NCERT Solutions for Class 12 Chemistry Chapter 1

Q: What is a solution?

A solution is a homogeneous mixture composed of two or more substances. In a solution, a solute is dissolved in a solvent. The solute is typically present in a smaller amount, while the solvent is present in a larger amount.

Q: How to get NCERT solutions for Class 12 chemistry Ch 1?

You can get NCERT solutions for Class 12 Chemistry Chapter 1 from various educational websites and online learning platforms. They are also available as free downloadable PDFs, which can be accessed anytime for easy study and revision.

Q: What is Raoult's Law?

Raoult's Law states that the vapor pressure of a solvent in a solution is directly proportional to the mole fraction of the solvent in that solution.

Q: What is the difference between molarity and molality?

Molarity (M) refers to the number of moles of solute per liter of solution, whereas molality (m) is the number of moles of solute per kilogram of solvent. The key difference is that molarity uses the total volume of the solution, while molality considers only the mass of the solvent, making molality independent of temperature.

Q: What is meant by concentration of a solution?

Concentration of a solution indicates how much solute is present in a given amount of solvent or solution. It can be expressed in various ways, such as molarity, molality, mass percentage, or parts per million.

Q: What are NCERT Solutions for Class 12 Chemistry Chapter 1?

NCERT Solutions for Class 12 Chemistry Chapter 1 are detailed answers to all questions provided in the textbook. They explain concepts of the solid state, crystal structures, and related numerical problems in a simple, step-by-step manner to help students understand and practise effectively for exams.

Q: What are NCERT Solutions for Class 12 Chemistry Chapter 1?

NCERT Solutions for Class 12 Chemistry Chapter 1 are detailed answers to all questions provided in the textbook. They explain concepts of the solid state, crystal structures, and related numerical problems in a simple, step-by-step manner to help students understand and practise effectively for exams.

NCERT Solutions for Class 12 Chemistry Chapter 1 - Solutions

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NCERT Solutions for Class 12 Chemistry Chapter 1 - Solutions

NCERT Solutions for Class 12 Chapter 1

Have you ever stirred salt or sugar into water and noticed it completely disappear? This simple observation demonstrates the formation of a solution - a homogeneous mixture of two or more substances. Solutions play a vital role in explaining fascinating scientific phenomena: What causes a solute to dissolve in a solvent? Why does adding salt to water increase its boiling point? How does antifreeze prevent car engines from freezing during winter?

The Class 12 Chemistry NCERT Chapter - Solutions dives deep into these concepts, covering essential topics like the nature of solutions, solubility, colligative properties, Raoult’s Law, and how solutions affect the boiling and freezing points of liquids. Mastering these topics helps students build a strong foundation for board exams, competitive tests, and real-life applications in industries like pharmaceuticals, environmental science, and chemical engineering.

These NCERT Solutions for Class 12 Chemistry provide simple explanations for every question in the textbook. They are designed to help students develop analytical and critical thinking skills. To further enhance logical reasoning and problem-solving ability, carefully selected HOTS questions are also included. Step by step solutions for all exercises are provided below, students can scroll down to access them.

NCERT Solutions for Class 12 Chemistry Chapter 1: Download PDF

Students can download the class 12 chemistry chapter 1 solutions question answer pdf for free. These solutions of NCERT are designed to help you understand the fundamental concepts and solve textbook questions with ease. They give simple, easy answers that help you understand tough topics.

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NCERT Solution for Class 12 Chapter 1 (Intext Questions)

Here are detailed and accurate solutions class 12 question answer for Intext Questions.They help students understand the concepts and make it easier to revise important topics for exams. These NCERT Solutions for Class 12 will help in better concept clarity and exam preparation.

Question 1.1 Calculate the mass percentage of benzene $(C_{6} H_{6})$ and carbon tetrachloride $(C C l_{4})$ if $22 g$ of benzene is dissolved in $122 g$ of carbon tetrachloride.

Answer:

We know that solute and solvent forms solution.

So mass percentage of benzene (solute) :-

$= \frac{22}{22 + 122} \times 100$

= $\frac{22}{144} \times 100$

$= 15.28 %$

Similarly mass percentage of CCl₄ :-

$= \frac{122}{22 + 122} \times 100$

$= \frac{122}{144} \times 100$

$= 84.72 %$

Question 1.2 Calculate the mole fraction of benzene in solution containing $30 %$ by mass in carbon tetrachloride.

Answer :

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 CCl₄ .

Moles of CCl $4 = \frac{Given mass}{Molar mass}$

$= \frac{70}{154} = 0.4545 mol$

$(M o l a r m a s s o f C C l_{4} = 12 + 4 (35.5))$

Similarly moles of benzene :

$= \frac{30}{78}$

$= 0.3846$

(Molar mass of benzene $= 6 (12) + 6 (1))$

So mole fraction of benzene is given :

$= \frac{0.3846}{0.3846 + 0.4545}$

$= 0.458$

Question 1.3(a) Calculate the molarity of each of the following solutions:

$30 g$ of $C o (N O_{3})_{2} . 6 H_{2} O$ in $4.3 L$ of solution

Answer :

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

So moles of solute :

$= \frac{Given mass}{Molar mass}$

$= \frac{30}{291} = 0.103$

Since molar mass of $Co {({NO}_{3})}_{2} \cdot 6 H_{2} O = 59 + 2 (14 + 316) + 6 (16 + 12)$

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

Now, $M o l a r i t y = \frac{N o . o f m o l e s o f s o l u t e}{V o l u m e o f s o l u t i o n (l)}$

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

Question 1.3(b) Calculate the molarity of each of the following solutions:

$30 m L$ of $0.5 M$ $H_{2} S O_{4}$ diluted to $500 m L$ .

Answer :

By conservation of moles we can write :

M₁V₁ = M₂V₂

Given that M₁= 0.5 M and V₁ = 30 ml ; V₂ = 500 ml

$M_{2} = \frac{M_{1} V_{1}}{V_{2}} = 0.03 M$

Question 1.4 Calculate the mass of urea $(N H_{2} C O N H_{2})$ required in making $2.5 k g$ of $0.25 m o l a l$ aqueous solution.

Answer :

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

So moles of urea will be :

$M o l e s = \frac{G i v e n m a s s}{M o l a r m a s s} = \frac{x}{60} m o l e s$

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

$= \frac{\frac{x}{60}}{2.5 - 0.001 x} = 0.25$

we get x = 37

Thus mass of urea required = 37 g.

Question 1.5 Calculate

(a) molality

(b) molarity and

of KI if the density of $20 %$ (mass/mass) aqueous KI is $1.202 g m L^{- 1}$ .

Answer :

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 :

$= \frac{20}{166}$ $(M o l a r m a s s = 39 + 127 = 166 g m o l^{- 1})$

(a) Molality :-

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

$= \frac{\frac{20}{166}}{0.08} = 1.506 m$ .

(b) Molarity :-

$D e n s i t y = \frac{M a s s}{V o l u m e}$

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

$= \frac{100}{1.202} = 83.19 mL$

Molarity $= \frac{Moles}{Volume (l)}$

$= \frac{\frac{20}{166}}{83.19 \times 10^{- 3}} = 1.45 M$

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

So, mol fraction of KI :-

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

Page- 9

Question 1.6 $H_{2} S,$ a toxic gas with rotten egg-like smell, is used for the qualitative analysis. If the solubility of $H_{2} S,$ in water at STP is $0.195 m .$ calculate Henry’s law constant.

Answer :

For finding Henry's constant we need to know about the mole fraction of H₂S.

Solubility of H₂S in water is given to be 0.195 m .

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

$M o l e s o f w a t e r := \frac{1000}{18} = 55.55 m o l e s$

So $x_{H_{2} S} = M o l e f r a c t i o n o f H_{2} S$

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

At STP conditions, pressure = 1 atm or 0.987 bar

Equation is : $p_{H_{2} S} = K_{h} \times x_{H_{2} S}$

So we get :

$K_{h} = \frac{0.987}{0.0035} = 282 b a r$

Question 1.7 Henry’s law constant for $C O_{2}$ in water is $1.67 \times 10^{8} P a$ at $298 K .$ Calculate the quantity of $C O_{2}$ in $500 m L$ of soda water when packed under $2.5 a t m$ $C O_{2}$ pressure at $298 K .$

Answer :

We know that ,

$p = k_{h} \times x$

Pressure of CO₂= 2.5 atm

We know that : $1 a t m = 1.01 \times 10^{5} P a$

So, Pressure of CO₂ = $2.53 \times 10^{5}$ Pa

By Henry Law we get,

$x = \frac{p}{k_{h}}$

$= \frac{2.53 \times 10^{5}}{1.67 \times 10^{8}}$

$= 1.52 \times 10^{- 3}$

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, $x_{H_{2} O} = \frac{n_{C O_{2}}}{n_{H_{2} O} + n_{C O_{2}}} \approx \frac{n_{C O_{2}}}{n_{H_{2} O}}$

So, moles of CO₂= 0.042 mol

Using relation of mole and given mass, we get

Mass of CO₂ = 1.848 g.

Page-15

Question 1.8 The vapour pressure of pure liquids A and B are $450$ and $700 m m H g$ respectively, at $350 K$ . Find out the composition of the liquid mixture if total vapour pressure is $600 m m H g$ . Also find the composition of the vapour phase.

Answer :

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, $P_{A}^{\circ} = 450 m m o f H g; P_{B}^{\circ} = 700 m m o f H g$

Using Raoult’s law ,

$p_{t o t a l} = p_{A}^{\circ} x_{A} + p_{B}^{\circ} (1 - x_{A})$

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 :

$P_{A} = p_{A}^{\circ} x_{A}$

= 450(0.4) = 180 mm of Hg

$P_{B} = p_{B}^{\circ} x_{B}$

= 700(0.6) = 420 mm of Hg

$M o l e f r a c t i o n o f l i q u i d A = \frac{P_{A}}{P_{A} + P_{B}}$

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

And mole fraction of liquid B = 0.70

Page - 23

Question 1.9 Vapour pressure of pure water at $298 K$ is $23.8 m m H g .$ $50 g$ of urea $(N H_{2} C O N H_{2})$ is dissolved in $850 g$ of water. Calculate the vapour pressure of water for this solution and its relative lowering.

Answer :

Given that vapour pressure of pure water, $p_{w}^{\circ} = 23.8 m m o f H g$

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 :

$\frac{p_{w}^{\circ} - p_{w}}{p_{w}^{\circ}} = \frac{n_{2}}{n_{1} + n_{2}}$

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.

Question 1.10 Boiling point of water at $750 m m H g$ is ${99.63}^{\circ} C$ . How much sucrose is to be added to $500 g$ of water such that it boils at $100^{\circ} C .$

Answer :

Here we will use the formula :

$Δ T_{b} = \frac{K_{b} \times 1000 \times w_{2}}{M_{2} \times w_{1}}$

Elevation in temperature = 100 - 99.63 = 0.37

K b = 0.52 ; $M o l a r m a s s o f s u c r o s e = 11 (12) + 22 (1) + 11 (16) = 342 g m o l^{- 1}$

Putting all values in above formula, we get :

$w_{2} = \frac{0.37 \times 342 \times 500}{0.52 \times 1000} = 121.67 g$

Thus 121.67 g of sucrose needs to be added.

Question 1.11 Calculate the mass of ascorbic acid (Vitamin C, $C_{6} H_{8} O_{6}$ ) to be dissolved in $75 g$ of acetic acid to lower its melting point by ${1.5}^{\circ} C$ . $K_{f} = 3.9 K k g m o l^{- 1}$

Answer :

Elevation in melting point = 1.5 degree celsius.

Here we will use the following equation :

$Δ T_{b} = \frac{K_{b} \times 1000 \times w_{2}}{M_{2} \times w_{1}}$

Putting given values in the above equation :

$w_{2} = \frac{1.5 \times 176 \times 75}{3.9 \times 1000} = 5.08 g$

Thus 5.08 ascorbic acid is needed for required condition.

Question 1.12 Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving $1.0 g$ of polymer of molar mass $185, 000$ in $450 m L$ of water at $37^{\circ} C .$

Answer :

We know that :

$O s m o t i c P r e s s u r e = Π = \frac{n}{v} R T$

We are given with :-

$M o l e s o f p o l y m e r = \frac{1}{185000}$

Volume, V = 0.45 L

Thus osmotic pressure :

$= \frac{\frac{1}{185000} \times 8.314 \times 10^{3} \times 310}{0.45} = 30.98 P a$

NCERT Solutions for Class 12 Chemistry Chapter 1 (Exercise Questions)

Here are detailed and accurate NCERT solutions for Exercise Questions of Class 12 Chemistry Chapter 1. These solutions will help in better concept clarity and exam preparation.

Question.1.1 Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.

Answer :

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 :-

Question 1.2 Give an example of a solid solution in which the solute is a gas.

Answer:

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

Question 1.3(i) Define the following terms:

Mole fraction

Answer :

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

i.e., $M o l e f r a c t i o n = \frac{N u m b e r o f m o l e s i n a c o m p o n e n t}{T o t a l n u m b e r o f m o l e s i n a l l c o m p o n e n t s}$

Question 1.3(ii) Define the following terms:

Molality

Answer :

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

i.e., $M o l a l i t y = \frac{N u m b e r o f m o l e s o f s o l u t e}{M a s s o f s o l v e n t i n K g}$

It is independent of temperature.

Question 1.3(iii) Define the following terms:

Molarity

Answer :

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

i.e., $M o l a r i t y = \frac{N o . o f m o l e s o f s o l u t e}{V o l u m e o f s o l u t i o n i n l i t r e}$

It depends on temperature because volume is dependent on temperature.

Question 1.3(iv) Define the following terms:

Mass percentage.

Answer :

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

i.e., $M a s s p e r c e n t a g e = \frac{M a s s o f a c o m p o n e n t}{T o t a l m a s s o f s o l u t i o n} \times 100$

Question 1.4 Concentrated nitric acid used in laboratory work is $68 %$ nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is $1.504 g m L^{- 1} ?$

Answer :

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

So moles of 68 g HNO₃:-

$= \frac{68}{63}$

= 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

Question 1.5 A solution of glucose in water is labelled as $10 %$ w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is $1.2 g m L^{- 1},$ then what shall be the molarity of the solution?

Answer :

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 (C₆H₁₂O₆) = $180 g m o l^{- 1}$

So moles of glucose are :

$\frac{10}{180} = 0.056 m o l$

$M o l e s o f w a t e r = \frac{90}{18} = 5 m o l$

$M o l a l i t y = \frac{0.056}{0.09} = 0.62 m$

Mole fraction :-

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

Molarity :- Volume of 100 g solution :

$= \frac{100}{1.2} = 83.3 m L$

$M o l a r i t y = \frac{0.056}{83.33 \times 10^{- 3}} = 0.67 M$

Question 1.6 How many mL of $0.1 M H C l$ are required to react completely with $1 g$ mixture of $N a_{2} C O_{3}$ and $N a H C O_{3}$ containing equimolar amounts of both?

Answer :

Total amount of mixture of Na₂CO₃ and NaHCO₃= 1 g.

Let the amount of Na₂CO₃be x g.

So the amount of NaHCO₃will be equal to (1 - x) g.

$M o l a r m a s s o f N a_{2} C O_{3} = 106; m o l a r m a s s o f N a H C O_{3} = 84$

Now it is given that it is an equimolar mixture.

So, Moles of Na₂CO₃= Moles of NaHCO₃

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

or x = 0.558 g

So $M o l e s o f N a_{2} C O_{3} = \frac{0.558}{106} = 0.00526$

and $M o l e s o f N a H C O_{3} = \frac{1 - 0.558}{84} = 0.0053$

$\begin{matrix} 2 HCl (a q) + {Na}_{2} {CO}_{3} (a q) ⟶ H_{2} O (t) + {CO}_{2} (g) + 2 NaCl (a q) \\ {NaHCO}_{3} + HCl ⟶ NaCl + H_{2} {CO}_{3} \end{matrix}$

It is clear that for 1 mol of Na₂CO₃ 2 mol of HCl is required, similarly for 1 mol of NaHCO₃ 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 1 L of solution.

Thus required volume :

$= \frac{0.01578}{0.1} = 0.1578 l = 157.8 m L$

Question 1.7 A solution is obtained by mixing $300 g$ of $25 %$ solution and $400 g$ of $40 %$ solution by mass. Calculate the mass percentage of the resulting solution.

Answer :

According to question we have 2 solute,

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

$\frac{25}{100} \times 300 = 75 g$

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

$\frac{40}{100} \times 400 = 160 g$

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

Thus mass percentage of solute is :

$= \frac{235}{700} \times 100 = 33.5 %$

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

Question 1.8 An antifreeze solution is prepared from $222.6 g$ of ethylene glycol $(C_{2} H_{6} O_{2})$ and $200 g$ of water. Calculate the molality of the solution. If the density of the solution is $1.072 g m L^{- 1}$ then what shall be the molarity of the solution?

Answer :

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

Moles of ethylene glycol :

$= \frac{222.6}{62} = 3.59 m o l$

We know that :

$M o l a l i t y = \frac{M o l e s o f e t h y l e n e g l y c o l}{M a s s o f w a t e r} \times 100$

$= \frac{3.59}{200} \times 100 = 17.95 m$

Now for molarity :-

Total mass of solution = 200 + 222.6 = 422.6 g

Volume of solution

$= \frac{422.6}{1.072} = 394.22 m L$

So molarity :-

$= \frac{3.59}{394.22} \times 1000 = 9.11 M$

Question 1.9(i) A sample of drinking water was found to be severely contaminated with chloroform $(C H C l_{3})$ supposed to be a carcinogen. The level of contamination was $15 p p m$ (by mass):

express this in percent by mass

Answer :

We know that 15 ppm means 15 parts per million.

Required percent by mass :

$= \frac{M a s s o f c h l o r o f o a m}{T o t a l m a s s} \times 100$

$= \frac{15}{10^{6}} \times 100 = 1.5 \times 10^{- 3} %$

Question 1.9(ii) A sample of drinking water was found to be severely contaminated with chloroform $(C H C l_{3})$ supposed to be a carcinogen. The level of contamination was $15 p p m$ (by mass):

determine the molality of chloroform in the water sample.

Answer :

Moles of chloroform :

$= \frac{15}{119.5} = 0.1255 m o l$

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

So molality will be :

$= \frac{0.1255}{10^{6}} \times 1000 = 1.255 \times 10^{- 4} m$

Question 1.10 What role does the molecular interaction play in a solution of alcohol and water?

Answer :

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.

Question 1.11 Why do gases always tend to be less soluble in liquids as the temperature is raised?

Answer:

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.

Question 1.12 State Henry’s law and mention some important applications.

Answer :

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 the 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 CO₂ 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 at 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.

Question 1.13 The partial pressure of ethane over a solution containing $6.56 \times 10^{- 3} g$ of ethane is 1 bar. If the solution contains $5.00 \times 10^{- 2} g$ of ethane, then what shall be the partial pressure of the gas?

Answer :

Using Henry's Law we can write,

$m = k . P$

Putting value in this equation, we get :

$6.56 \times 10^{- 3} = k \times 1$

So, the magnitude of k is $6.56 \times 10^{- 3}$ .

Now, we will again use the above equation for $m = 5.0 \times 10^{- 2} g$ .

So the required partial pressure is :-

$p = \frac{m}{k} = \frac{5.0 \times 10^{- 2}}{6.56 \times 10^{- 3}}$

or $p = 7.62 b a r$

Question 1.14 What is meant by positive and negative deviations from Raoult's law and how is the sign of $Δ_{m i x} H$ related to positive and negative deviations from Raoult's law?

Answer :

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., $Δ_{M i x} H \neq 0$ and $Δ_{M i x} V \neq 0$ . 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.

Question 1.15 An aqueous solution of $2 %$ non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?

Answer :

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 :

$M_{2} = \frac{2 \times 18 \times 1.013}{0.009 \times 98} = 41.35 g m o l^{- 1}$

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

Question 1.16 Heptane and octane form an ideal solution. At $373 K$ , the vapour pressures of the two liquid components are $105.2 k P a$ and $46.8 k P a$ respectively. What will be the vapour pressure of a mixture of $26.0 g$ of heptane and $35 g$ of octane?

Answer :

Vapour pressure of heptane = $p_{h}^{\circ} = 105.2 K P a$

and vapour pressure of octane = $p_{o}^{\circ} = 46.8 K P a$

Firstly, we will find moles of heptane and octane so that we can find the 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 :- $p_{h} = 0.456 \times 105.2 = 47.97 K P a$

(ii) of octane :- $p_{o} = 0.544 \times 46.8 = 25.46 K P a$

So total pressure of solution = $p_{h} + p_{o}$

= 47.97 + 25.46 = 73.43 KPa

Question 1.17 The vapour pressure of water is $12.3 k P a$ at $300 K .$ Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.

Answer :

It is asked the vapour pressure of a 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 $p = 12.083 K P a$

Thus the vapour pressure of the solution is 12.083 KPa

Question 1.18 Calculate the mass of a non-volatile solute $(m o l a r m a s s 40 g m o l^{- 1})$ which should be dissolved in $114 g$ octane to reduce its vapour pressure to $80 %$ .

Answer :

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

After adding solute to octane, the vapour pressure becomes :

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

Moles of octane :

$= \frac{114}{114} = 1$ $(M o l a r m a s s o f o c t a n e = 8 (12) + 18 (1) = 114 g m o l^{- 1})$

Using Raoult's law we get :

$\frac{p_{o}^{\circ} - p}{p_{o}^{\circ}} = x_{2}$

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

or $w = 10 g$

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

Question 1.19(i) A solution containing 30 g of non-volatile solute exactly in $90 g$ of water has a vapour pressure of $2.8 k P a$ at $298 K .$ . Further, $18 g$ of water is then added to the solution and the new vapour pressure becomes $2.9 k P a$ at $298 K .$ . Calculate:

the molar mass of the solute

Answer :

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

Let the molar mass of the solute be M.

Initially, we had 30 g of solute and 90 g of water.

Moles of water :

$\frac{90}{18} = 5 m o l$

By Raoult's law we have:-

$\frac{p_{w}^{\circ} - p}{p_{w}^{\circ}} = \frac{n_{2}}{n_{1} + n_{2}}$

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

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

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

Moles of H₂O :

$\frac{90 + 18}{18} = 6 m o l$

Putting this in above equation we obtain:-

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

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

From equation (i) and (ii) we get

M = 23 u

So the molar mass of solute is 23 units.

Question 1.19(ii) A solution containing $30 g$ of non-volatile solute exactly in $90 g$ of water has a vapour pressure of $2.8 k P a$ at $298 K$ Further, $18 g$ of water is then added to the solution and the new vapour pressure becomes $2.9 k P a$ at $298 K .$ Calculate

vapour pressure of water at 298 K.

Answer :

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

We had :-

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

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.

Question 1.20 A $5 %$ solution (by mass) of cane sugar in water has freezing point of $271 K .$ Calculate the freezing point of $5 %$ glucose in water if freezing point of pure water is $273.15 K .$

Answer :

It is given that the 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 - $Δ T_{f} = k_{f} \times m$

or $k_{f} = \frac{Δ T_{f}}{m} = 13.99 K K g m o l^{- 1}$

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 :

$= \frac{.0278}{0.095} = 0.2926 m o l k g^{- 1}$

So, we can find the elevation in freezing point:

$Δ T_{f} = k_{f} \times m$

$= 13.99 \times 0.2926 = 4.09 K$

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

Question 1.21 Two elements A and B form compounds having formula AB₂ and AB₄ . When dissolved in 20 g of benzene $(C_{6} H_{6}),$ $1 g$ of AB₂ lowers the freezing point by $2.3 K$ whereas $1.0 g$ of AB₄ lowers it by $1.3 K$ . The molar depression constant for benzene is $5.1 K k g m o l^{- 1} .$ Calculate atomic masses of A and B.

Answer :

In this question we will use the formula :

$Δ T_{f} = k_{f} \times m$

Firstly for compound AB₂ :-

$M_{B} = \frac{K_{f} \times W_{b} \times 1000}{w_{A} \times Δ T_{f}}$

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

Similarly for compound AB₄ :-

$M_{B} = \frac{5.1 \times 1 \times 1000}{20 \times 1.3} = 196 g / m o l$

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.

Question 1.22 At $300 K,$ 36 g of glucose present in a litre of its solution has an osmotic pressure of $4.98 b a r .$ If the osmotic pressure of the solution is $1.52 b a r s$ at the same temperature, what would be its concentration?

Answer :

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

$\frac{Π_{1}}{C_{1}} = \frac{Π_{2}}{C_{2}}$ $(Π = C . R . T; R T = \frac{Π}{C})$

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 $C_{2} = \frac{1.52 \times 36}{4.98 \times 180} = 0.061 M$

Hence the required concentration is 0.061 M.

Question 1.23(i) Suggest the most important type of intermolecular attractive interaction in the following pairs.

n-hexane and n-octane

Answer :

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

Question 1.23(ii) Suggest the most important type of intermolecular attractive interaction in the following pairs.

$I_{2} a n d C C l_{4}$

Answer :

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

Question 1.23(iii) Suggest the most important type of intermolecular attractive interaction in the following pairs.

$N a C l O_{4} a n d W a t e r$

Answer :

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

Question 1.23(iv) Suggest the most important type of intermolecular attractive interaction in the following pairs.

$m e t h a n o l a n d a c e t o n e$

Answer :

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

Question 1.23(v) Suggest the most important type of intermolecular attractive interaction in the following pairs.

acetonitrile $(C H_{3} C N)$ and acetone $(C_{3} H_{6} O) .$

Answer :

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

Question 1.24 Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain.

Cyclohexane, KCl, CH₃OH, CH₃CN.

Answer :

The order will be : Cyclohexane > CH₃CN > CH₃OH > KCl

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

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

Question 1.25(i) Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water?

phenol

Answer :

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.

Question 1.25(ii) Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water?

toluene

Answer :

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)

Question 1.25(iii) Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water?

formic acid

Answer :

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

Question 1.25(iv) Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water?

ethylene glycol

Answer :

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.

Question 1.25(v) Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water?

chloroform

Answer :

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

Question 1.25(vi) Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water?

pentanol

Answer :

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

Question .1.26 If the density of some lake water is $1.25 g m L^{- 1}$ and contains $92 g$ of $N a^{+} i o n s$ per kg of water, calculate the molality of $N a^{+} i o n s$ in the lake.

Answer :

We know that, Molality :

$M o l a l i t y = \frac{M o l e s o f s o l u t e}{M a s s o f s o l v e n t i n K g}$

So, for moles of solute we have :

$M o l e s o f N a^{+} = \frac{92}{23} = 4$

Thus, molality :

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

Molality of Na + ions is 4m.

Question 1.27 If the solubility product of $C u S$ is $6 \times 10^{- 16}$ , calculate the maximum molarity of $C u S$ in aqueous solution.

Answer :

We are given, $K_{s p} = 6 \times 10^{- 16}$

The dissociation equation of CuS is given by :-

$CuS ⇌ {Cu}^{2 +} + S^{2 -}$

So, the equation becomes :- $K_{s p} = C u^{2 +} \times S^{2 -}$

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

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

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

Question 1.28 Calculate the mass percentage of aspirin $(C_{9} H_{8} O_{4})$ in acetonitrile $(C H_{3} C N)$ when $6.5 g$ of $C_{9} H_{8} O_{4}$ is dissolved in $450 g$ of $C H_{3} C N$ .

Answer :

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

We know that :

$M a s s p e r c e n t a g e = \frac{M a s s o f s o l u t e}{M a s s o f s o l u t i o n} \times 100$

So, $M a s s p e r c e n t a g e = \frac{6.5}{456.5} \times 100 = 1.42 %$

Thus the mass percentage of aspirin is $1.42 %$

Question 1.29 Nalorphene $(C_{19} H_{21} N O_{3})$ , similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose of nalorphene generally given is $1.5 m g .$ Calculate the mass of $1.5 - 10^{- 3} m$ aqueous solution required for the above dose.

Answer :

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 :

$\frac{1.5 \times 10^{- 3}}{311} = 4.82 \times 10^{- 6} m o l e s$

Molality :

$= \frac{N o . o f m o l e s o f s o l u t e}{M a s s o f s o l v e n t i n k g}$

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

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

So the required weight of water is 3.2 g.

Question 1.30 Calculate the amount of benzoic acid $(C_{6} H_{5} C O O H)$ required for preparing $250 m L$ of $0.15 M$ solution in methanol.

Answer :

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

We are given with the molarity of a solution.

$M o l a r i t y = \frac{M o l e s o f s o l u t e}{V o l u m e o f s o l u t i o n i n l i t r e}$

or $0.15 = \frac{M o l e s o f s o l u t e}{\frac{250}{1000}}$

or $M o l e s o f s o l u t e = \frac{0.15 \times 250}{1000} = 0.0375 m o l$

So mass of benzoic acid :

$= M o l e s o f b e n z o i c a c i d \times M o l a r m a s s$

$= 0.0375 \times 122 = 4.575 g$

Hence the required amount of benzoic acid is 4.575 g.

Question 1.31 The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly.

Answer :

We know that depression in the 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.

Question 1.32 Calculate the depression in the freezing point of water when $10 g$ of $C H_{3} C H_{2} C H C I C O O H$ is added to $250 g$ of water.

$K_{a} = 1.4 \times 10^{- 3}$ , $K_{f} = 1.86 K k g m o l^{- 1}$

Answer :

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

1643795600066

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

or $K_{a} = C a^{2}$ $(∵ a << 1)$

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 CH₃ CH₂CHClCOOH.

So, moles =

$\frac{10}{122.5} = 0.0816 m o l$

Thus, molality of the solution :

$= \frac{0.0816 \times 1000}{250} = 0.3265 m$

Now we will use :

$Δ T_{f} = i K_{f} m$

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

Question 1.33 $19.5 g$ of $C H_{2} F C O O H$ is dissolved in $500 g$ of water. The depression in the freezing point of water observed is ${1.0}^{\circ} C$ . Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid.

Answer :

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

So moles of CH₂FCOOH :

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

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

Now, we know that : $Δ T_{f} = i K_{f} m$

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

Now for dissociation constant :-

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

and, $K_{a} = \frac{C a^{2}}{1 - a}$

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

$K_{a} = 3 \times 10^{- 3}$

Question 1.34 Vapour pressure of water at $293 K$ is $17.535 m m H g .$ Calculate the vapour pressure of water at $293 K$ when $25 g$ of glucose is dissolved in $450 g$ of water.

Answer :

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

Moles of glucose :

$= \frac{25}{180} = 0.139 m o l$ $(M o l a r m a s s o f g l u c o s e = 6 (12) + 12 (1) + 6 (16) = 180 g m o l^{- 1})$

and moles of water :

$= \frac{450}{18} = 25 m o l$ $(M o l a r m a s s o f w a t e r = 18 g m o l^{- 1})$

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 $p = 17.44 m m o f H g$

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

Question 1.35 Henry’s law constant for the molality of methane in benzene at $298 K$ is $4.27 \times 10^{5} m m H g$ . Calculate the solubility of methane in benzene at $298 K$ under $760 m m H g .$

Answer :

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

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

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

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

Question 1.36 $100 g$ of liquid A $(m o l a r m a s s 140 g m o l^{- 1})$ was dissolved in $1000 g$ of liquid B $(m o l a r m a s s 180 g m o l^{- 1})$ . The vapour pressure of pure liquid B was found to be $500 T o r r .$ Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure the solution is $475 T o r r .$

Answer :

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 $P_{t o t a l} = P_{A}^{\circ} x_{A} + P_{B}^{\circ} x_{B}$

or $475 = P_{A}^{\circ} \times 0.114 + 500 \times 0.886$

or $P_{A}^{\circ} = 280.7 t o r r$

Thus vapour pressure in solution due to A = $P_{A}^{\circ} x_{A}$

$= 280.7 \times 0.114 = 32 t o r r$

Question 1.37 Vapour pressures of pure acetone and chloroform at $328 K$ are $741.8 m m H g$ and $638.8 m m H g$ respectively. Assuming that they form ideal solution over the entire range of composition, plot ptotal, pchloroform, and pacetone as a function of xacetone. The experimental data observed for different compositions of mixture is:

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

Answer :

1594728504498

it has negative deviation from the ideal solution.

Question 1.38 Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at $300 K$ are $50.71 m m H g$ and $32.06 m m H g$ respectively. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with $100 g$ toluene.

Answer :

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

No. of moles of benzene :

$= \frac{80}{78} = 1.026 m o l$

and the no. of moles of toluene :

$= \frac{100}{92} = 1.087 m o l$ .

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 $= 50.71 \times 0.486 + 32.06 \times 0.514 = 24.65 + 16.48$

or $= 41.13 m m o f H g$

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

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

Question 1.39 The air is a mixture of a number of gases. The major components are oxygen and nitrogen with approximate proportion of $20 %$ is to $79 %$ by volume at $298 K$ . The water is in equilibrium with air at a pressure of $10 a t m .$ At $298 K$ . if the Henry’s law constants for oxygen and nitrogen at $298 K$ are $3.30 \times 107 m m$ and $6.51 \times 107 m m$ respectively, calculate the composition of these gases in water

Answer :

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 :

$\frac{20}{100} \times 7600 = 1520 m m o f H g$

and partial pressure of nitrogen :

$\frac{79}{100} \times 7600 = 6004 m m o f H g$

Now, by Henry's Law :

$P = K_{h} . x$

For oxygen :

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

For nitrogen :

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

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

Question 1.40 Determine the amount of $C a C l_{2}$ $(i = 2.47)$ dissolved in $2.5 l i t r e$ of water such that its osmotic pressure is $0.75 a t m$ at $27^{\circ} C$ .

Answer :

We know that osmotic pressure :

$Π = i (\frac{n}{v}) R T$

or $Π = i (\frac{w}{M v}) R T$

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

So the value of w can be found.

$w = \frac{0.75 \times 111 \times 2.5}{2.47 \times 0.0821 \times 300}$ $(M = 1 \times 40 + 2 \times 35.5 = 111 g m o l^{- 1})$

$= 3.42 g$

Hence 3.42 g CaCl₂ is required.

Question 1.41 Determine the osmotic pressure of a solution prepared by dissolving $25 m g$ of $K_{2} S O_{4}$ in $2 l i t r e$ of water at $25^{\circ} C,$ assuming that it is completely dissociated.

Answer :

Dissociation of K 2 SO 4 is as follows :-

$K_{2} {SO}_{4 (s)} \to 2 K_{(aq)}^{+} + {SO}_{4} 2 - (aq)$ 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.

$Π = i C R T$

Putting all the values :-

$Π = \frac{3 \times 25 \times 10 - 3 \times 0.082 \times 298}{174 \times 2}$

$= 5.27 \times 10^{- 3} a t m$

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Question: $HA (aq) ⇌ H^{+} (aq) + A^{-} (aq)$

The freezing point depression of a 0.1 m aqueous solution of a monobasic weak acid HA is ${0.20}^{\circ} C$ . The dissociation constant for the acid is
Given :
$K_{f} (H_{2} O) = 1.8 K kg {mol}^{- 1}$ , molality $\equiv$ molarity

(1). $1.38 \times 10^{- 3}$

(2). $1.1 \times 10^{- 2}$

(3). $1.90 \times 10^{- 3}$

(4). $1.89 \times 10^{- 1}$

Answer:

$\begin{aligned} Δ T_{f} = i K_{f} m \\ i = \frac{Δ T_{f}}{K_{f} \cdot m} \\ i = \frac{0.20}{1.8 \times 0.1} = 1.11 \\ i = 1.11 \\ α = \frac{i - 1}{n - 1} (for HA, n = 2) \\ α = \frac{1.11 - 1}{1} = 0.11 \\ K_{a} = \frac{c α^{2}}{1 - α} = \frac{0.1 \times (0.11)^{2}}{1 - 0.11} = 1.38 \times 10^{- 3} \end{aligned}$

Hence, the correct answer is option (1).

Question: Which of the following binary mixture does not show the behaviour of minimum boiling azeotropes ?

(1) $H_{2} O + {CH}_{3} {COC}_{2} H_{5}$

(2) $C_{6} H_{5} OH + C_{6} H_{5} {NH}_{2}$

(3) ${CS}_{2} + {CH}_{3} {COCH}_{3}$

(4) ${CH}_{3} OH + {CHCl}_{3}$

Answer:

The solution showing positive deviation from Raoult's law will form a minimum boiling azeotrope. Phenol + Aniline shows negative deviation, so they will not form a minimum boiling azeotrope.

A binary mixture of $C_{6} H_{5} OH$ and $C_{6} H_{5} {NH}_{2}$ exhibits negative deviation from Raoult's law. This means that the vapor pressure of the solution is lower than the vapor pressures of the pure components, $C_{6} H_{5} OH$ and $C_{6} H_{5} {NH}_{2}$ . Consequently, the boiling point of the solution is higher than the boiling points of the pure substances. Therefore, this mixture forms a maximum boiling azeotrope.

Hence, the correct answer is option (2).

Question: Given below are two statements :
Statement (I) : Molal depression constant $K_{f}$ is given by $\frac{M_{1} R T_{f}}{Δ S_{f u s}}$ , where symbols have their usual meaning.
Statement (II) : $K_{f}$ for benzene is less than the $K_{f}$ for water.
In the light of the above statements, choose the most appropriate answer from the options given below :

(1) Statement I is incorrect but Statement II is correct

(2) Both Statement I and Statement II are incorrect

(3) Both Statement I and Statement II are correct

(4) Statement I is correct but Statement II is incorrect

Answer:

Statement-I
Molar depression constant $k_{f} = \frac{M_{i} {RT}_{f}^{2}}{Δ H_{fius}}$

$\begin{aligned} k_{f} = \frac{M_{1} {RT}_{f}}{[\frac{Δ H_{fis}}{T_{f}}]} \\ k_{f} = \frac{M_{1} {RT}_{f}}{Δ S_{fus}} \end{aligned}$

Hence statement-I is correct
but $k_{f}$ for benzene $= 5.12 \frac{\circ C}{molal}$
$k_{f}$ for water $= 1.86 \frac{\circ C}{molal}$ Hence statement- II is incorrect

Hence, the correct answer is option (4).

CBSE Class 12th Syllabus: Subjects & Chapters

Select your preferred subject to view the chapters

Approaches to Solve Problems of Chapter 1 Chemistry Class 12

Sometimes, problems related to class 12 chemistry chapter 1 solutions seem difficult, but once we understand the basic formulas, it becomes very easy to solve all the questions related to Solutions. We can follow the steps given below to solve the questions based on the Solutions:

1. Understand the types of solutions

The first step is to identify the solute and solvent and learn to classify the solution based on the states of components like solid in liquid, gas in liquid, etc. Also, learn to apply the concentration formulas
Molarity (M), Molality (m), Mole fraction (χ), Mass %, Volume %

(i). Molarity

$M = \frac{moles of solute}{volume of solution in liters}$

(ii). Molality

$m = \frac{moles of solute}{mass of solvent in kg}$

(iii). Mole Fraction

$X_{A} = \frac{moles of component A}{total moles of solution}$

2. Learn important formulas

Important formulas from chapter 1 solutions class 12 chemistry are given below. These formulas help students to solve questions easily.

Henry's Law: $P = k_{H} \cdot x$ for gas solubility.
Raoult's Law- For ideal solutions, use $p_{A} = x_{A} \cdot p_{A}^{0}$ .
Apply to vapor pressure problems.

3. Colligative Properties

Identify which property is involved ( $Δ Tf, Δ Tb, π, Δ P$ ).
Use-

(i). Elevation in Boiling Point : $Δ T_{b} = i \cdot K_{b} \cdot m$

(ii). Depression in Freezing Point: $Δ T_{f} = i \cdot K_{f} \cdot m$

(iii). Osmotic Pressure: $π = i \cdot C \cdot R \cdot T$

4. Van’t Hoff Factor (i)

Learn to determine if solute dissociates or associates. Adjust formulas using ? for non-ideal cases.

5. Practice Questions

While solving questions from class 12 chemistry chapter 1 solutions the numericals, keep units consistent and understand what each symbol represents. You can also draw diagrams for clarity also focus on ideal vs non-ideal solutions, positive/negative deviations, and azeotropes. Students can also access the NCERT exemplar solutions for better understanding.

Topics of NCERT Syllabus Class 12 Chemistry Solutions

The NCERT Solutions for Class 12 Chemistry Chapter 1 Solutions cover all the important topics included in the syllabus in a well structured manner. They help students understand each concept clearly and practise questions topic-wise for better exam preparation. Topics covered in this chapter are given below:

1.1 Types Of Solution

1.2 Expression Of Concentration Of Solution

1.3 Solubility

1.3.1 Solubility of a Solid in a Liquid
1.3.2 Solubility of a Gas in a Liquid

1.4 Vapour Pressure Of Liquid Solutions

1.4.1 Vapour Pressure of Liquid-Liquid Solutions
1.4.2 Raoult's Law as a special case of Henry's Law
1.4.3 Vapour Pressure of Solutions of Solids in Liquids

1.5 Ideal and Non-ideal Solutions

1.5.1 Ideal Solutions
1.5.2 Non-ideal Solutions

1.6 Colligative Properties and Determination of Molar Mass

1.6.1 Relative Lowering of Vapour Pressure
1.6.2 Elevation of Boiling Point
1.6.3 Depression of Freezing Point
1.6.4 Osmosis and Osmotic Pressure
1.6.5 Reverse Osmosis and Water Purification

1.7 Abnormal Molar Masses

What Extra Should Students Study Beyond the NCERT for JEE/NEET?

Here a comparison table highlighting what to study beyond NCERT for JEE. While NCERT books build a strong foundation, students should also refer to advanced reference books and practice materials. Solving class 12 chemistry chapter 1 solutions question answer can help strengthen problem-solving skills and improve exam performance.

Concept	JEE	NCERT
Solution	✅	✅
Concentration terms	✅	✅
Vapour Pressure	✅	✅
Factors on which Vapour Pressure depends	✅	✅
Vapour Pressure of Solution Containing Two Volatile Liquids	✅	✅
Vapour Pressure of solution containing non-volatile solute	✅	✅
Ideal Solution	✅	✅
Examples of Ideal Solution	✅	✅
Non-Ideal Solution showing negative deviation from Raoult's law	✅	✅
Non-Ideal Solution showing positive deviation from Raoult's law	✅	✅
Azeotropic mixture	✅	✅
Relation between Raoult's law and Dalton's law	✅	✅
Elevation in Boiling point	✅	✅
Depression in Freezing point	✅	✅
Osmosis and Osmotic Pressure	✅	✅
Reverse Osmosis	✅	✅
Isotonic, Hypertonic, Hypotonic Solutions	✅	✅
Van't Hoff Factor (i) or Abnormal Colligative Property	✅	✅
Calculation of Extent of Dissociation in Electrolytic Solution	✅	❌
Calculation of Extent of Association in an Electrolytic Solution	✅	❌
Solubility and Henry's Law	✅	✅
Real life examples of Hery's Law	✅	✅

What Students Learn from NCERT Solutions for Class 12 Chemistry Chapter 1 Solutions

These class 12 chemistry chapter 1 solutions help students understand the key concepts of the solid state in a simple and structured way. Given below some points on key learning from this chapter

Using these solutions students can understand the fundamental concepts of the solid state.
Here they can learn about types of solids and their properties.
Properties like unit cells, crystal lattices, and packing efficiency are explained very well in these solutions class 12 question answer
These solutions help students to calculate the density and voids in crystals
Imperfections in solids, coordination number and geometrical arrangements are explained in these solutions very well.

NCERT solutions for class 12 chemistry

Along with NCERT Solutions for Class 12 Chemistry Chapter 1 Solutions also refer to chapter-wise solutions for the entire Class 12 Chemistry syllabus:

NCERT solutions for class 12 Chemistry chapter 1 Solutions

NCERT Solutions for class 12 Chemistry chapter 2 Electrochemistry

NCERT solutions for class 12 Chemistry chapter 3 Chemical Kinetics

NCERT solutions for class 12 Chemistry chapter 4 The d and f block elements

NCERT solutions for class 12 Chemistry chapter 5 Coordination compounds

NCERT Solutions for class 12 Chemistry chapter 6 Haloalkanes and Haloarenes

NCERT solutions for class 12 Chemistry chapter 7 Alcohols, Phenols, and Ethers

NCERT Solutions for class 12 Chemistry chapter 8 Aldehydes, Ketones and Carboxylic Acids

NCERT solutions for class 12 Chemistry chapter 9 Amines

NCERT solutions for class 12 Chemistry chapter 10 Biomolecules

NCERT Exemplar Class 12 Solutions

The NCERT Exemplar Class 12 solutions provide questions with detailed answers to enhance conceptual understanding. The hyperlinks of NCERT exemplar of class 12 are given below:

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: Other Subjects

Along with NCERT Solutions for Class 12 Chemistry, students can also check solutions for other subjects that are equally important for board exams and competitive exams like JEE and NEET.

NCERT Solutions for class 12 biology

NCERT solutions for class 12 maths

NCERT solutions for class 12 chemistry

NCERT Solutions for class 12 physics

NCERT Books and NCERT Syllabus

The NCERT books and syllabus provide a clear and structured framework for learning all subjects. The hyperlinks of the NCERT Books and NCERT syllabus are given below:

NCERT Books Class 12 Chemistry

NCERT Syllabus Class 12 Chemistry

NCERT Books Class 12

NCERT Syllabus Class 12

Frequently Asked Questions (FAQs)

Q: How to get NCERT solutions for Class 12 chemistry Ch 1?

You can get NCERT solutions for Class 12 Chemistry Chapter 1 from various educational websites and online learning platforms. They are also available as free downloadable PDFs, which can be accessed anytime for easy study and revision.

Q: What are NCERT Solutions for Class 12 Chemistry Chapter 1?

NCERT Solutions for Class 12 Chemistry Chapter 1 are detailed answers to all questions provided in the textbook. They explain concepts of the solid state, crystal structures, and related numerical problems in a simple, step-by-step manner to help students understand and practise effectively for exams.

Q: What are NCERT Solutions for Class 12 Chemistry Chapter 1?

Q: What is a solution?

A solution is a homogeneous mixture composed of two or more substances. In a solution, a solute is dissolved in a solvent. The solute is typically present in a smaller amount, while the solvent is present in a larger amount.

Q: What is Raoult's Law?

Raoult's Law states that the vapor pressure of a solvent in a solution is directly proportional to the mole fraction of the solvent in that solution.

Q: What is meant by concentration of a solution?

Concentration of a solution indicates how much solute is present in a given amount of solvent or solution. It can be expressed in various ways, such as molarity, molality, mass percentage, or parts per million.

Q: What is the difference between molarity and molality?

Molarity (M) refers to the number of moles of solute per liter of solution, whereas molality (m) is the number of moles of solute per kilogram of solvent. The key difference is that molarity uses the total volume of the solution, while molality considers only the mass of the solvent, making molality independent of temperature.

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Questions related to CBSE Class 12th

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Have a question related to CBSE Class 12th ?

12th Exam 2025Kitin tareekh ko hoga

Hello,

The date of 12 exam is depends on which board you belongs to . You should check the exact date of your exam by visiting the official website of your respective board.

Hope this information is useful to you.

Read Complete Answer

Nidhi Raghuwanshi

21 Sep'25

12th biology question paper 2023-2025

Hello,

Class 12 biology questions papers 2023-2025 are available on cbseacademic.nic.in , and other educational website. You can download PDFs of questions papers with solution for practice. For state boards, visit the official board site or trusted education portal.

Hope this information is useful to you.

Read Complete Answer

Nidhi Raghuwanshi

21 Sep'25

In March 2025, I passed the CBSE 12th board and gave KCET. I am taking a drop year. I will reappear in the CBSE board exam in the upcoming 2026 exam in February as a private candidate to increase my PCM aggregate, as KCET gives 50% weightage to PCM aggregate, and I will also give the KCET exam. So, to calculate rank, do they use the board marksheet from 2025 or 2026? Also for each subject will best out of two marks accross both the marksheets be considered? #kcet

Hello Pruthvi,

Taking a drop year to reappear for the Karnataka Common Entrance Test (KCET) is a well-defined process. As a repeater, you are fully eligible to take the exam again to improve your score and secure a better rank for admissions.

The main procedure involves submitting a new application for the KCET through the official Karnataka Examinations Authority (KEA) website when registrations open for the next academic session. You must pay the required application fee and complete all formalities just like any other candidate. A significant advantage for you is that you do not need to retake your 12th board exams. Your previously secured board marks in the qualifying subjects will be used again. Your new KCET rank will be calculated by combining these existing board marks with your new score from the KCET exam. Therefore, your entire focus during this year should be on preparing thoroughly for the KCET to achieve a higher score.

For more details about the KCET Exam preparation, CLICK HERE.

I hope this answer helps you. If you have more queries, feel free to share your questions with us, and we will be happy to assist you.

Thank you, and I wish you all the best in your bright future.

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Ark Pandey

20 Sep'25

I'm studying scince in karnatakastate board, can i take commerce in cbse for 12th next year?

Yes, you can switch from Science in Karnataka State Board to Commerce in CBSE for 12th. You will need a Transfer Certificate from your current school and meet the CBSE school’s admission requirements. Since you haven’t studied Commerce subjects like Accountancy, Economics, and Business Studies, you may need to catch up before or during 12th. Not all CBSE schools accept direct admission to 12th from another board, so some may ask you to join Class 11 first. Make sure to check the school’s rules and plan your subject preparation.

Read Complete Answer

Abhilasha Agrawal

18 Sep'25

12th class cbse boards most important que ans ans in hindi midum subject hindi impotent question jo exam main aaynge wo btai

Hello

For the 12th CBSE Hindi Medium board exam, important questions usually come from core chapters like “Madhushala”, “Jhansi ki Rani”, and “Bharat ki Khoj”.
Questions often include essay writing, letter writing, and comprehension passages. Grammar topics like Tenses, Voice Change, and Direct-Indirect Speech are frequently asked.
Students should practice poetry questions on themes and meanings. Important questions also cover summary writing and translation from Hindi to English or vice versa.
Previous years’ question papers help identify commonly asked questions.
Focus on writing practice to improve handwriting and presentation. Time management during exams is key to answering all questions effectively.

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JEE Test Series

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

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

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.

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

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

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

NCERT Solutions for Class 12 Chemistry Chapter 1 - Solutions

NCERT Solutions for Class 12 Chemistry Chapter 1: Download PDF

NCERT Solution for Class 12 Chapter 1 (Intext Questions)

NCERT Solutions for Class 12 Chemistry Chapter 1 (Exercise Questions)

Related eBooks & Sample Papers

Class 12 Chemistry NCERT Chapter 1: Higher Order Thinking Skills (HOTS) Questions

Approaches to Solve Problems of Chapter 1 Chemistry Class 12

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What Students Learn from NCERT Solutions for Class 12 Chemistry Chapter 1 Solutions

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