NCERT Exemplar Class 12 Chemistry Solutions Chapter 2 Solutions

# NCERT Exemplar Class 12 Chemistry Solutions Chapter 2 Solutions

Edited By Sumit Saini | Updated on Sep 16, 2022 04:11 PM IST | #CBSE Class 12th

NCERT exemplar Class 12 Chemistry solutions chapter 2 studies the formation of different types of solutions made out of combinations of two or more pure substances. This NCERT chapter mainly focuses on different kinds of liquid solutions focusing on binary compounds, their properties, and their formation and also the factors affecting them. NCERT exemplar Class 12 Chemistry chapter 2 solutions provide for different methods of showing the concentration of different solutions quantitatively as well as qualitatively. It also provides for various combinations of solid, liquid, and gas acting as solute and solvent along with examples for the same.

NCERT exemplar Class 12 Chemistry solutions chapter 2 also provides mathematical formulas for calculating the percentage, mole fraction, molarity, morality of different compounds with the help of illustrations. Class 12 Chemistry NCERT exemplar solutions chapter 2 also gives an insight into the concept of solubility of different compounds, the factors affecting the solubility such as temperature, mole fraction, and partial pressure with reference to Henry’s Law.
Also, check - NCERT Solutions for Class 12, other subjects

## Question:1

Which of the following units is useful in relating the concentration of a solution with its vapour pressure?
(i) mole fraction
(ii) parts per million
(iii) mass percentage
(iv) molality

The answer is the option (i) Vapour pressure and the concentration (mole fraction) of solution are related. Raoult’s law states that the vapour pressure $(p_{A})$ of a component is directly proportional to the mole fraction $(x_{A})$ of the component.
$p_{x}\propto x_{A}$
where, $x_{A}=\frac{n_{A}}{n_{A}+n_{B}}$

Question:2

On dissolving sugar in water at room temperature solution feels cool to touch. Under which of the following cases dissolution of sugar will be most rapid?
(i) Sugar crystals in cold water.
(ii) Sugar crystals in hot water.
(iii) Powdered sugar in cold water.
(iv) Powdered sugar in hot water.

The answer is the option (iv).

Dissolution is an endothermic process, which is why the solution is cool to touch. Also, powdered sugars have a higher surface area than sugar crystals which further promotes dissolution.

Question:3

At equilibrium the rate of dissolution of a solid solute in a volatile liquid solvent is __________.
(i) less than the rate of crystallisation
(ii) greater than the rate of crystallisation
(iii) equal to the rate of crystallisation
(iv) zero

The answer is the option (iii). The rate of dissolution of solid solute is equal to the rate of crystallisation at equilibrium.

Question:4

A beaker contains a solution of a substance ‘A’. Precipitation of substance ‘A’ takes place when a small amount of ‘A’ is added to the solution. The solution is _________.
(i) saturated
(ii) supersaturated
(iii) unsaturated
(iv) concentrated

The answer is the option (ii). Adding a small amount of solute in an already saturated solution will make the solution supersaturated and will lead to the precipitation of the solute.

Question:5

The maximum amount of a solid solute that can be dissolved in a specified amount of a given liquid solvent does not depend upon ____________.
(i) Temperature
(ii) Nature of solute
(iii) Pressure
(iv) Nature of solvent

The answer is the option (iii). Solids don’t compress significantly on application of pressure and thus have no impact to its solubility in solvent.

Question:6

Low concentration of oxygen in the blood and tissues of people living at high altitude is due to ____________.
(i) low temperature
(ii) low atmospheric pressure
(iii) high atmospheric pressure
(iv) both low temperature and high atmospheric pressure

The answer is the option (ii). Due to lower atmospheric pressure at high altitude, oxygen concentration is less in the blood and tissues of the people living there.

Question:7

Considering the formation, breaking and strength of hydrogen bond, predict which of the following mixtures will show a positive deviation from Raoult’s law?
(i) Methanol and acetone.
(ii) Chloroform and acetone.
(iii) Nitric acid and water.
(iv) Phenol and aniline

The answer is the option (i). Methanol has hydrogen bonds. When acetone is mixed with it, some of the bonds break weakening the methanol-methanol interaction. This is why they exhibit a positive deviation.

Question:8

Colligative properties depend on ____________.
(i) the nature of the solute particles dissolved in solution.
(ii) the number of solute particles in solution.
(iii) the physical properties of the solute particles dissolved in solution.
(iv) the nature of solvent particles

The answer is the option (ii). Number of solute particles determines its colligative properties.

Question:9

Which of the following aqueous solutions should have the highest boiling point?
(i) $1.0 M NaOH$
(ii) $1.0 M Na_2SO_4$
(iii) $1.0 M NH_4NO_3$
(iv) $1.0 M KNO_3$

The answer is the option (ii). As $1.0 M Na_2SO_4$ furnishes the maximum number of ions, it should have the highest boiling poin

Question:10

The unit of ebulioscopic constant is _______________.
(i) $K kg mol^{-1} or K (molality)^{-1}$
(ii) $mol kg K^{-1} or K^{-1}(molality)$
(iii) $kg mol^{-1} K^{-1} or K^{-1}(molality)^{-1}$
(iv) $K mol kg^{-1} or K (molality)$

The answer is the option (i)
$K_{b}=\frac{\Delta T_{b}}{m}=\frac{K}{mol kg^{-1}}$
The unit of ebullioscopic constant is $K kg mol^{-1} or K (molality)^{-1}$

Question:11

In comparison to a 0.01 M solution of glucose, the depression in freezing point of a 0.01 M $MgCl_2$ solution is _____________.
(i) the same

The answer is the option (iii). Colligative properties only depend on the number of solute. While 0.01 M solution of glucose does not ionize, 0.01 M $MgCl_2$solution ionizes to produce a single $Mg^{2+}$ion and $2 Cl^{-}$ ions.

Question:12

An unripe mango placed in a concentrated salt solution to prepare pickle, shrivels because _____________.
(i) it gains water due to osmosis.
(ii) it loses water due to reverse osmosis.
(iii) it gains water due to reverse osmosis.
(iv) it loses water due to osmosis.

The answer is the option (iv). Due to osmosis, Water inside mango (lower concentration) moves outside in a concentrated salt solution (higher concentration).

Question:13

At a given temperature, osmotic pressure of a concentrated solution of a substance _____________.
(i) is higher than that at a dilute solution.
(ii) is lower than that of a dilute solution.
(iii) is same as that of a dilute solution.
(iv) cannot be compared with an osmotic pressure of a dilute solution.

The answer is the option (i).
At a given temperature, osmotic pressure of a concentrated solution of a substance is higher than that at a dilute solution.

Question:14

Which of the following statements is false?
(i) Two different solutions of sucrose of the same molality prepared in different solvents will have the same depression in freezing point.
(ii) The osmotic pressure of a solution is given by the equation $\pi = CRT$ (where C is the molarity of the solution).
(iii) Decreasing order of osmotic pressure for 0.01 M aqueous solutions of barium chloride, potassium chloride, acetic acid and sucrose is $BaCl_2> KCl > CH_3COOH > sucrose.$
(iv) According to Raoult’s law, the vapour pressure exerted by a volatile component of a solution is directly proportional to its mole fraction in the solution.

The answer is the option (i). Colligative properties depend on molarity of solute and the solvent. Two different solvents (with same molality of sucrose) will have different freezing points

Question:15

The values of Van’t Hoff factors for KCl, NaCl and $K_2SO_4$, respectively, are
_____________.

(i) 2, 2 and 2
(ii) 2, 2 and 3
(iii) 1, 1 and 2
(iv) 1, 1 and 1

The answer is the option (ii). While KCl and NaCl ionize to give 2 ions,$K_2SO_4$ gives 3 ions. So, van’t Hoff factors for KCl, NaCl and K2SO4 are 2,2 and 3 respectively

Question:16

Which of the following statements is false?
(i) Units of atmospheric pressure and osmotic pressure are the same.
(ii) In reverse osmosis, solvent molecules move through a semipermeable membrane from a region of lower concentration of solute to a region of higher concentration.
(iii) The value of molal depression constant depends on the nature of the solvent.
(iv) Relative lowering of vapour pressure is a dimensionless quantity.

The answer is the option (ii). Solvent molecules move from a solution containing high concentration of solute to one with a low concentration of solute.

Question:17

Value of Henry’s constant KH ____________.
(i) increases with increase in temperature.
(ii) decreases with increase in temperature.
(iii) remains constant.
(iv) first increases then decrease.

The answer is the option (i). There is an increase in the value of Henry’s constant with increase in temperature.

Question:18

The value of Henry’s constant KH is _____________.
(i) greater for gases with higher solubility.
(ii) greater for gases with lower solubility.
(iii) constant for all gases.
(iv) not related to the solubility of gases.

The answer is the option (ii). Gases having a lower solubility have a higher $K_H$.

Question:19

Consider Figure and mark the correct option.
(i) water will move from the side (A) to side (B) if a pressure lower than the osmotic pressure is applied on piston (B).
(ii) water will move from the side (B) to side (A) if a pressure greater than the osmotic pressure is applied on piston (B).
(iii) water will move from the side (B) to side (A) if a pressure equal to osmotic pressure is applied on piston (B).
(iv) water will move from the side (A) to side (B) if pressure equal to osmotic pressure is applied on the piston (A).

The answer is the option (ii). If the piston above B applies a pressure exceeding the osmotic pressure, water will move from concentrated solution (side B) to Fresh Water (side A).

Question:20

We have three aqueous solutions of NaCl labelled as ‘A’, ‘B’ and ‘C’ with concentrations 0.1M, 0.01M and 0.001M, respectively. The value of Van’t Hoff factor for these solutions will be in the order______.
(i) iA< iB< iC
(ii) iA> iB> iC
(iii) iA= iB= iC
(iv) iA<iB> iC

The answer is the option (iii). Due to complete dissociation of NaCl, the van’t Hoff factor will be the same for the three solutions.

Question:21

On the basis of information given below mark the correct option.
Information:

(A) In bromoethane and chloroethane mixture intermolecular interactions of A–A and B–B type are nearly the same as A–B type interactions.
(B) In ethanol and acetone mixture A–A or B–B type intermolecular interactions are stronger than A–B type interactions.
(C) In chloroform and acetone mixture A–A or B–B type intermolecular interactions are weaker than A–B type interactions.
(i) Solution (B) and (C) will follow Raoult’s law.
(ii) Solution (A) will follow Raoult’s law.
(iii) Solution (B) will show a negative deviation from Raoult’s law.
(iv) Solution (C) will show a positive deviation from Raoult’s law

The answer is the option (ii). In a Bromoethane (A) and Chloroethane (B) mixture, the interactions A-A, B-B and A-B are nearly equal, so the solution will be nearly ideal and thus follow Raoult’s law.

Question:22

Two beakers of capacity 500 mL were taken. One of these beakers, labeled as “A”, was filled with 400 mL water whereas the beaker labelled “B” was filled with 400 mL of 2 M solution of NaCl. At the same temperature, both the beakers were placed in closed containers of the same material and same capacity as shown in Figure.
At a given temperature, which of the following statement is correct about the vapour pressure of pure water and that of NaCl solution.

(i) the vapour pressure in container (A) is more than that in a container (B).
(ii) the vapour pressure in container (A) is less than that in the container (B).
(iii) vapour pressure is equal in both containers.
(iv) the vapour pressure in container (B) is twice the vapour pressure in container (A).

The answer is the option (i). Vapour pressure of container (B) is lower as NaCl being a non-volatile solute reduces the vapour pressure.

Question:23

If two liquids A and B form minimum boiling azeotrope at some specific composition then _______________.
(i) A–B interactions are stronger than those between A–A or B–B.
(ii) vapour pressure of solution increases because more number of molecules of liquids A and B can escape from the solution.
(iii) vapour pressure of solution decreases because less number of molecules of only one of the liquids escape from the solution.
(iv) A–B interactions are weaker than those between A–A or B–B.

(i) A–B interactions are stronger than those between A–A or B–B.

Question:24

4L of 0.02 M aqueous solution of NaCl was diluted by adding one litre of water. The molality of the resultant solution is _____________.
(i) 0.004
(ii) 0.008
(iii) 0.012
(iv) 0.016

The answer is the option (iv)

$M=\frac{n}{}V=0.02 =\frac{n}{4} \;or\; n=0.08$
$M= \frac{n}{ \text{(Mass of water in Kg)} }= \frac{0.08}{}5 =0.016$

Question:25

On the basis of information given below mark the correct option.
Information: On adding acetone to methanol some of the hydrogen bonds between methanol molecules break.

(i) At specific composition, methanol-acetone mixture will form minimum boiling azeotrope and will show positive deviation from Raoult’s law.
(ii) At specific composition, methanol-acetone mixture forms maximum boiling azeotrope and will show positive deviation from Raoult’s law.
(iii) At specific composition, methanol-acetone mixture will form minimum boiling azeotrope and will show negative deviation from Raoult’s law.
(iv) At specific composition, methanol-acetone mixture will form maximum boiling azeotrope and will show negative deviation from Raoult’s law.

The answer is the option (ii) At specific composition methanol-acetone mixture forms maximum boiling azeotrope and will show positive deviation from Raoult’s law.

Question:26

KH value for Ar(g), CO2(g), HCHO (g) and CH4(g) are 40.39, 1.67, $1.83\times 10^{-5}$ and 0.413 respectively.
Arrange these gases in the order of their increasing solubility.

(i) $HCHO < CH_4< CO_2 < Ar$
(ii) $HCHO < CO_2< CH_4< Ar$
(iii) $Ar < CO_2< CH_4< HCHO$
(iv) $Ar < CH_4 < CO_2 < HCHO$

The answer is the option (iii) Solubility decreases with increasing value of KH. The increasing order of solubility of the gases is as follows: $Ar < CO_2< CH_4< HCHO$

## NCERT Exemplar Class 12 Chemistry Solutions Chapter 2: MCQ (Type 2)

Question:27

Which of the following factor(s) affect the solubility of a gaseous solute in the a fixed volume of liquid solvent?
(a) nature of solute (b) temperature (c) pressure

(i) (a) and (c) at constant T
(ii) (a) and (b) at constant P
(iii) (b) and (c) only
(iv) (c) only

The answer is the option (i, ii) solubility of a gaseous solute depends on nature of solute, pressure and temperature.

Question:28

Intermolecular forces between two benzene molecules are near to same strength as those between two toluene molecules. For a mixture of benzene and toluene, which of the following are not true?
(i) $\Delta _{mix} H = zero$
(ii) $\Delta _{mix} V = zero$
(iii) These will form a minimum boiling azeotrope.
(iv) These will not form the ideal solution.

The answer is the option (iii, iv) As interactions in a Benzene and Toluene solution are nearly identical, they will form an ideal solution.
$\Delta _{mix} H = 0$, $\Delta _{mix} V =0$

Question:29

Relative lowering of vapour pressure is a colligative property because _____________.
(i) It depends on the concentration of a non-electrolyte solute in a solution and does not depend on the nature of the solute molecules.
(ii) It depends on a number of particles of electrolyte solute in a solution and does not depend on the nature of the solute particles.
(iii) It depends on the concentration of a non-electrolyte solute in solution as well as on the nature of the solute molecules.
(iv) It depends on the concentration of an electrolyte or nonelectrolyte solute in solution as well as on the nature of solute molecules.

The answer is the option (i, ii) Nature of solute particles is not a factor in determining the relative lowering of vapour pressure. It is only affected by the number of solute particles (non-electrolyte and electrolyte solutes)

Question:30

Van’t Hoff factor i is given by the expression _____________.
(i) i = Normal molar mass/Abnormal molar mass
(ii) i = Abnormal molar mass/ Normal molar mass
(iii) i= Observed colligative property/calculated colligative property
(iv) i = Calculated colligative property/ Observed colligative property

The answer is the option (i,iii)
$i=\frac{\text{Normal molar mass}}{\text{Abnormal molar mass}}\; or\; i=\frac{\text{Observed colligative property}}{\text{Calculated colligative property}}$

Question:31

Isotonic solutions must have the same _____________.
(i) solute
(ii) density
(iii) elevation in boiling point
(iv) depression in freezing point

The answer is the option (ii, iii)

Isotonic solutions must have the same
(ii) density
(iii) elevation in boiling point

Question:32

Which of the following binary mixtures will have the same composition in the liquid and vapour phase?
(i) Benzene – Toluene
(ii) Water-Nitric acid
(iii) Water-Ethanol
(iv) n-Hexane – n-Heptane

The answer is the option (ii, iii) Water-Nitric acid and Water-Ethanol solutions have the same composition in liquid and vapour phases

Question:33

In isotonic solutions ________________.
(i) solute and solvent both are same.
(ii) osmotic pressure is the same.
(iii) solute and solvent may or may not be same.
(iv) a solute is always the same solvent may be different.

The answer is the option (ii, iii) Osmotic pressure is the same for isotonic solutions.

Question:34

The answer is the option (i,iv) for ideal solution.
$p_1 \propto x_1$ and $p_2 \propto y_2$

Question:35

Colligative properties are observed when _____________.
(i) a non-volatile solid is dissolved in a volatile liquid.
(ii) a non-volatile liquid is dissolved in another volatile liquid.
(iii) a gas is dissolved in non-volatile liquid.
(iv) a volatile liquid is dissolved in another volatile liquid.

The answer is the option (i, ii) Colligative properties can be observed on dissolution of a non-volatile solid or liquid in a volatile liquid

## NCERT Exemplar Class 12 Chemistry Solutions Chapter 2: Short Answer Type

Question:36

As the composition of both the liquid and vapour phase is same and both components are coming in the distillate, this implies that they have formed an azeotropic mixture.

Question:37

Boiling point of water increases on addition of NaCl, as it is a non-volatile solute and, hence, reduces the vapour pressure of water and increase thew boiling point. Unlike NaCl, Methyl alcohol is more volatile than water and thus increases the vapour pressure of water and reduces the boiling point.

Question:38

“Like dissolves like” means that a solute dissolves in a solvent if their intermolecular interactions are similar. In other words, we can say that polar solutes can dissolve in polar solvents and non-polar solutes can dissolve in non-polar solvents.

Question:39

$\text{Molarity}=\frac{\text{number of moles of solute}}{\text{Volume of solution in litres} }$
Since, molarity is dependent on volume of the solution, it changes with temperature. The other concentration terms like mass percentage, ppm, mole fraction, molality involve moles or mass of components which are independent of temperature.

Question:40

According to Henry’s law:
$p=K_H\times x$
$K_H=\frac{p}{}x$
Solubility of gas in a liquid decreases with increasing Henry’s law constant.

Question:41

For a given pressure, as the temperature decreases, the solubility of oxygen in water increases. This means that cold sea water has higher concentration of oxygen and aquatic species thrive more in cold water.

Question:42

(a)

1. Deep sea divers require compressed air for breathing under water, which has both $N_2$ and $O_2$. In atmospheric conditions, $N_2$ isn’t soluble in blood, but at higher pressure (as you go deeper into the ocean, the pressure increases) it starts dissolving. When the diver comes back up, the pressure decreases and $N_2$ comes out of the body, leaving behind bubbles in the blood stream, which restrict the flow of blood affecting nerve impulses. In worse cases, these bubbles can burst the capillaries and not let $O_2$ reach the tissues. This condition, known as bends, is very painful and can even be life-threatening.

2. In lower pressure (at higher altitudes), $O_2$ exhibits a lower partial pressure, resulting in lower oxygen concentration in blood and tissues. This condition is known as anoxia and leads to climbers becoming weak and losing ability to think clearly.

(b) $CO_2$ isn’t very soluble in soft drinks at atmospheric pressure. So, to dissolve $CO_2$ in soft drinks, soda bottles are sealed under high pressure. However, when the bottle is opened, the pressure decreases suddenly and excess $CO_2$ fizzes out.

Question:43

In pure liquid water, molecules of water cover the entire surface. When a non-volatile solute is dissolved in water, it replaces part of the surface water molecules. As a result, the number of solvent molecules which can escape also gets reduced and thus, the vapour pressure of the solution reduces.

Question:44

Salt is spread over snow covered roads to lower water’s freezing point enough to melt the ice (snow). This leads to the snow melting away and clearing the road.

Question:45

What is a “semi-permeable membrane”?

Semipermeable membrane are membranes which only permit the flow of solvent molecules, but not the solute molecules. Only the solvent molecules move across the semipermeable membrane during osmosis and Reverse osmosis.

Question:46

For carrying out reverse osmosis cellulose acetate, potassium ferrocyanide, etc. are used as semipermeable membrane.

## NCERT Exemplar Class 12 Chemistry Solutions Chapter 2: Matching Type

Question:47

Match the items given in Column I and Column II.

 Column I Column II (i) Saturated solution (a) Solution having the same osmotic pressure at a given temperature as that of the given solution (ii) Binary solution (b) A solution whose osmotic pressure is less than that of another. (iii) Isotonic solution (c) Solution with two components. (iv)Hypotonic solution (d) A solution which contains the maximum amount of solute that can be dissolved in a given amount of solvent at a given temperature. (v) Solid solution (e) A solution whose osmotic pressure is more than another. (vi) Hypertonic solution (f) A solution in the solid phase
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(i —> d), (ii —> c); (iii—> a); (iv —> b), (v —> f); (vi —> e)
i.Saturated solution: It is a solution where no more of the solute can be dissolved at the given temperature and pressure.
ii.Binary solution: It is a solution with two components.
iii.Isotonic solution: It is a solution having same osmotic pressure at a given temperature as that of given solution.
iv.Hypotonic solution: It is a solution whose osmotic pressure is lesser than that of a given solution.
v,Solid solution: It is a solution in solid phase.
vi.Hypertonic solution: It is a solution whose osmotic pressure is greater than that of a given solution

Question:48

Match the items given in Column I with the type of solutions given in Column II.

 Column I Column II (i) Soda water (a) A solution of the gas in solid (ii) Sugar solution (b) A solution of the gas in gas (iii) German silver (c) A solution of solid in liquid (iv) Air (d) A solution of solid in solid (v) Hydrogen gas in palladium (e) A solution of the gas in liquid (f) A solution of liquid in solid

(i → e), (ii → c), (iii → d); (iv → b), (v → a)
i. Soda water: A solution of gas in $(CO_2)$ liquid(soft drink)
ii. Sugar solution: A solution of solid (sugar) in liquid (water)
iii. German silver: It is an alloy of Cu, Zn and Ni and a solid solution of solid in solid.
iv. Air: A solution of gas igas. Air is a mixture of various gases.
v. Hydrogen gas in palladium: Hydrogen gas in palladium is used as a reducing agent and is an example of solution of gas in solid.

Question:49

Match the laws given in Column I with expressions given in Column II.

 Column I Column II (i) Raoult’s law (a)$\Delta T_f= K_fm$ (ii) Henry’s law (b) $\pi = CRT$ (iii) Elevation of boiling point (c) $p = x_{1}P_1^0+ x_{2}P_2^0$ (iv) Depression in freezing point (d) $\Delta T_b= K_bm$ (v) Osmotic pressure (e) $p = K_H.x$

(i) → (c) (ii) → (e) (iii) → (d) (iv) → (a) (v) → (b)
i.Raoult’s law: $p = x_{1}P_1^0+ x_{2}P_2^0$
ii. Henry’s law: $p = K_H.x$
iii. Elevation of boiling point: $\Delta T_b= K_bm$
iv. Depression in freezing point: $\Delta T_f= K_fm$
v. Osmotic pressure: $\pi = CRT$

Question:50

Match the terms given in Column I with expressions given in Column II.

 Column I Column II (i) Mass percentage (a) $\frac{Number \;of\; moles\; of\; the\; solute\; component}{Volume\; of \;solution \;in \; litres}$ (ii) Volume percentage (b)$\frac{Number \;of \;moles\; of\; a\; component}{Total\; number\; of\; moles\; of\; all\; the\; components}$ (iii) Mole fraction (c) $\frac{Volume\; of\; the\; solute\; component\; in \;solution}{Total\; volume \;of solution\;}\times 100$ (iv) Molality (d) $\frac{Mass\; of\; the\; solute\; component\; in \;solution}{Total \;mass \;of \;the\; solution}\times 100$ (v) Molarity (e) $\frac{Number\; of \;moles \;of\; the\; solute\; components}{Mass\; of \;solvent \;in\; kilograms}\times 100$

(i → d), (ii → c), (iii →b), (iv → e), (v →a)

## NCERT Exemplar Class 12 Chemistry Solutions Chapter 2: Assertion and Reason Type

Question:51

The answer is the option (a) Molarity depends on volume, which changes with temperature. Hence, molarity also changes with change in temperature.

Question:52

The answer is the option (d) Addition of a volatile solute to a volatile solvent increases vapour pressure and decreases boiling point. Methyl alcohol and water interaction are an example of the same.

## NCERT Exemplar Class 12 Chemistry Solutions Chapter 2: Long Answer Type

Question:55

1. $\text{w/w(Mass percentage)}=\frac{\text{Mass of component in the solution}}{\text{Total mass of the solution}}\times 100$

2. $\text{V/V(Volume percentage)}=\frac{\text{Volume of component in the solution}}{\text{Total volume of the solution}}\times 100$

3. $\text{w/V (mass by Volume percentage)}=\frac{\text{Mass of the solute}}{\text{Total volume of the solution}}\times 100$

4. $\text{ppm (parts per million)}=\frac{\text{Number of parts of component}}{\text{Total number of parts of all components}}\times 10^{6}$

5. $\text{x( Mole fraction)}=\frac{\text{Number of moles of component}}{\text{Total number of moles of all components}}$

6. $\text{M (Molarity)}=\frac{\text{Moles of solute}}{\text{Volume of the solution in litre}}$

7. $\text{m (Molality)}=\frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}$

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Mass percentage, ppm, mole fraction and molality are only dependent on mass or number of moles making them independent of temperature changes.

Question:56

$p=p_A^ox_A+p_B^ox_B$
Where, $p_A^o,p_B^o$ are the vapour pressure of pure components (A) and (B)
And $x_{A}\text{and }x_{B}$ are the mole fractions of the components in the solution.

(b) On dissolving in water, the vapour pressure lowers. Vapour pressure of solution is:
$p=p^ox_A$
Where $x_{A}$=mole fraction of solvent
$p^{0}$=vapour pressure of pure solvent
p=vapour pressure of solution
Similarly,
$\frac{\Delta p}{p^o}=x_B$
$\Delta p=p^o-p$
$x_{B}$=mole fraction of solute

Question:57

Ideal solutions: They obey Raoult’s law irrespective of the concentration. For an ideal solution,
$\Delta _{mix}H=0, \Delta _{mix}V=0$
$\text{A-B interactions}\approx \text{A-A interactions and B-B interactions}$
Non-ideal solutions: They don’t follow Raoult’s law over the entire range of concentration
Positive deviations: Vapour pressure is higher than the calculated values
$\Delta _{mix}H=+ve, \Delta _{mix}V=+ve$
A-B interactions < A-A interactions and B-B interactions
Negative deviations: Vapour pressure is lower than the calculated values
$\Delta _{mix}H=-ve, \Delta _{mix}V=-ve$
A-B interactions>A-A interactions and B-B interactions

Question:58

Azeotropes are solutions which have the same compositions of components in both the liquid and vapour phase and boil at a constant temperature. The components cannot be separated by fractional distillation as they have the same boiling point. There are two types of azeotropes

1. Minimum boiling azeotropes: They show large positive deviation from Raoult’s law (A-B interactions are weaker than A-A and B-B interactions), e.g., ethanol-water mixture

2. Maximum boiling azeotropes: They show large negative deviation from Raoult’s law (A-B interactions are stronger than A-A and B-B interactions) e.g., solution having composition 68% HNO3 and 32% water by mass.

Question:59

Raisins swell in water due to osmosis. Water moves from a place of lower concentration (Water holder) to higher concentration (Raisin) through the skin of raisin which acts as a semipermeable membrane.

Applications of the phenomenon

1. Water moves from soil to plant roots partly due to osmosis.

2. Preservation of meat against bacterial action by adding salt.

3. Preservation of fruits against bacterial action by adding sugar. Bacterium in canned fruit loses water through the process of osmosis, shrivels, and dies.

4. Reverse osmosis is used for desalination of water.

Question:60

Some of the biological and industrial importance of osmosis are as follows-

1. Osmosis is responsible for the circulation of water to all the body parts of animals

2. Water moves from soil to plant roots partly due to osmosis. Concentration of cell sap inside the root hair cells is higher compared to that of water present in the soil.

3. Water circulation inside the plant body from root to treetop is also because of osmosis.

4. Osmosis helps in growth of plants and germination of seeds.

5. Due to endosmosis, Red Blood Cells burst when they are placed in water.

6. Osmosis controls various functions of plants, e.g., stretching of leaves and flowers, opening, and closing of flowers.

7. Salt and sugar in pickles prevent the growth of bacteria and fungi by osmosis and thus, act as preservatives.

8. Endosmosis is responsible for the swelling of dead bodies under water.

9. Dried fruits and vegetables swell and return to their original form when placed in water. It is also due to the endosmosis of water.

10. Edema: Tissues become puffy when a person intakes an excess amount of salt.

Question:61

This can be achieved as under:

1. Take a mineral acid solution and put an egg inside it and leave it for 2 hours. Most of the outer shell will dissolve. Remove any remaining part with your fingers.

2. Take a saturated solution (hypertonic) and lace the egg in it for 3 hours. Egg’s size reduces as the egg shrivels due to osmosis.

3. Place the egg in a bottle with a narrow neck and fill water (hypotonic) in it. The egg will regain its shape.

Question:62

Abnormal molecular masses are shown by the compounds which dissociate/associate in the solvent.

1. Association: Colligative properties depend on the number of particles in a solution. Certain compounds like benzoic acid or ethanoic acid dimerise in benzene due to hydrogen bonding resulting in the reduction of the number of particles and thus, solutes show lower colligative property.

2. Dissociation: Similarly, certain compounds like electrolytes (NaCl, KCl) etc dissociate into ions, increasing the number of particles, and thus, show higher value of colligative property.

Van’t Hoff introduced a factor to account for association or dissociation, known as Van’t Hoff factor.
$i=\frac{\text{Expected molar mass}}{\text{Abnormal molar mass}}$
$=\frac{\text{Observed colligative property}}{\text{Calculated colligative property}}$
$=\frac{\text{Total number of moles of particles after association or disaasociation}}{\text{Total number of moles of particles before association or disaccioation}}$

NCERT exemplar Class 12 Chemistry solutions chapter 2 provided here for the NCERT Books are very useful and detailed from the point of view of aiding practice, preparation and working for Board exams as well as the JEE Main exams. Students can also use NCERT exemplar Class 12 Chemistry solutions chapter 2 PDF download as and when needed.

## Main Subtopics Covered in NCERT Exemplar Class 12 Chemistry Solutions Chapter 2 Inverse Trigonometric Functions

• Types of Solutions
• Expressing Concentration of Solutions
• Solubility
• Solubility of a Solid in a Liquid
• Solubility of a Gas in a Liquid
• Vapour Pressure of Liquid Solutions
• Vapour Pressure of Liquid-Liquid Solutions
• Raoult’s Law as a special case of Henry’s Law
• Vapour Pressure of Solutions of Solids in Liquids
• Ideal and Non- ideal Solutions
• Ideal Solutions
• Non-ideal Solutions
• Colligative Properties and Determination of Molar Mass
• Relative Lowering of Vapour Pressure
• Elevation of Boiling Point
• Depression of Freezing Point
• Osmosis and Osmotic Pressure
• Reverse Osmosis and Water Purification
• Abnormal Molar Masses

## NCERT Exemplar Class 12 Chemistry Solutions Chapter 2 Inverse Trigonometric Functions-What will the students learn from?

The students will be able to understand the formation of different types of solutions and their properties that can be easily related to various solutions around us. They will also be able to understand and express the concentration of different solutions with the help of different methods provided by going through NCERT exemplar solutions for Class 12 Chemistry chapter 2. The students will learn to calculate the concentration, percentage, morality, molality, etc. of different compounds and also the factors affecting them by solving NCERT exemplar Class 12 Chemistry solutions chapter 2. It will also give an insight into the solubility of different solutions and the effect of temperature on these solutions. The students will also learn about Henry’s law and Raoult’s law to establish a relationship between different properties of solid and knowledge about colligative properties of solid.

## NCERT Exemplar Class 12 Chemistry Solutions

 Chapter 1 Solid State Chapter 3 Electrochemistry Chapter 4 Chemical Kinetics Chapter 5 Surface Chemistry Chapter 6 General Principles and Processes of Isolation of Elements Chapter 7 The p-Block Elements Chapter 8 The d & f Block Elements Chapter 9 Coordination Compounds Chapter 10 Haloalkanes and Haloarenes Chapter 11 Alcohols, Phenols, and Ethers Chapter 12 Aldehydes, Ketones, and Carboxylic Acids Chapter 13 Amines Chapter 14 Biomolecules Chapter 15 Polymers Chapter 16 Chemistry in Everyday Life

## NCERT Class 12 Chemistry Chapter 2 Solutions Important Topics To Cover For Exams

NCERT exemplar Class 12 Chemistry solutions chapter 2 also establishes different relationships between different properties of different combinations and their reliability on one another such as Raoult’s law that helps to determine the vapour pressure of distinguished components of any solution with the help of Dalton’s law of partial pressure, etc. for liquid-liquid and solid-liquid solutions.

NCERT exemplar Class 12 Chemistry chapter 2 solutions also covers a brief description on various properties of solutions depending on the number of solute particles also known as colligative properties with relation to the boiling point, vapour pressure, mole fraction, freezing point, osmotic pressure, temperature and much more.

Class 12 Chemistry NCERT exemplar solutions chapter 2 also talks about various azeotropes formed from such solutions and the process of water purification through reverse osmosis.

### NCERT Exemplar Class 12 Solutions

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### Check NCERT Solutions for Class 12 Chemistry

 Chapter 1 The Solid State Chapter 2 Solutions Chapter 3 Electrochemistry Chapter 4 Chemical Kinetics Chapter 5 Surface chemistry Chapter 6 General Principles and Processes of isolation of elements Chapter 7 The P-block elements Chapter 8 The d and f block elements Chapter 9 Coordination compounds Chapter 10 Haloalkanes and Haloarenes Chapter 11 Alcohols, Phenols, and Ethers Chapter 12 Aldehydes, Ketones and Carboxylic Acids Chapter 13 Amines Chapter 14 Biomolecules Chapter 15 Polymers Chapter 16 Chemistry in Everyday life

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### Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

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

Thank you and wishing you all the best for your bright future.

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

• No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
• You have to appear for the 2025 12th board exams.
• Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
• Aim to register before late October to avoid extra fees.
• Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

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

Hello Aspirant , Hope your doing great . As per your query , your eligible for JEE mains in the year of 2025 , Every candidate can appear for the JEE Main exam 6 times over three consecutive years . The JEE Main exam is held two times every year, in January and April.

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