## NCERT Solutions for Class 12 Chemistry Chapter 2 – Free PDF Download

**NCERT Solutions for Class 12 Chemistry Chapter 2 Solutions** provides accurate and adequate Chemistry knowledge. The NCERT Solutions for Class 12 Chemistry can be utilised by the students to prepare for the examination and to solve the exercise questions of the Class 12 Chemistry Chapter 2. All the solutions are created according to the latest CBSE syllabus and guidelines.

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### Chemistry Class 12 NCERT Solutions Chapter 2 Solutions – Important Questions

**Q 2.1) If 22 g of benzene is dissolved in 122 g of carbon tetrachloride, determine the mass percentage of carbon tetrachloride (CCl _{4}) and benzene (C_{6}H_{6}).**

**Answer 2.1:**

Mass percentage of Benzene (C_{6}H_{6}) =

=

=

= 15.28%

Mass percentage of Carbon Tetrachloride (CCl_{4}) =

=

=

= 84.72%

**Q 2.2) If benzene in solution containing 30% by mass in carbon tetrachloride, calculate the mole fraction of benzene.**

**Answer 2.2:**

Assume the mass of benzene be 30 g in the total mass of the solution of 100 g.

Mass of CCl_{4} = (100 − 30) g

= 70 g

Molar mass of benzene (C_{6}H_{6}) = (6 × 12 + 6 × 1) g

= 78 g

Therefore, Number of moles of C_{6}H_{6} =

= 0.3846 mol

Molar mass of CCl_{4} = 1 x 12 + 4 x 355 = 154 g

Therefore, Number of moles of CCl_{4} =

= 0.4545 mol

Thus, the mole fraction of C_{6}H_{6} is given as:

=

= 0.458

**Q 2.3) Determine the molarity of each of the solutions given below:**

**(a) 30 g of Co(NO) _{3}. 6H_{2}O in 4.3 L of solution**

**(b) 30 mL of 0.5 M H _{2}SO_{4} diluted to 500 mL.**

**Answer 2.3:**

We know that,

Molarity =

(a) Molar mass of Co(NO)_{3}. 6H_{2}O = 59 + 2 (14 + 3 x 16) + 6 x 18 = 291 g

Therefore, Moles of Co(NO)_{3}. 6H_{2}O =

= 0.103 mol

Therefore, molarity =

= 0.023 M

(b) Number of moles present in 1000 mL of 0.5 M H_{2}SO_{4} = 0.5 mol

Therefore, Number of moles present in 30 mL of 0.5 M H_{2}SO_{4} =

= 0.015 mol

Therefore, molarity =

= 0.03 M

**Q 2.4) To make 2.5 kg of 0.25 molar aqueous solution, determine the mass of urea (NH _{2}CONH_{2}) that is required.**

**Answer 2.4:**

Molar mass of urea (NH2CONH2) = 2(1 × 14 + 2 × 1) + 1 × 12 + 1 × 16 = 60 g

0.25 molar aqueous solution of urea means:

1000 g of water contains 0.25 mol = (0.25 × 60) g of urea = 15 g of urea

That is,

(1000 + 15) g of solution contains 15 g of urea

Therefore, 2.5 kg (2500 g) of solution contains =

= 36.95 g

= 37 g of urea (approx.)

Hence, mass of Urea required is 37 g.

**Q 2.5) If 1.202 g $mL^{-1}$ is the density of 20% aqueous KI, determine the following:**

**(a) Molality of KI**

**(b) Molarity of KI**

**(c) Mole fraction of KI**

**Answer 2.5:**

(a) Molar mass of KI = 39 + 127 = 166 g

20% aqueous solution of KI means 20 g of KI is present in 100 g of solution.

That is,

20 g of KI is present in (100 – 20) g of water = 80 g of water

Therefore, molality of the solution =

=

= 1.506 m

= 1.51 m (approx.)

(b) It is given that the destiny of the solution = 1.202

Volume of 100 g solution =

=

= 83.19 mL

=

Therefore, molarity of the solution =

= 1.45 M

(c) Moles of KI =

Moles of water =

Therefore, mole =

Fraction of KI =

= 0.0263

**Q 2.6) Calculate Henry’s law constant when the solubility of H _{2}S (a toxic gas with rotten egg like smell) in water at STP is 0.195 m **

**Answer 2.6:**

It is given that the solubility of H_{2}S in water at STP is 0.195 m, i.e., 0.195 mol of H_{2}S is dissolve in 1000 g of water.

Moles of water =

= 55.56 mol

Therefore, Mole fraction of H_{2}S, x =

=

= 0.0035

At STP, pressure (p) = 0.987 bar

According to Henry’s law: p =

=>

=

= 282 bar

**Q 2.7) 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 2.7:**

Total amount of solute present in the mixture is given by,

= 75 + 160

= 235 g

Total amount of solution = 300 + 400 = 700 g

Therefore, mass percentage of the solute in the resulting solution =

= 33.57%

And, mass percentage of the solvent in the resulting solution is:

= (100 – 33.57) %

= 66.43%

**Q 2.8) The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.**

**Answer 2.8:**

It is given that:

According to Raoult’s law:

Therefore, total pressure,

=>

=>

=>

=> 600 = (450 – 700) x_{A} + 700

=> –100 = –250x_{A}

=> x_{A} = 0.4

Therefore,

Now,

= 450 x 0.4 = 180 mm of Hg

= 700 x 0.6 = 420 mm of Hg

Now, in the vapour phase: Mole fraction of liquid A =

=

=

= 0.30

And, mole fraction of liquid B = 1 – 0.30 = 0.70

**Q 2.9) Find the vapor pressure of water and its relative lowering in the solution which is 50 g of urea (NH _{2}CONH_{2}) dissolved in 850 g of water. (Vapor pressure of pure water at 298 K is 23.8 mm Hg)**

**Answer 2.9:**

It is given that vapour pressure of water,

Weight of water taken,

Weight of urea taken,

Molecular weight of water,

Molecular weight of urea,

Now, we have to calculate vapour pressure of water in the solution. We take vapour pressure as p_{1}.

Now, from Raoult’s law, we have:

=>

=>

=>

=>

=>

Hence, the vapour pressure of water in the given solution is 23.4 mm of Hg and its relative lowering is 0.0173.

**Q 2.10) How much of sucrose is to be added to 500 g of water such that it boils at 100°C if the molar elevation constant for water is 0.52 K kg mol ^{-1} and the boiling point of water at 750 mm Hg is 99.63°C?**

**Answer 2.10:**

Here, elevation of boiling point

= 0.37 K

Mass of water,

Molar mass of sucrose (C_{12}H_{22}O_{11}),

= 342 g

Molar elevation constant, K_{b} = 0.52 K kg

We know that:

=>

=

= 121.67 g (approximately)

Hence, the amount of sucrose that is to be added is 121.67 g

**Q 2.11) To lower the melting point of 75 g of acetic acid by 1.5 ^{0}C, how much mass of ascorbic acid is needed to be dissolved in the solution where K_{t} = 3.9 K kg **

**Answer 2.11:**

Mass of acetic acid (w_{1}) = 75 g

Molar mass of ascorbic acid (C_{6}H_{8}O_{6}), M_{2} = 6 x 12 + 8 x 1 + 6 x 16 = 176 g

Lowering the melting point

We know that:

=>

=

= 5.08 g (approx)

Hence, the amount of ascorbic acid needed to be dissolved is 5.08 g.

**Q 2.12) If a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at 37°C, calculate the osmotic pressure in Pascal exerted by it?**

**Answer 2.12:**

It is given that:

Volume of water (V) = 450 mL = 0.45 L

Temperature (T) = 37 + 273 = 310 K

Number of moles of the polymer, n =

We know that:

Osmotic pressure,

=

= 30.98 Pa

= 31 Pa (approx)

**Q 2.13) The partial pressure of ethane over a solution containing 6.56 × 10 ^{–3} g of ethane is 1 bar. If the solution contains 5.00 × 10^{–2} g of ethane, then what shall be the partial pressure of the gas?**

**Answer 2.13:**

Molar mass of ethane (C_{2}H_{6}) = 2 x 12 + 6 x 1 = 30 g

Therefore, number of moles present in

=

Let ‘x’ be the number of moles of the solvent, According to Henry’s law,

=> 1 bar =

=> 1 bar =

=>

Number of moles present in

=

According to Henry’s law,

=

=

= 0.764 bar

Hence, partial pressure of the gas shall be 0.764 bar.

**Q 2.14) What is meant by positive and negative deviations from Raoult’s law and how is the sign of $\Delta_{sol} H$ related to positive and negative deviations from Raoult’s law?**

**Answer 2.14: **

According to Raoult’s law, the partial vapour pressure of each volatile component in any solution is directly proportional to its mole fraction. The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solutions. The solutions that do not obey Raoult’s law (non-ideal solutions) have vapour pressures either higher or lower than that predicted by Raoult’s law. If the vapour pressure is higher, then the solution is said to exhibit positive deviation, and if it is lower, then the solution is said to exhibit negative deviation from Raoult’s law.

**Vapour pressure of a two-component solution showing positive deviation from Raoult’s law**

**Vapour pressure of a two-component solution showing negative deviation from Raoult’s law**

In the case of an ideal solution, the enthalpy of the mixing of the pure components for forming the solution is zero.

In the case of solutions showing positive deviations, absorption of heat takes place.

In the case of solutions showing negative deviations, evolution of heat takes place.

**Q 2.15) 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 2.15:**

Vapour pressure of the solution at normal boiling point,

Vapour pressure of pure water at normal boiling point,

Mass of solute, w_{2} = 2 g

Mass of solvent (water), M_{1} = 18 g

According to Raoult’s law,

=>

=>

=>

= 41.35 g

Hence, 41.35 g

**Q 2.16) Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?**

**Answer 2.16:**

Vapour pressure of heptanes,

Vapour pressure of octane,

We know that,

Molar mass of heptanes (C_{7}H_{16}) = 7 x 12 + 16 x 1 = 100 g

Therefore, number of moles of heptane =

Molar mass of octane (C_{8}H_{18}) = 8 x 12 + 18 x 1 = 114 g

Therefore, number of moles of octane =

Mole fraction of heptane,

And, mole fraction of octane,

Now, partial pressure of heptane,

= 0.456 x 105.2

= 47.97 kPa

Partial pressure of octane,

= 0.544 x 46.8

= 25.46 kPa

Hence, vapour pressure of solution,

= 47.97 + 25.46

= 73.43 kPa

**Q 2.17) The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.**

**Answer 2.17:**

1 molal solution means 1 mol of the solute is present in 100 g of the solvent (water).

Molar mass of water = 18 g

Therefore, number of moles present in 1000 g of water =

= 55.56 mol

Therefore, mole fraction of the solute in the solution is

It is given that,

Vapour pressure of water,

Applying the relation,

=>

=> 12.3 – p_{1} = 0.2177

=> p_{1} = 12.0823

= 12.08 kPa (approx)

Hence, the vapour pressure of the solution is 12.08 kPa.

**Q 2.18) Calculate the mass of a non-volatile solute (molar mass 40 g mol ^{–1}) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%?**

**Answer 2.18:**

Let

Then, after dissolving the non-volatile solute the vapour pressure of octane is

Molar mass of solute, M_{2} = 40 g

Mass of octane, w_{1} = 114 g

Molar mass of octane, (C_{8}H_{18}), M_{1} = 8 x 12 + 18 x 1 = 114 g

Applying the relation,

=>

=>

=> 0.2 =

=> w_{2} = 8 g

Hence, the required mass of the solute is 8 g.

**Q 2.19) A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate: (i) molar mass of the solute (ii) vapour pressure of water at 298 K.**

**Answer 2.19:**

(i) Let, the molar mass of the solute be M g

Now, the number of moles of solvent (water),

And, the number of moles of solute,

p_{1} = 2.8 kPa

Applying the relation:

=>

=>

=>

=>

=>

=>

=>

After the addition of 18 g of water:

Again applying the relation:

=>

=>

=>

=>

=>

=>

=>

Dividing equation (i) by (ii), we have:

=>

=>

=> 87M + 435 = 84M + 504

=> 3M = 69

=> M = 23 g

Therefore, 23 g

(ii) Putting the value of ‘M’ in equation (i), we have:

=>

=>

Hence, 3.53 kPa is the vapour pressure of water at 298 K.

**Q 2.20) A 5% solution (by mass) of cane sugar in water has freezing point of 271K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K**

**Answer 2.20:**

Molar mass of sugar (C_{12}H_{22}O_{11}) = 12 x 12 + 22 x 1 + 11 x 16 = 342 g

5% solution (by mass) of cane sugar in water means 5 g of cane sugar is present in (100 − 5)g

= 95 g of water.

Now, number of moles of cane sugar =

Therefore, molality of the solution,

Applying the relation,

=>

=

= 13.99 K kg

Molar mass of glucose (C_{6}H_{12}O_{6}) = 6 x 12 + 12 x 1 + 6 x 16 = 180 g

5% glucose in water means 5 g of glucose is present in (100 − 5) g = 95 g of water.

Therefore, number of moles of glucose =

Therefore, molality of the solution, m =

= 0.2926 mol

Applying the relation:

=

= 4.09 K (approx)

Hence, the freezing point of 5 % glucose solution is (273.15 – 4.09) K = 269.06 K.

**Q 2.21) Two elements A and B form compounds having formula AB _{2} and AB_{4}. When dissolved in 20 g of benzene (C_{6}H_{6}), 1 g of AB_{2} lowers the freezing point by 2.3 K whereas 1.0 g of AB_{4} lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg mol^{–1}. Calculate atomic masses of A and B.**

**Answer 2.21:**

We know that,

Then,

= 110.87 g

= 196.15 g

Now, we have the molar masses of AB_{2} and AB_{4} as 110.87 g

Let the atomic masses of A and B be x and y respectively.

Now, we can write:

x + 2y = 110.87 …(i)

x + 4y = 196.15 …(ii)

Subtracting equation (i) from (ii), we have

2y = 85.28

=> y = 42.64

Putting the value of ‘y’ in equation (1), we have:

x + 2 (42.64) = 110.87

=> x = 25.59

Hence, the atomic masses of A and B are 25.59 u and 42.64 u respectively.

**Q 2.22) At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?**

**Answer 2.22:**

Given:

T = 300 K

n = 1.52 bar

R = 0.083 bar L

Applying the relation, n = CRT

=> C =

=

= 0.061 mol

Since the volume of the solution is 1 L, the concentration of the solution would be 0.061 M.

** ****Q 2.23) Suggest the most important type of intermolecular attractive interaction in
the following pairs.
(i) n-hexane and n-octane
(ii) I**

_{2}and CCl

_{4}(iii) NaClO

_{4}and water (iv) methanol and acetone (v) acetonitrile (CH

_{3}CN) and acetone (C

_{3}H

_{6}O).

**Answer 2.23:**

(i) Van der Wall’s forces of attraction.

(ii) Van der Wall’s forces of attraction.

(iii) Ion-dipole interaction.

(iv) Dipole-dipole interaction.

(v) Dipole-dipole interaction.

**Q 2.24) Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, KCl, CH _{3}OH, CH_{3}CN.**

**Answer 2.24:**

n-octane is a non-polar solvent. Therefore, the solubility of a non-polar solute is more than that of a polar solute in the n-octane.

The order of increasing polarity is:

Cyclohexane < CH_{3}CN < CH_{3}OH < KCl

Therefore, the order of increasing solubility is:

KCl < CH_{3}OH < CH_{3}CN < Cyclohexane

**Q 2.25) Amongst the following compounds, identify which are insoluble, partially
soluble and highly soluble in water?
(i) phenol (ii) toluene (iii) formic acid
(iv) ethylene glycol (v) chloroform (vi) pentanol.**

**Answer 2.25:**

(i) Phenol (C_{6}H_{5}OH) has the polar group −OH and non-polar group –C_{6}H_{5}. Thus, phenol is partially soluble in water.

(ii) Toluene (C_{6}H_{5}−CH_{3}) has no polar groups. Thus, toluene is insoluble in water.

(iii) Formic acid (HCOOH) has the polar group −OH and can form H-bond with water.

Thus, formic acid is highly soluble in water.

(iv)Ethylene glycol has a polar −OH group and can form H−bond. Thus, it is highly soluble in water.

(v) Chloroform is insoluble in water.

(vi)Pentanol (C_{5}H_{11}OH) has a polar −OH group, but it also contains a very bulky nonpolar −C_{5}H_{11} group. Thus, pentanol is partially soluble in water.

**Q 2.26) If the density of some lake water is 1.25g mL ^{–1} and contains 92 g of Na^{+} ions per kg of water, calculate the molarity of Na^{+} ions in the lake**

**Answer 2.26:**

Number of moles present in 92 g of Na^{+} ions =

= 4 mol

Therefore, molality of Na^{+} ions in the lake =

= 4 m

**Q 2.27) If the solubility product of CuS is 6 × 10 ^{–16}, calculate the maximum molarity of
CuS in aqueous solution.**

**Answer 2.27: **

Solubility product of CuS,

Let s be the solubility of CuS in mol L^{-1}.

Now, s s

= s x s

= s^{2}

Then, we have,

=>

=

Hence,

**Q 2.28) Calculate the mass percentage of aspirin (C _{9}H_{8}O_{4}) in acetonitrile (CH_{3}CN) when**

6.5 g of C_{9}H_{8}O_{4} is dissolved in 450 g of CH_{3}CN.

**Answer 2.28:**

6.5 g of C_{9}H_{8}O_{4} is dissolved in 450 g of CH_{3}CN.

Then, total mass of the solution = (6.5 + 450) g = 456.5 g

Therefore, mass percentage of C_{9}H_{8}O_{4} =

= 1.424%

**Q 2.29) Nalorphene (C _{19}H_{21}NO_{3}), similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose of nalorphene generally given is 1.5 mg. Calculate the mass of 1.5 × 10^{–3} m aqueous solution required for the above dose.**

**Answer 2.29:**

The molar mass of nalorphene (C_{19}H_{21}NO_{3}) = 19 x 12 + 21 x 1 + 1 x 14 + 3 x 16 = 311 g mol^{-1}

In

1 kg (1000 g) of water contains

= 0.4665 g

Therefore, total mass of the solution = (1000 + 0.4665) g = 1000.4665 g

This implies that the mass of the solution containing 0.4665 g of nalorphene is 1000.4665 g.

Therefore, the mass of the solution containing 1.5 mg of nalorphene is:

= 3.22 g

Hence, 3.22 g is the required mass of aqueous solution.

**Q 2.30) Calculate the amount of benzoic acid (C _{6}H_{5}COOH) required for preparing 250**

mL of 0.15 M solution in methanol.

**Answer 2.30:**

0.15 M solution of benzoic acid in methanol means,

1000 mL of solution contains 0.15 mol of benzoic acid

Therefore, 250 mL of solution contains

= 0.0375 mol of benzoic acid

Molar mass of benzoic acid (C_{6}H_{5}COOH) = 7 x 12 + 6 x 1 + 2 x 16 = 122 g mol^{-1}

Hence, required benzoic acid = 0.0375 mol x 122 g mol^{-1} = 4.575 g

**Q 2.31) 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 2.31:**

Among H, Cl, and F, H is least electronegative while F is most electronegative. Then, F can withdraw electrons towards itself more than Cl and H. Thus, trifluoroacetic acid can easily lose H^{+} ions i.e., trifluoroacetic acid ionizes to the largest extent. Now, the more ions produced, the greater is the depression of the freezing point. Hence, the depression in the freezing point increases in the order:

Acetic acid < trichloroacetic acid < trifluoroacetic acid

**Q 2.32) Calculate the depression in the freezing point of water when 10 g of CH _{3}CH_{2}CHClCOOH is added to 250 g of water. K_{a} = 1.4 × 10^{–3}, K_{f} = 1.86 K kg mol^{–1 }**

**Answer 2.32:**

Molar mass of CH_{3}CH_{2}CHCICOOH = 15 + 14 + 13 + 35.5 + 12 + 16 + 16 + 1

= 122.5 g mol^{-1}

Therefore, No. of moles present in 10 g of CH_{3}CH_{2}CHCICOOH =

= 0.0816 mol

It is given that 10 g of CH_{3}CH_{2}CHCICOOH is added to 250 g of water.

Therefore, Molality of the solution, CH_{3}CH_{2}CHCICOOH =

= 0.3264 mol kg^{-1}

Let ’a’ be the degree of dissociation of CH_{3}CH_{2}CHCICOOH.

CH_{3}CH_{2}CHCICOOH undergoes dissociation according to the following equation:

=

Since a is very small with respect to 1,

Now,

=>

=>

=

= 0.0655

Again,

Total moles of equilibrium = 1 – a + a + a = 1 + a

=

= 1 + 0.0655

= 1.0655

Hence, the depression in the freezing point of water is given as:

=

= 0.65 K

**Q 2.33) 19.5 g of CH _{2} FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.00 C. Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid**

**Answer 2.33:**

Given:

w_{1} = 500 g

w_{2} = 19.5 g

K_{f} = 1.86 K kg

We know that:

=