And mole fraction of liquid B = 0.70

Given that vapour pressure of pure water, $p^{\circ}_w = 23.8\ mm\ of\ Hg$

Moles of water :

$= \frac{850}{18} = 47.22$

Moles of urea :

$= \frac{50}{60} = 0.83$

Let the vapour pressure of water be p w .

By Raoult's law, we get :

$\frac{p^{\circ}_w - p_w}{p^{\circ}_w} = \frac{n_2}{n_1\ + n_2}$

or $\frac{23.8 - p_w}{23.8} = \frac{0.83}{47.22\ + 0.83}=0.0173$

or p w = 23.4 mm of Hg.

Relative lowering :- Hence, the vapour pressure(v.P) of water in the solution = 23.4 mm of Hg

and its relative lowering = 0.0173.

Here we will use the formula :

$\Delta T_b = \frac{K_b\times1000\times w_2}{M_2\times w_1}$

Elevation in temperature = 100 - 99.63 = 0.37

K b = 0.52 ; $Molar\ mass\ of\ sucrose = 11(12) + 22(1) + 11(16) = 342\ g\ mol^{-1}$

Putting all values in above formula, we get :

$w_2 = \frac{0.37\times342\times500}{0.52\times1000}= 121.67\ g$

Thus 121.67 g of sucrose needs to be added.

Elevation in melting point = 1.5 degree celsius.

Here we will use the following equation :

$\Delta T_b = \frac{K_b\times 1000\times w_2}{M_2\times w_1}$

Putting given values in the above equation :

$w_2 = \frac{1.5\times176\times75}{3.9\times1000} = 5.08\ g$

Thus 5.08 ascorbic acid is needed for required condition.

We know that :

$Osmotic\ Pressure = \Pi = \frac{n}{v}RT$

We are given with :-

$Moles\ of\ polymer = \frac{1}{185000}$

Volume, V = 0.45 L

Thus osmotic pressure :

$= \frac{\frac{1}{185000}\times8.314\times10^3\times310}{0.45} = 30.98\ Pa$

NCERT solutions for class 12 chemistry chapter 2 Solutions : Exercises

Solution :- A solution is a homogeneous mixture of two or more non-reacting substances. It has two components :- solute and solvent.

Types of solutions are given below :-

Solution of hydrogen in palladium is such an example in which solute is a gas and solvent is solid.

Q.2.3(i) Define the following terms:

Mole fraction

Mole fraction is defined as the ratio of number of moles of a component and total number of moles in all components.

i.e., $Mole\ fraction = \frac{Number\ of\ moles\ in\ a\ component }{Total\ number\ of\ moles\ in\ all\ components}$

Q.2.3(ii) Define the following terms:

Molality

It is defined as the number of moles of solute dissolved per kg (1000g) of solvent

i.e., $Molality = \frac{Number\ of\ moles\ of\ solute}{Mass\ of\ solvent\ in\ Kg}$

It is independent of temperature.

Q2.3(iii) Define the following terms:

Molarity

Molarity is defined as number of moles of solute dissolved per litre(or 1000ml) of solution.

i.e., $Molarity = \frac{No.\ of\ moles\ of\ solute}{Volume\ of\ solution\ in\ litre}$

It depends on temperature because volume is dependent on temperature.

Q2.3(iv) Define the following terms:

Mass percentage.

Mass percentage is defined as the percentage ratio of mass of one component to the total mass of all the components.

i.e., $Mass\ percentage = \frac{Mass\ of\ a\ component}{Total\ mass\ of\ solution}\times100$

According to given question, in 100 g of solution 68 g is nitric acid and rest is water.

So moles of 68 g HNO 3 :-

$=\ \frac{68}{63} = 1.08$

Density of solution is given to be 1.504.

So volume of 100 g solution becomes :-

$=\ \frac{100}{1.504} = 66.49\ mL$

Thus, molarity of nitric acid is :

$Molarity = \frac{1.08}{\frac{66.49}{1000}} = 16.24\ M$

According to question, $10\%$ mass percentage means in 100 g of solution 10 g glucose is dissolved in 90 g water.

Molar mass of glucose (C 6 H 12 O 6 ) = $180\ g\ mol^{-1}$

So moles of glucose are :

$\frac{10}{180} = 0.056 mol$

$Moles\: of\: water = \frac{90}{18} = 5 mol$

$Molality = \frac{0.056}{0.09} = 0.62\ m$

Mole fraction :-

$\frac{0.056}{0.056+5} = 0.011$

Molarity :- Volume of 100 g solution :

$=\frac{100}{1.2} =83.3\ mL$

$Molarity = \frac{0.056}{83.33\times10^{-3}} =0.67\ M$

Total amount of mixture of Na 2 CO 3 and NaHCO 3 = 1 g.

Let the amount of Na 2 CO 3 be x g.

So the amount of NaHCO 3 will be equal to (1 - x) g.

$Molar\ mass\ of\ Na_2CO_3 = 106\ ;\ molar\ mass\ of\ NaHCO_3 = 84$

Now it is given that it is an equimolar mixture.

So, Moles of Na 2 CO 3 = Moles of NaHCO 3 .

or $\frac{x}{106} = \frac{1-x}{84}$

or x = 0.558 g

So $Moles\ of \ Na_2CO_3 = \frac{0.558}{106} = 0.00526$

and $Moles\ of \ NaHCO_3 = \frac{1 - 0.558}{84} = 0.0053$

It is clear that for 1 mol of Na 2 CO 3 2 mol of HCl is required, similarly for 1 mol of NaHCO 3 1 mol of HCl is required.

So number of moles required of HCl = 2(0.00526) + 0.0053 = 0.01578 mol

It is given that molarity of HCl is 0.1 which means 0.1 mol of HCl in 1l of solution.

Thus required volume :

$= \frac{0.01578}{0.1} = 0.1578\ l = 157.8\ mL$

According to question we have 2 solute,

Solute 1. : $25\%$ of 300 g gives :

$\frac{25}{100}\times300 = 75\ g$

Solute 2. : $40\%$ of 400 g gives :

$\frac{40}{100}\times400 = 160\ g$

So total amount of solute = 75 + 160 = 235 g.

Thus mass percentage of solute is :

$= \frac{235}{700}\times100 = 33.5\%$

and mass percentage of water $= 100 - 33.5 = 66.5\%$

For finding molality we need to find the moles of ethylene glycol.

Moles of ethylene glycol :

$= \frac{222.6}{62}= 3.59\ mol$

We know that :

$Molality = \frac{Moles\ of\ ethylene\ glycol}{Mass\ of\ water}\times100$

$= \frac{3.59}{200}\times100 = 17.95\ m$

Now for molarity :-

Total mass of solution = 200 + 222.6 = 422.6 g

Volume of solution

$= \frac{422.6}{1.072}= 394.22\ mL$

So molarity :-

$= \frac{3.59}{394.22}\times1000= 9.11\ M$

express this in percent by mass

We know that 15 ppm means 15 parts per million.

Required percent by mass :

$= \frac{Mass\ of\ chlorofoam}{Total\ mass}\times100$

$= \frac{15}{10^6}\times100 = 1.5\times10^{-3}\%$

determine the molality of chloroform in the water sample.

Moles of chloroform :

$=\frac{15}{119.5} = 0.1255\ mol$

Mass of water is $10^6$ . (Since contamination is 15 ppm)

So molality will be :

$=\frac{0.1255}{10^6}\times1000 = 1.255\times10^{-4}\ m$

Both alcohol and water individually have strong hydrogen bonds as their force of attraction. When we mix alcohol with water they form solution due to the formation of hydrogen bonds but they are weaker as compared to hydrogen bonds of pure water or pure alcohol.

Thus this solution shows a positive deviation from the ideal behaviour.

It is known that dissolution of gas in a liquid is an exothermic process. So, by Le Chatelier principle we know that equilibrium shifts backwards as we increase temperature in case of exothermic process. Thus gases always tend to be less soluble in liquids as the temperature is raised.

According to Henry's law at a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of solution or liquid.

i.e., p = k h x, Here k h is Henry’s law constant.

Some of its applications are as follows:-

(a) We can increase the solubility of CO 2 in soft drinks, the bottle is sealed under high pressure.

(b) To avoid bends (due to blockage of capillaries) and the toxic effects of high concentrations of nitrogen in the blood, the tanks used by scuba divers are filled with air diluted with helium (11.7% helium, 56.2% nitrogen and 32.1% oxygen).

(c) The partial pressure of oxygen is less at high altitudes than that at the ground level. This leads to low concentrations of oxygen in the blood and tissues climbers. Due to low blood oxygen, climbers become weak and unable to think clearly which are symptoms of a condition known as anoxia.

Using Henry's Law we can write,

$m = k.P$

Putting value in this equation, we get :

$6.56\times10^{-3} = k\times 1$

So, the magnitude of k is $6.56\times10^{-3}$ .

Now, we will again use the above equation for $m = 5.0\times10^{-2}\ g$ .

So the required partial pressure is :-

$p = \frac{m}{k} = \frac{5.0\times10^{-2}}{6.56\times10^{-3}}$

or $p = 7.62\ bar$

Positive and negative deviation: - A non-ideal solution is defined as a solution which does not obey Raoult’s law over the entire range of concentration i.e., $\Delta _{Mix}H \neq 0$ and $\Delta _{Mix}V \neq 0$ . The vapour pressure of these solutions is either higher or lower than that expected by Raoult’s law. If vapour pressure is higher, the solution shows a positive deviation and if it is lower, it shows a negative deviation from Raoult’s law.

Enthalpy relation to positive and negative deviation can be understood from the following example:-

Consider a solution made up of two components - A and B. In the pure state the intermolecular force of attraction between them are A-A and B-B. But when we mix the two, we get a binary solution with molecular interaction A-B.

If A-B interaction is weak than A-A and B-B then enthalpy of reaction will be positive thus reaction will tend to move in a backward direction. Hence molecules in binary solution will have a higher tendency to escape. Thus vapour pressure increases and shows positive deviation from the ideal behaviour.

Similarly, for negative deviation, A-B interaction is stronger than that of A-A and B-B.

It is given that $2\%$ of aq. solution. This means 2 g of non-volatile solute in 98 g of H 2 O.

Also the vapour of water at normal boiling point = 1.013 bar.

Using Raoult's law :

$\frac{p_w^{\circ} - p}{p_w^{\circ}} = \frac{w_2.M_1}{w_1.M_2}$

So we get :

$M_2 = \frac{2\times18\times1.013}{0.009\times98} = 41.35\ g\ mol^{-1}$

Thus the molar mass of non-volatile solute is 41.35 unit.

Vapour pressure of heptane = $p_h^{\circ} = 105.2\ KPa$

and vapour pressure of octane = $p_o^{\circ} = 46.8\ KPa$

Firstly we will find moles of heptane and octane so that we can find vapour pressure of each.

Molar mass of heptane = 7(12) + 16(1) = 100 unit.

and molar mass of octane = 8(12) + 18(1) = 114 unit.

So moles of heptane :

$\frac{26}{100} = 0.26$

and moles of octane :

$\frac{35}{114} = 0.31$

Mole fraction of heptane = 0.456 and mole fraction of octane = 0.544

Now we will find the partial vapour pressure:-

(i) of heptane :- $p_h = 0.456\times105.2 = 47.97\ KPa$

(ii) of octane :- $p_o = 0.544\times46.8 = 25.46\ KPa$

So total pressure of solution = $p_h+p_o$

= 47.97 + 25.46 = 73.43 KPa

It is asked the vapour pressure of 1 molal solution which means 1 mol of solute in 1000 g H 2 O.

Moles in 1000g of water = 55.55 mol. (Since the molecular weight of H 2 O is 18)

Mole fraction of solute :

$\frac{1}{1+55.55} = 0.0177$

Applying the equation :

$\frac{p_w^{\circ} - p}{p_w^{\circ}} = x_2$

or $\frac{12.3 - p}{12.3} = 0.0177$

or $p = 12.083\ KPa$

Thus the vapour pressure of the solution is 12.083 KPa

Let the initial vapour pressure of octane = $p_o^{\circ}$ .

After adding solute to octane, the vapour pressure becomes :

$=\frac{80}{100}\times p_o^{\circ} = 0.8p_o^{\circ}$

Moles of octane :

$= \frac{114}{114} = 1$ $\left ( Molar\ mass\ of\ octane = 8(12) + 18(1) = 114\ g\ mol^{-1} \right )$

Using Raoult's law we get :

$\frac{p_o^{\circ} - p}{p_o^{\circ}} = x_2$

or $\frac{p_o^{\circ} -0.8p_o^{\circ} }{p_o^{\circ}} = \frac{\frac{W}{40}}{\frac{W}{40}+1}$

or $w = 10\ g$

Thus required mass of non-volatile solute = 10g.

molar mass of the solute

In this question we will find molar mass of solute by using Raoult's law .

Let the molar mass of solute is M.

Initially we have 30 g solute and 90 g water.

Moles of water :

$\frac{90}{18} = 5\ mol$

By Raoult's law we have :-

$\frac{p_w^{\circ} - p}{p_w^{\circ}} = \frac{n_2}{n_1+n_2}$

or $\frac{p_w^{\circ} - 2.8}{p_w^{\circ}} = \frac{\frac{30}{M}}{5+\frac{30}{M}} = \frac{30}{5M+30}$

or $\frac{p_w^{\circ} }{2.8} = \frac{5M+30}{5M}$ ------------------------------ (i)

Now we have added 18 g of water more, so the equation becomes:

Moles of H 2 O :

$\frac{90+18}{18} = 6\ mol$

Putting this in above equation we obtain :-

$\frac{p_w^{\circ} - 2.9}{p_w^{\circ}} = \frac{\frac{30}{M}}{6+\frac{30}{M}} = \frac{30}{6M+30}$

or $\frac{p_w^{\circ} }{2.9} = \frac{6M +30}{6M}$ -----------------------------------(ii)

From equation (i) and (ii) we get

M = 23 u

So the molar mass of solute is 23 units.

vapour pressure of water at 298 K.

In the previous part we have calculated the value of molar mass the Raoul's law equation.

$\frac{p_w^{\circ} }{2.8} = \frac{5M+30}{5M}$

Putting M = 23 u in the above equation we get,

$\frac{p_w^{\circ} }{2.8} = \frac{5(23)+30}{5(23) } =\frac{145}{115}$

or $p_w^{\circ} =3.53$

Thus vapour pressure of water = 3.53 kPa.

It is given that freezing point of pure water is 273.15 K.

So, elevation of freezing point = 273.15 - 271 = 2.15 K

$5\%$ solution means 5 g solute in 95 g of water.

Moles of cane sugar :

$= \frac{5}{342} = 0.0146$

Molality :

$= \frac{0.0146}{0.095} = 0.1537$

We also know that - $\Delta T_f = k_f \times m$

or $k_f = \frac{\Delta T_f}{m} = 13.99\ K\ Kg\ mol^{-1}$

Now we will use the above procedure for glucose.

$5\%$ of glucose means 5 g of gluocse in 95 g of H 2 O.

Moles of glucose :

$\frac{5}{180} = 0.0278$

Thus molality :

$= \frac{.0278}{0.095} = 0.2926\ mol\ kg^{-1}$

So, we can find the elevation in freezing point:

$\Delta T_f = k_f \times m$

$= 13.99 \times 0.2926 = 4.09\ K$

Thus freezing point of glucose solution is 273.15 - 4.09 = 269.06 K.

In this question we will use the formula :

$\Delta T_f = k_f \times m$

Firstly for compound AB 2 :-

$M_B = \frac{K_f\times W_b \times 1000}{w_A\times \Delta T_f}$

or $= \frac{5.1\times 1 \times 1000}{20\times 2.3} = 110.87\ g/mol$

Similarly for compound AB 4 :-

$M_B= \frac{5.1\times 1 \times 1000}{20\times 1.3} = 196\ g/mol$

If we assume atomic weight of element A to be x and of element B to be y, then we have :-

x + 2y = 110.87 ----------------- (i)

x + 4y = 196 ----------------- (ii)

Solving both the equations, we get :-

x = 25.59 ; y = 42.6

Hence atomic mass of element A is 25.59u and atomic mass of element B is 42.6u.

According to given conditions we have same solution under same temperature. So we can write :

$\frac{\Pi _1}{C_1} = \frac{\Pi _2}{C_2}$ $\left ( \Pi = C.R.T\ ; RT = \frac{\Pi }{C} \right )$

So, if we put all the given values in above equation, we get

$\frac{4.98}{\frac{36}{180}} = \frac{1.52}{C_2}$

or $C_2 = \frac{1.52\times36}{4.98\times180} = 0.061 M$

Hence the required concentration is 0.061 M.

n-hexane and n-octane

Since both the compounds are alkanes so their mixture has van der Waal force of attraction between compounds.

The binary mixture of these compounds has van der Waal force of attraction between them.

The given compounds will have ion-dipole interaction between them.

Methanol has -OH group and acetone has ketone group. So there will be hydrogen bonding between them.

acetonitrile $\inline (CH_{3}CN)$ and acetone $\inline (C_{3}H_{6}O).$

They will have dipole-dipole interaction since both are polar compounds.

Cyclohexane, KCl, CH3OH, CH3CN.

The order will be : Cyclohexane > CH 3 CN > CH 3 OH > KCl

In this, we have used the fact that like dissolves like.

Since cyclohexane is an alkane so its solubility will be maximum.

phenol

We know the fact that like dissolves in like.

Since phenol is had both polar and non-polar group so it is partially soluble in water.

toluene

Since toluene is a non-polar compound i.e., it doesn't have any polar group so it is insoluble in water. (because water is a polar compound)

formic acid

Since the -OH group in formic acid (polar) can form H-bonds with water thus it is highly soluble in water.

ethylene glycol

Ethylene glycol is an organic compound but is polar in nature. Also, it can form H-bonds with water molecules, thus it is highly soluble in water.

chloroform

Chloroform is a non-polar compound so it is insoluble in water.

pentanol

Pentanol has both polar and non-polar groups so it is partially soluble in water.

We know that, Molality :

$Molality = \frac{Moles\ of\ solute}{Mass\ of\ solvent\ in\ Kg}$

So, for moles of solute we have :

$Moles\ of\ Na^+ = \frac{92}{23} = 4$

Thus, molality :

$= \frac{4}{1} = 4$

Molality of Na + ions is 4m.

We are given, $K_{sp} = 6\times10^{-16}$

The dissociation equation of CuS is given by :-

So, the equation becomes :- $K_{sp} = Cu^{2+}\times {S^{2-}}$

or $K_{sp} = s\times s = s^2$

or $s = 2.45\times 10^{-8}\ M$

Thus maximum molarity of solution is $2.45\times 10^{-8}\ M$ .

Total mass of solution = Mass of aspirin + Mass of acetonitrile = 6.5 + 450 = 456.5 g.

We know that :

$Mass\ percentage = \frac{Mass\ of\ solute }{Mass\ of\ solution}\times100$

So, $Mass\ percentage = \frac{6.5 }{456.5}\times100 = 1.42\%$

Thus the mass percentage of aspirin is $1.42\%$

We are given with molality of the solution, so we need to find the moles of Nalorphene.

Molar mass of nalorphene = 19(12) + 21(1) + 1(14) + 3(16) = 311u.

So moles of nalorphene :

$\frac{1.5\times10^{-3}}{311} = 4.82\times 10^{-6}\ moles$

Molality :

$= \frac{No.\ of\ moles\ of\ solute}{Mass\ of\ solvent\ in\ kg}$

or $1.5\times 10^{-3} = \frac{4.82\times 10^{-6}}{w}$

or $w = 3.2\times 10^{-3}\ Kg$

So the required weight of water is 3.2 g.

Molar mass of benzoic acid = 7(12) + 6(1) + 2(16) = 122u.

We are given with the molarity of solution.

$Molarity = \frac{Moles\ of\ solute}{Volume\ of\ solution\ in\ litre}$

or $0.15 = \frac{Moles\ of\ solute}{\frac{250}{1000}}$

or $Moles\ of\ solute= \frac{0.15\times 250}{1000} = 0.0375\ mol$

So mass of benzoic acid :

$= Moles\ of\ benzoic\ acid \times Molar\ mass$

$=0.0375\times 122 = 4.575\ g$

Hence the required amount of benzoic acid is 4.575 g.

We know that depression in freezing point of water will depend upon the degree of ionisation.

The degree of ionisation will be highest in the case of trifluoroacetic acid as it is most acidic among all three.

The order of degree of ionisation on the basis if acidic nature will be:- Trifluoroacetic acid > Trichloroacetic acid > Acetic acid.

So the depression in freezing point will be reverse of the above order.

Firstly we will find the Vant's Hoff factor the dissociation of given compound.

So we can write, $K_a = \frac{(Ca\times Ca)}{C(1-a)}$

or $K_a = Ca^2$ $(\because a << 1)$

or $a = \sqrt{\frac{K_a}{C}}$

Putting values of K a and C in the last result, we get :

$a = 0.0655$

At equilibrium i = 1 - a + a + a = 1 + a = 1.0655

Now we need to find the moles of the given compound CH 3 CH 2 CHClCOOH.

So, moles =

$\frac{10}{122.5} = 0.0816\ mol$

Thus, molality of the solution :

$= \frac{0.0816\times1000}{250} = 0.3265\ m$

Now we will use :

$\Delta T_f = i\ K_f\ m$

or $= 1.065 \times 1.86 \times 0.3265 = 0.6467\ K$

Firstly we need to calculate molality in order to get vant's hoff factor.

So moles of CH 2 FCOOH :

$\frac{19.5}{78} = 0.25$

We need to assume volume of solution to be nearly equal to 500 mL. (as 500 g water is present)

Now, we know that : $\Delta T_f = i\ K_f\ m$

or $i = \frac{1}{0.93} = 1.0753$

Now for dissociation constant :-

$a = i - 1 = 1.0753 - 1 = 0.0753$

and, $K_a = \frac{Ca^2}{1-a}$

Put values of C and a in the above equation, we get :

$K_a = 3\times10^{-3}$

Firstly we will find number of moles of both water and glucose.

Moles of glucose :

$= \frac{25}{180} = 0.139\ mol$ $(Molar\ mass\ of\ glucose = 6(12)+12(1)+6(16) = 180\ g\ mol^{-1})$

and moles of water :

$= \frac{450}{18} = 25\ mol$ $(Molar\ mass\ of\ water = 18\ g\ mol^{-1})$

Now,

$\frac{p_w^{\circ} - p}{p_w^{\circ}} = \frac{n_g}{n_g + n_w}$

or $\frac{17.535 - p}{17.535} = \frac{0.139}{0.139+ 25}$

or $p = 17.44\ mm\ of\ Hg$

Thus vapour pressure of water after glucose addition = 17.44 mm of Hg

We know that : $P = k\times C$

We are given value of P and k, so C can be found.

$C = \frac{760}{4.27\times10^5} = 178\times 10^{-5}$

Hence solubility of methane in benzene is $178\times 10^{-5}$ .

For calculating partial vapour pressure we need to calculate mole fractions of components.

So number of moles of liquid A :

$= \frac{100}{140} = 0.714$

and moles of liquid B :

$= \frac{1000}{180} = 5.556$

Mole fraction of A (x A ) :

$= \frac{0.714}{0.714+5.556} = 0.114$

and mole fraction of B (x B ) :

$= \frac{5.556}{0.714+5.556} = 0.866$

Now, P total = P A + P B

or $P_{total} = P_A^{\circ}x_A\ + P_B^{\circ}x_B$

or $475 = P_A^{\circ}\times0.114\ + 500\times0.886$

or $P_A^{\circ} = 280.7\ torr$

Thus vapour pressure in solution due to A = $P_A^{\circ}x_A$

$= 280.7 \times0.114 = 32\ torr$

 $\inline 100 \times x_{acetone}$ 0 11.8 23.4 36.0 50.8 58.2 64.5 72.1 $\inline p_{acetone}/mm\; Hg$ 0 54.9 110.1 202.4 322.7 405.9 454.1 521.1 $\inline p_{chloroform}/mm\; Hg$ 632.8 548.1 469.4 359.7 257.7 193.6 161.2 120.7 $p_{total}$ 632.8 603 579.5 562.1 580.4 599.5 615.5 64.18

Plot this data also on the same graph paper. Indicate whether it has positive deviation or negative deviation from the ideal solution.

it has negative deviation from the ideal solution.

Firstly, we will find the no. of moles of the given compounds.

No. of moles of benzene :

$=\frac{80}{78} = 1.026\ mol$

and the no. of moles of toluene :

$=\frac{100}{92} = 1.087\ mol$ .

Now we will find mol fraction of both:-

Mole fraction of benzene :-

$=\frac{1.026}{1.026+1.087} = 0.486$

and mole fraction of toluene :

$=1 - 0.486 = 0.514$

Now,

P total = P b + P t

or $= 50.71\times0.486\ + 32.06\times0.514\ = 24.65 +16.48$

or $= 41.13\ mm\ of\ Hg$

Hence mole fraction of benzene in vapour phase is given by :

$= \frac{24.65}{41.13} = 0.60$

We have been given that the water is in equilibrium with air at a pressure of 10 atm or 7600 mm of Hg.

So the partial pressure of oxygen :

$\frac{20}{100}\times7600 = 1520\ mm\ of\ Hg$

and partial pressure of nitrogen :

$\frac{79}{100}\times7600 = 6004\ mm\ of\ Hg$

Now, by Henry's Law :

$P = K_h.x$

For oxygen :

$x = \frac{1520}{3.30\times10^{7}} = 4.61\times 10^{-5}$

For nitrogen :

$x = \frac{6004}{6.51\times10^{7}} = 9.22\times 10^{-5}$

Hence the mole fraction of nitrogen and oxygen in water is $9.22\times 10^{-5}$ and $4.61\times 10^{-5}$ respectively.

We know that osmotic pressure :

$\Pi = i\ (\frac{n}{v})\ R\ T$

or $\Pi = i\ (\frac{w}{M\ v})\ R\ T$

We have been given the values of osmotic pressure, V, i and T.

So the value of w can be found.

$w = \frac{0.75\times111\times2.5}{2.47\times0.0821\times300}$ $(M = 1\times40 + 2\times 35.5 = 111\ g\ mol^{-1})$

$= 3.42\ g$

Hence 3.42 g CaCl 2 is required.

Dissociation of K 2 SO 4 is as follows :-

It is clear that 3 ions are produced, so the value of i will be 3.

Molecular weight of K 2 SO 4 = 2(39) + 1(32) + 4(16) = 174u.

$\Pi = i\ C\ R\ T$

Putting all the values :-

$\Pi =\frac{3 \times 25 \times 10-3 \times 0.082 \times 298}{174\times2}$

$= 5.27\times10^{-3}\ atm$

More About Class 12 Chemistry Chapter 2 Solutions

NCERT Class 12 Chemistry chapter 2 solutions mainly discuss questions based on liquid solutions and their properties. The NCERT solutions for Class 12 Chemistry chapter 2 Solutions also cover other questions based on important concepts like types of solutions, Raoult's law and Henry's law, the concentration of solutions in different units, solubility, the vapour pressure of liquid solutions, ideal and non-ideal solutions, colligative properties, determination of molar mass and abnormal molar masses.

 $Molality(m)=\frac{Number \:of \:moles \:of \:solute}{Weight\:of\:solvent(kg)}$$Molarity(N)=\frac{Number \:of \:moles \:of \:solute}{Volume\:of\:solution(kg)}$$Normality(N)=\frac{Number \:of \:gram\:equivalent \:of \:solute}{Volume\:of\:solution(kg)}$

Topics and Sub-topics of NCERT Class 12 Chemistry Chapter 2 Solutions-

2.1Types of Solutions

2.2 Expressing Concentration of Solutions

2.3 Solubility

2.4 Vapour Pressure of Liquid Solutions

2.5 Ideal and Non-ideal Solutions

2.6 Colligative Properties and Determination of Molar Mass

2.7 Abnormal Molar Masses

This chapter of class 12 NCERT solutions is the second chapter of NCERT Class 12. It basically introduces basic concepts related to concentration, molarity, molality, mole fraction, Raoult's law, henry law, vapour pressure, colligative properties etc. NCERT solutions for Class 12 Chemistry Chapter 2 is an extension of chapter 1 class 11 NCERT also known as some basic concepts of chemistry. Class 12 NCERT solutions is very useful in the subsequent chapters like thermodynamics and Equilibrium etc. Ch 2 Chemistry Class 12 is very easy if basic concepts are understood well. Students can score decent marks in this chapter as most of the questions are form colligative properties and vapour pressure concepts which are easy to comprehend in the examination. Apart from NCERT, students can refer class notes for Chemistry Class 12 Chapter 2 to revise and score well in the final board examination as well as competitive exams.

Class 12 NCERT solutions has good amount of weightage in exams like NEET and JEE as well. Class 12 Chemistry Chapter 2 solutions is must read for the 12th class students. Most of the concepts like Molarity, molality, mole fraction, percentage composition etc. have been discussed in class 11 only, hence it is not very difficult to grasp the subsequent topics Henrys law, Colligative properties. Hence it is generally recommended to study class 11 chapters well before entering class 12th as heavy portion of syllabus is interlinked. Ch 2 Chemistry Class 12 solutions will take 8-10 hours to complete if all the concepts of 11th class are well read.

### NCERT Exemplar Class 12 Solutions

 NCERT Exemplar Class 12 Chemistry Solutions NCERT Exemplar Class 12 Mathematics Solutions NCERT Exemplar Class 12 Biology Solutions NCERT Exemplar Class 12 Physics Solutions

NCERT solutions for class 12 chemistry

 Chapter 1 Chapter 2 Solutions Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16

NCERT solutions for class 12 subject wise

Benefits of NCERT Solutions for Class 12 Chemistry Chapter 2 Solutions

• Hope you have understood well with the help of the free NCERT class 12 Chemistry chapter 2 solutions provided here.
• After completing NCERT solutions for class 12 chemistry chapter 2 Solutions, students will be able to differentiate between the types of solutions characteristics of ideal and non-ideal solutions, define solubility and colligative properties, understand abnormal molar mass.
• NCERT class 12 Chemistry chapter 2 pdf which you read here will also help you in building your concepts as well as a strong base in the subject. These will also help you in various competitive exams.
• With the help of these NCERT class 12 Chemistry chapter 2 solutions pdf download, you will be able to write answers well.

### Also Check NCERT Books and NCERT Syllabus here:

Happy Learning!!

1. What are the important topics of this chapter?
• Henry’s law
• Solubility of gases in liquids
• Colligative properties
• Raoult's law
• Relative lowering of vapour pressure
• Elevation of boiling point
• Osmotic pressure
• Abnormal molecular mass
• Van't Hoff factor
2. Where can I find complete solutions of NCERT syllabus Class 12 Chemistry?
3. What is the weightage of NCERT book Class 12 Chemistry chapter 2 in JEE Mains?

4 marks questions can be expected. Practice JEE Main previous year papers to understand the question type asked.

4. What is the weightage of NCERT class 12 Chemistry chapter 2 in NEET?

Weightage of solutions is 5% . For good score follow NCERT textbook and solve NEET previous year papers.

## Upcoming School Exams

Application Date:16 March,2023 - 29 September,2023

#### International Olympiad of English Language

Application Date:16 March,2023 - 29 September,2023

Application Date:16 March,2023 - 29 September,2023

Application Date:16 March,2023 - 29 September,2023

#### National Means Cum-Merit Scholarship

Application Date:16 August,2023 - 09 October,2023

## Certifications By Top Providers

Full-Stack Web Development
Via Masai School
Master Data Management for Beginners
Via TCS iON Digital Learning Hub
Via State Bank of India
Edx
1110 courses
Coursera
783 courses
Udemy
328 courses
Futurelearn
136 courses
IBM
86 courses

## Explore Top Universities Across Globe

University of Essex, Colchester
Wivenhoe Park Colchester CO4 3SQ
University College London, London
Gower Street, London, WC1E 6BT
The University of Edinburgh, Edinburgh
Old College, South Bridge, Edinburgh, Post Code EH8 9YL
University of Nottingham, Nottingham
University Park, Nottingham NG7 2RD
Lancaster University, Lancaster
Bailrigg, Lancaster LA1 4YW
Bristol Baptist College, Bristol
The Promenade, Clifton Down, Bristol BS8 3NJ
MCAT Exam Pattern 2023 (Section-wise) - Check Here!
4 minSep 14, 2023 14:09 PM IST
5 minSep 14, 2023 13:09 PM IST
What is SAT
3 minSep 14, 2023 13:09 PM IST
New SAT Format 2023 - What You Need to Know
4 minSep 14, 2023 13:09 PM IST

### Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Hello student,

Hope you are in good health, so the answer to your question is as follows:-

First of all please get registered to delhi university admissions and fill the application fees.

Then try to calculate your best 4 according to the course .

As you have not mentioned which category you belong to , I am assuming you to be a general category student. So ,suppose  if your best four came out to be  somewhere like 89 , so if I see the last years cutoff for every BSC course the last cutoff comes out  to be 89 , 87 where you can get bsc home science and bsc life science according to last year cutoff , but it always the chance of matter  every year , cutoff always surprise us so you have to look every cutoff and when you find that you are getting the cutoff for that particular course do visit immediately to that college with all the correct papers and do the admission right  away .

For further details you can go through the following link:-

Hope this helps!

Hello pallavi,
Your percentage is decent for good college in du university for  your desired stream..

Check this college predictor to check in which colleges the chances of your admission are high

https://university.careers360.com/delhi-university-college-predictor

Best of luck!!!
Hello dear,
Your percentage is decent for good college in du university for  your desired stream..

Check this college predictor to check in which colleges the chances of your admission are high

https://university.careers360.com/delhi-university-college-predictor

Best of luck!!!

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

Good Luck

Hello dear,

If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.

As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.

Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.

Believe in Yourself! You can make anything happen

All the very best.

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9

Database professionals use software to store and organise data such as financial information, and customer shipping records. Individuals who opt for a career as data administrators ensure that data is available for users and secured from unauthorised sales. DB administrators may work in various types of industries. It may involve computer systems design, service firms, insurance companies, banks and hospitals.

4 Jobs Available
##### Bio Medical Engineer

The field of biomedical engineering opens up a universe of expert chances. An Individual in the biomedical engineering career path work in the field of engineering as well as medicine, in order to find out solutions to common problems of the two fields. The biomedical engineering job opportunities are to collaborate with doctors and researchers to develop medical systems, equipment, or devices that can solve clinical problems. Here we will be discussing jobs after biomedical engineering, how to get a job in biomedical engineering, biomedical engineering scope, and salary.

4 Jobs Available
##### GIS Expert

GIS officer work on various GIS software to conduct a study and gather spatial and non-spatial information. GIS experts update the GIS data and maintain it. The databases include aerial or satellite imagery, latitudinal and longitudinal coordinates, and manually digitized images of maps. In a career as GIS expert, one is responsible for creating online and mobile maps.

3 Jobs Available
##### Database Architect

If you are intrigued by the programming world and are interested in developing communications networks then a career as database architect may be a good option for you. Data architect roles and responsibilities include building design models for data communication networks. Wide Area Networks (WANs), local area networks (LANs), and intranets are included in the database networks. It is expected that database architects will have in-depth knowledge of a company's business to develop a network to fulfil the requirements of the organisation. Stay tuned as we look at the larger picture and give you more information on what is db architecture, why you should pursue database architecture, what to expect from such a degree and what your job opportunities will be after graduation. Here, we will be discussing how to become a data architect. Students can visit NIT Trichy, IIT Kharagpur, JMI New Delhi

3 Jobs Available
##### Geothermal Engineer

Individuals who opt for a career as geothermal engineers are the professionals involved in the processing of geothermal energy. The responsibilities of geothermal engineers may vary depending on the workplace location. Those who work in fields design facilities to process and distribute geothermal energy. They oversee the functioning of machinery used in the field.

3 Jobs Available
##### Geotechnical engineer

The role of geotechnical engineer starts with reviewing the projects needed to define the required material properties. The work responsibilities are followed by a site investigation of rock, soil, fault distribution and bedrock properties on and below an area of interest. The investigation is aimed to improve the ground engineering design and determine their engineering properties that include how they will interact with, on or in a proposed construction.

The role of geotechnical engineer in mining includes designing and determining the type of foundations, earthworks, and or pavement subgrades required for the intended man-made structures to be made. Geotechnical engineering jobs are involved in earthen and concrete dam construction projects, working under a range of normal and extreme loading conditions.

3 Jobs Available
##### Graphic Designer

Within the graphic design and graphic arts industry, a graphic designer is a specialist who designs and builds images, graphic design, or visual effects to develop a piece of artwork. In career as graphic designer, individuals primarily generate the graphics for publishing houses and printed or electronic digital media like pamphlets and commercials. There are various options for industrial graphic design employment. Graphic design career includes providing numerous opportunities in the media industry.

3 Jobs Available
##### Cartographer

How fascinating it is to represent the whole world on just a piece of paper or a sphere. With the help of maps, we are able to represent the real world on a much smaller scale. Individuals who opt for a career as a cartographer are those who make maps. But, cartography is not just limited to maps, it is about a mixture of art, science, and technology. As a cartographer, not only you will create maps but use various geodetic surveys and remote sensing systems to measure, analyse, and create different maps for political, cultural or educational purposes.

3 Jobs Available
##### Risk Management Specialist

Individuals who opt for a career as a risk management specialist are professionals who are responsible for identifying risks involved in business that may include loss of assets, property, personnel or cash flow. Credit risk manager responsibilities are to identifies business opportunities and eliminates issues related to insurance or safety that may cause property litigation. A risk management specialist is responsible for increasing benefits.

4 Jobs Available
##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
##### Securities Broker

A career as a securities broker is filled with excitement and plenty of responsibilities. One cannot afford to miss out on the details. These types of brokers explain to their clients the complex details related to the securities or the stock market. Choosing to become a securities broker is a good career choice especially due to the liberalization as well as economic growth. There are several companies and organizations in India which hire a securities broker. If you are also thinking of making a career in this field then continue reading the article, it will answer all your questions related to the field.

3 Jobs Available
##### Bank Probationary Officer (PO)

A career as Bank Probationary Officer (PO) is seen as a promising career opportunity and a white-collar career. Each year aspirants take the Bank PO exam. This career provides plenty of career development and opportunities for a successful banking future. If you have more questions about a career as  Bank Probationary Officer (PO), what is probationary officer or how to become a Bank Probationary Officer (PO) then you can read the article and clear all your doubts.

3 Jobs Available
##### Credit Manager

Credit Management refers to the process of granting credit, setting the terms it’s granted on, recovering the credit when it’s due, and confirming compliance with the organization's credit policy, among other credit-related operations. Individuals who opt for a career as Credit Manager should have hands-on experience with accounting software, a solid understanding of lending procedures, excellent analytical skills with the ability to create and process financial spreadsheets, negotiation skills, and a bachelor’s or master’s degree in a field relevant to finance or accounting. Ultimately, Credit Management job is to help organizations minimize bad debts and increase revenues from the loan.

3 Jobs Available
##### Investment Banker

An Investment Banking career involves the invention and generation of capital for other organizations, governments, and other entities. Individuals who opt for a career as Investment Bankers are the head of a team dedicated to raising capital by issuing bonds. Investment bankers are termed as the experts who have their fingers on the pulse of the current financial and investing climate. Students can pursue various Investment Banker courses, such as Banking and Insurance, and Economics to opt for an Investment Banking career path.

3 Jobs Available
##### Insurance Analyst

In the career as an insurance analyst, one can monitor the choices the customers make about which insurance policy options best suit their requirements. They research and make recommendations that have a real impact on the financial well-being of a client down the road. Insurance companies are helping people prepare themselves for the long term. Insurance Analysts find the documents of the claim and perform a thorough investigation, like travelling to places where the incident has occurred, gathering evidence, and working with law enforcement officers.

3 Jobs Available
##### Finance Executive

A career as a Finance Executive requires one to be responsible for monitoring an organisation's income, investments and expenses to create and evaluate financial reports. His or her role involves performing audits, invoices, and budget preparations. He or she manages accounting activities, bank reconciliations, and payable and receivable accounts.

3 Jobs Available
##### Transportation Planner

A career as Transportation Planner requires technical application of science and technology in engineering, particularly the concepts, equipment and technologies involved in the production of products and services. In fields like land use, infrastructure review, ecological standards and street design, he or she considers issues of health, environment and performance. A Transportation Planner assigns resources for implementing and designing programmes. He or she is responsible for assessing needs, preparing plans and forecasts and compliance with regulations.

3 Jobs Available
##### Naval Architect

A Naval Architect is a professional who designs, produces and repairs safe and sea-worthy surfaces or underwater structures. A Naval Architect stays involved in creating and designing ships, ferries, submarines and yachts with implementation of various principles such as gravity, ideal hull form, buoyancy and stability.

2 Jobs Available
##### Welding Engineer

Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues.

2 Jobs Available
##### Field Surveyor

Are you searching for a Field Surveyor Job Description? A Field Surveyor is a professional responsible for conducting field surveys for various places or geographical conditions. He or she collects the required data and information as per the instructions given by senior officials.

2 Jobs Available
##### Highway Engineer

Highway Engineer Job Description: A Highway Engineer is a civil engineer who specialised in planning and building thousands of miles of roads that support connectivity and allow transportation across the country. He or she ensures that traffic management schemes are effectively planned concerning economic sustainability and successful implementation.

2 Jobs Available
##### Conservation Architect

A Conservation Architect is a professional responsible for conserving and restoring buildings or monuments having a historic value. He or she applies techniques to document and stabilise the object’s state without any further damage. A Conservation Architect restores the monuments and heritage buildings to bring them back to their original state.

2 Jobs Available
##### Safety Manager

A Safety Manager is a professional responsible for employee’s safety at work. He or she plans, implements and oversees the company’s employee safety. A Safety Manager ensures compliance and adherence to Occupational Health and Safety (OHS) guidelines.

2 Jobs Available

A Team Leader is a professional responsible for guiding, monitoring and leading the entire group. He or she is responsible for motivating team members by providing a pleasant work environment to them and inspiring positive communication. A Team Leader contributes to the achievement of the organisation’s goals. He or she improves the confidence, product knowledge and communication skills of the team members and empowers them.

2 Jobs Available
##### Orthotist and Prosthetist

Orthotists and Prosthetists are professionals who provide aid to patients with disabilities. They fix them to artificial limbs (prosthetics) and help them to regain stability. There are times when people lose their limbs in an accident. In some other occasions, they are born without a limb or orthopaedic impairment. Orthotists and prosthetists play a crucial role in their lives with fixing them to assistive devices and provide mobility.

6 Jobs Available
##### Veterinary Doctor

A veterinary doctor is a medical professional with a degree in veterinary science. The veterinary science qualification is the minimum requirement to become a veterinary doctor. There are numerous veterinary science courses offered by various institutes. He or she is employed at zoos to ensure they are provided with good health facilities and medical care to improve their life expectancy.

5 Jobs Available
##### Pathologist

A career in pathology in India is filled with several responsibilities as it is a medical branch and affects human lives. The demand for pathologists has been increasing over the past few years as people are getting more aware of different diseases. Not only that, but an increase in population and lifestyle changes have also contributed to the increase in a pathologist’s demand. The pathology careers provide an extremely huge number of opportunities and if you want to be a part of the medical field you can consider being a pathologist. If you want to know more about a career in pathology in India then continue reading this article.

5 Jobs Available
##### Gynaecologist

Gynaecology can be defined as the study of the female body. The job outlook for gynaecology is excellent since there is evergreen demand for one because of their responsibility of dealing with not only women’s health but also fertility and pregnancy issues. Although most women prefer to have a women obstetrician gynaecologist as their doctor, men also explore a career as a gynaecologist and there are ample amounts of male doctors in the field who are gynaecologists and aid women during delivery and childbirth.

4 Jobs Available
##### Ophthalmic Medical Technician

Ophthalmic technician careers are one of the booming careers option available in the field of healthcare. Being a part of this field as an ophthalmic medical technician can provide several career opportunities for an individual. With advancing technology the job of individuals who opt for a career as ophthalmic medical technicians have become of even more importance as he or she is required to assist the ophthalmologist in using different types of machinery. If you want to know more about the field and what are the several job opportunities, work environment, just about anything continues reading the article and all your questions shall be answered.

3 Jobs Available

People might think that a radiation therapist only spends most of his/her time in a radiation operation unit but that’s not the case. In reality, a radiation therapist’s job is not as easy as it seems. The job of radiation therapist requires him/her to be attentive, hardworking, and dedicated to his/her work hours. A radiation therapist is on his/her feet for a long duration and might be required to lift or turn disabled patients. Because a career as a radiation therapist involves working with radiation and radioactive material, a radiation therapist is required to follow the safety procedures in order to make sure that he/she is not exposed to a potentially harmful amount of radiation.

3 Jobs Available
##### Recreational Worker

A recreational worker is a professional who designs and leads activities to provide assistance to people to adopt a healthy lifestyle. He or she instructs physical exercises and games to have fun and improve fitness. A recreational worker may work in summer camps, fitness and recreational sports centres, nature parks, nursing care facilities, and other settings. He or she may lead crafts, sports, music, games, drama and other activities.

3 Jobs Available
##### Paediatrician

A career as paediatrician has emerged as one of India's most popular career choices. By choosing a career as paediatrician, not only in India but also overseas, one can find lucrative work profiles as demand for talented and professional paediatricians is increasing day by day. If you are passionate about children and have the patience to evaluate and diagnose their issues, you may have a good career as paediatricians. Paediatricians take care of children's physical, mental and emotional health from infancy to adolescence.

3 Jobs Available
##### Actor

For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs.

4 Jobs Available
##### Acrobat

Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

3 Jobs Available
##### Video Game Designer

Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages. Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

3 Jobs Available
##### Talent Agent

The career as a Talent Agent is filled with responsibilities. A Talent Agent is someone who is involved in the pre-production process of the film. It is a very busy job for a Talent Agent but as and when an individual gains experience and progresses in the career he or she can have people assisting him or her in work. Depending on one’s responsibilities, number of clients and experience he or she may also have to lead a team and work with juniors under him or her in a talent agency. In order to know more about the job of a talent agent continue reading the article.

If you want to know more about talent agent meaning, how to become a Talent Agent, or Talent Agent job description then continue reading this article.

3 Jobs Available

Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

3 Jobs Available
##### Screenwriter

Are you searching for a screenwriter job description? Individuals in the screenwriter career path are professionals who work in the TV and film industry. Screenwriter career description includes producing original stories with characters, plots and dialogues. A screenwriter develops content for visual media. Individuals in the screenwriter career path produce screenplays for long and short films, television, advertisements, and movies and games. They establish the dialogue, the characters, and the narrative of a screenplay. They may also turn a book into a screenplay for a television show or movie.

2 Jobs Available
##### Multimedia Specialist

A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications.

2 Jobs Available
##### Visual Communication Designer

Individuals who want to opt for a career as a Visual Communication Designer will work in the graphic design and arts industry. Every sector in the modern age is using visuals to connect with people, clients, or customers. This career involves art and technology and candidates who want to pursue their career as visual communication designer has a great scope of career opportunity.

2 Jobs Available
##### Copy Writer

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook.

5 Jobs Available
##### Editor

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

3 Jobs Available
##### Journalist

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available
##### News Anchor

A career as news anchor requires to be working closely with reporters to collect information, broadcast newscasts and interview guests throughout the day. A news anchor job description is to track the latest affairs and present news stories in an insightful, meaningful and impartial manner to the public. A news anchor in India needs to be updated on the news of the day. He or she even works with the news director to pick stories to air, taking into consideration the interests of the viewer.

3 Jobs Available
##### Publisher

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

3 Jobs Available
##### Vlogger

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. Ever since internet cost got reduced the viewership for these types of content has increased on a large scale. Therefore, the career as vlogger has a lot to offer. If you want to know more about the career as vlogger, how to become a vlogger, so on and so forth then continue reading the article. Students can visit Jamia Millia Islamia, Asian College of Journalism, Indian Institute of Mass Communication to pursue journalism degrees.

3 Jobs Available
##### Videographer

Careers in videography are art that can be defined as a creative and interpretive process that culminates in the authorship of an original work of art rather than a simple recording of a simple event. It would be wrong to portrait it as a subcategory of photography, rather photography is one of the crafts used in videographer jobs in addition to technical skills like organization, management, interpretation, and image-manipulation techniques. Students pursue Visual Media, Film, Television, Digital Video Production to opt for a videographer career path. The visual impacts of a film are driven by the creative decisions taken in videography jobs. Individuals who opt for a career as a videographer are involved in the entire lifecycle of a film and production.

2 Jobs Available
##### SEO Analyst

An SEO Analyst is a web professional who is proficient in the implementation of SEO strategies to target more keywords to improve the reach of the content on search engines. He or she provides support to acquire the goals and success of the client’s campaigns.

2 Jobs Available
##### Gemologist

A career as a gemologist is as magnificent and sparkling as gemstones. A gemologist is a professional who has knowledge and understanding of gemology and he or she applies the same knowledge in his everyday work responsibilities. He or she grades gemstones using various equipment and determines its worth. His or her other work responsibilities involve settling gemstones in jewellery, polishing and examining it.

4 Jobs Available
##### Production Manager

Production Manager Job Description: A Production Manager is responsible for ensuring smooth running of manufacturing processes in an efficient manner. He or she plans and organises production schedules. The role of Production Manager involves estimation, negotiation on budget and timescales with the clients and managers.

3 Jobs Available
##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
##### Merchandiser

A career as a merchandiser requires one to promote specific products and services of one or different brands, to increase the in-house sales of the store. Merchandising job focuses on enticing the customers to enter the store and hence increasing their chances of buying a product. Although the buyer is the one who selects the lines, it all depends on the merchandiser on how much money a buyer will spend, how many lines will be purchased, and what will be the quantity of those lines. In a career as merchandiser, one is required to closely work with the display staff in order to decide in what way a product would be displayed so that sales can be maximised. In small brands or local retail stores, a merchandiser is responsible for both merchandising and buying.

2 Jobs Available
##### Quality Systems Manager

A Quality Systems Manager is a professional responsible for developing strategies, processes, policies, standards and systems concerning the company as well as operations of its supply chain. It includes auditing to ensure compliance. It could also be carried out by a third party.

2 Jobs Available
##### Production Planner

Individuals who opt for a career as a production planner are professionals who are responsible for ensuring goods manufactured by the employing company are cost-effective and meets quality specifications including ensuring the availability of ready to distribute stock in a timely fashion manner.

2 Jobs Available
##### Procurement Manager

The procurement Manager is also known as  Purchasing Manager. The role of Procurement Manager is to source products and services for a company. Procurement Managers are involved in developing a purchasing strategy, including the company's budget and the supplies and as well as the vendors who can provide goods and services to the company. His or her ultimate goal is to bring the right products or services at the right time with cost-effectiveness.

2 Jobs Available
##### Metrologist

You might be googling Metrologist meaning. Well, we have an easily understandable Metrologist definition for you. A metrologist is a professional who stays involved in the measurement practices in varying industries including electrical and electronics. Metrologists are responsible for developing processes and systems for measuring objects and repairing electrical instruments. It also involves writing specifications of experimental electronic units.

2 Jobs Available
##### ITSM Manager

ITSM Manager is a professional responsible for heading the ITSM (Information Technology Service Management) or (Information Technology Infrastructure Library) processes. He or she ensures that operation management provides appropriate resource levels for problem resolutions. The ITSM Manager oversees the level of prioritisation for the problems, critical incidents, planned as well as proactive tasks.

3 Jobs Available
##### Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
##### IT Consultant

An IT Consultant is a professional who is also known as a technology consultant. He or she is required to provide consultation to industrial and commercial clients to resolve business and IT problems and acquire optimum growth. An IT consultant can find work by signing up with an IT consultancy firm, or they can work on their own as independent contractors and select the projects they want to work on.

2 Jobs Available
##### Data Architect

A Data Architect role involves formulating the organisational data strategy. It involves data quality, flow of data within the organisation and security of data. The vision of Data Architect provides support to convert business requirements into technical requirements.

2 Jobs Available
##### AI Data Analyst

An AI Data Analyst is responsible for procuring, preparing, cleansing and modelling data utilising machine learning models and new analytical methods. He or she designs and creates data reports in order to provide support to stakeholders to make better decisions.

2 Jobs Available
##### Automation Test Engineer

An Automation Test Engineer job involves executing automated test scripts. He or she identifies the project’s problems and troubleshoots them. The role involves documenting the defect using management tools. He or she works with the application team in order to resolve any issues arising during the testing process.

2 Jobs Available
##### UX Architect

A UX Architect is someone who influences the design processes and its outcomes. He or she possesses a solid understanding of user research, information architecture, interaction design and content strategy.

2 Jobs Available