# NCERT Solutions for Class 12 Chemistry Chapter 2 Solutions

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

NCERT Solutions for Class 12 Chemistry Chapter 2 Solutions provide accurate and adequate Chemistry knowledge. The NCERT Solutions for Class 12 Chemistry can be utilised by the students to prepare for the board 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 for 2022-23 and its 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 (CCl4) and benzene (C6H6).

Mass percentage of Benzene (C6H6) = $\frac{Mass\; of\; C_{6}H_{6}}{Total \;mass \;of \;the \;solution} v\times 100$

= $\frac{Mass\; of\; C_{6}H_{6}}{Mass\;of\;C_{6}H_{6} + Mass\;of\;CCl_{4}} \times 100$

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

= 15.28%

Mass percentage of Carbon Tetrachloride (CCl4) = $\frac{Mass\; of\; CCl_{4}}{Total \;mass \;of \;the \;solution} \times 100$

= $\frac{Mass\; of\; CCl_{4}}{Mass\;of\;C_{6}H_{6} + Mass\;of\;CCl_{4}} \times 100$

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

= 84.72%

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

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

Mass of CCl4 = (100 − 30) g

= 70 g

Molar mass of benzene (C6H6) = (6 × 12 + 6 × 1) g $mol^{-1}$

= 78 g $mol^{-1}$

Therefore, the number of moles of C6H6 = $\frac{30}{78}$ mol

= 0.3846 mol

Molar mass of CCl4 = 1 x 12 + 4 x 355 = 154 g $mol^{-1}$

Therefore, the number of moles of CCl4 = $\frac{70}{154}$ mol

= 0.4545 mol

Thus, the mole fraction of C6H6 is given as

$\frac{Number\;of\;moles\;of\;C_{6}H_{6}}{Number\;of\;moles\;of\;C_{6}H_{6} + Number\;of\;moles\;of\;CCl_{4}}$

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

= 0.458

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

(a) 30 g of Co(NO)3. 6H2O in 4.3 L of solution.

(b) 30 mL of 0.5 M H2SO4 diluted to 500 mL.

We know that,

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

(a) Molar mass of Co(NO)3. 6H2O = 59 + 2 (14 + 3 x 16) + 6 x 18 = 291 g $mol^{-1}$

Therefore, the moles of Co(NO)3. 6H2O = $\frac{30}{291}$ mol

= 0.103 mol

Therefore, molarity = $\frac{0.103\; mol}{4.3\; L}$

= 0.023 M

(b) Number of moles present in 1000 mL of 0.5 M H2SO4 = 0.5 mol

Therefore, the number of moles present in 30 mL of 0.5 M H2SO4 = $\frac{0.5\times 30}{1000}\; mol$

= 0.015 mol

Therefore, molarity = $\frac{0.015}{0.5\; L}\; mol$

= 0.03 M

Q 2.4) To make 2.5 kg of 0.25 molar aqueous solution, determine the mass of urea (NH2CONH2) that is required.

Molar mass of urea (NH2CONH2) = 2(1 × 14 + 2 × 1) + 1 × 12 + 1 × 16 = 60 g $mol^{-1}$

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 = $\frac{15\times 2500}{1000 + 15}\; g$

= 36.95 g

= 37 g of urea (approx.)

Hence, the 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

(a) Molar mass of KI = 39 + 127 = 166 g $mol^{-1}$

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 = $\frac{Moles\;of\;KI}{Mass\;of\;water\;in\;kg}$

= $\frac{\frac{20}{166}}{0.08}\; m$

= 1.506 m

= 1.51 m (approx.)

(b) It is given that the destiny of the solution = 1.202 $g\; mL^{-1}$

Volume of 100 g solution = $\frac{Mass}{Density}$

= $\frac{100\; g}{1.202\;g \; mL^{-1}}$

= 83.19 mL

= $83.19\times 10^{-3}\; L$

Therefore, the molarity of the solution = $\frac{\frac{20}{166}\; mol}{83.19\times 10^{-3}\; L}$

= 1.45 M

(c) Moles of KI = $\frac{20}{166}$ = 0.12 mol

Moles of water = $\frac{80}{18}$ = 4.44 mol

Therefore, mole = $\frac{Moles\; of\; KI}{Moles\; of\; KI + Moles\; of \; water}$

Fraction of KI = $\frac{0.12}{0.12 + 4.44}$

= 0.0263

Q 2.6) Calculate Henry’s law constant when the solubility of H2S (a toxic gas with a rotten egg-like smell) in water at STP is 0.195 m

It is given that the solubility of H2S in water at STP is 0.195 m, i.e., 0.195 mol of H2S is dissolved in 1000 g of water.

Moles of water = $\frac{1000\; g}{18\; g\; mol^{-1}}$

= 55.56 mol

Therefore, the mole fraction of H2S, x = $\frac{Moles\;of\;H_{2}S}{Moles\;of\;H_{2}S + Moles\;of\;water}$

= $\frac{0.195}{0.195 + 55.56}$

= 0.0035

At STP, pressure (p) = 0.987 bar

According to Henry’s law, p = $K_{H}$x

=> $K_{H} = \frac{P}{x}$

= $\frac{0.987}{0.0035}\; bar$

= 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.

The total amount of solute present in the mixture is given by,

$300\times \frac{25}{100} + 400\times \frac{40}{100}$

= 75 + 160

= 235 g

Total amount of solution = 300 + 400 = 700 g

Therefore, mass percentage of the solute in the resulting solution = $\frac{235}{700}\times 100$

= 33.57%

And the 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 the total vapour pressure is 600 mm Hg. Also, find the composition of the vapour phase.

It is given that

$P^{\circ}_{A}$ = 450 mm of Hg $P^{\circ}_{B}$ = 700 mm of Hg $P_{total}$ = 600 mm of Hg

According to Raoult’s law,

$P_{A} = P^{\circ}_{A}x_{A}$ $P_{B} = P^{\circ}_{B}x_{B} = P^{\circ}_{B}(1 – x_{A})$

Therefore, total pressure, $P_{total} = P_{A} + P_{B}$

=> $P_{total} = P^{\circ}_{A}x_{A} + P^{\circ}_{B}(1 – x_{A})$

=> $P_{total} = P^{\circ}_{A}x_{A} + P^{\circ}_{B} – P^{\circ}_{B}x_{A}$

=> $P_{total} = (P^{\circ}_{A} – P^{\circ}_{B})x_{A} + P^{\circ}_{B}$

=> 600 = (450 – 700) xA + 700

=> –100 = –250xA

=> xA = 0.4

Therefore, $x_{B} = 1 – x_{A}$ = 1 – 0.4 = 0.6

Now, $P_{A} = P^{\circ}_{A}x_{A}$

= 450 x 0.4 = 180 mm of Hg

$P_{B} = P^{\circ}_{B}x_{B}$

= 700 x 0.6 = 420 mm of Hg

Now, in the vapour phase, the mole fraction of liquid A = $\frac{P_{A}}{P_{A} + P_{B}}$

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

= $\frac{180}{600}$

= 0.30

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

Q 2.9) Find the vapour pressure of water and its relative lowering in the solution which is 50 g of urea (NH2CONH2) dissolved in 850 g of water (Vapor pressure of pure water at 298 K is 23.8 mm Hg).

It is given that vapour pressure of water, $P^{\circ}_{1}$ = 23.8 mm of Hg

Weight of water taken, $w_{1}$ = 850 g

Weight of urea taken, $w_{2}$ = 50 g

Molecular weight of water, $M_{1}$ = 18 g $mol^{-1}$

Molecular weight of urea, $M_{2}$ = 60 g $mol^{-1}$

Now, we have to calculate the vapour pressure of water in the solution. We take vapour pressure as p1.

Now, from Raoult’s law, we have

$\frac{P^{\circ}_{1} – P_{1}}{P^{\circ}_{1}} = \frac{n_{2}}{n_{1} + n_{2}}$

=> $\frac{P^{\circ}_{1} – P_{1}}{P^{\circ}_{1}} = \frac{\frac{w_{2}}{M_{2}}}{\frac{w_{1}}{M_{1}} + \frac{w_{2}}{M_{2}}}$

=> $\frac{23.8 – P_{1}}{23.8} = \frac{\frac{50}{60}}{\frac{850}{18} + \frac{50}{60}}$

=> $\frac{23.8 – P_{1}}{23.8} = \frac{0.83}{47.22 + 0.83}$

=> $\frac{23.8 – P_{1}}{23.8} = 0.0173$

=> $P_{1}$ = 23.4 mm of Hg

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?

Here, elevation of boiling point $\Delta T_{b}$ = (100 + 273) – (99.63 + 273)

= 0.37 K

Mass of water, $w_{1}$ = 500 g

Molar mass of sucrose (C12H22O11), $M_{2}$ = 11 x 12 + 22 x 1 + 11 x 16

= 342 g $mol^{-1}$

Molar elevation constant, Kb = 0.52 K kg $mol^{-1}$

We know that

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

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

= $\frac{0.37\times 342\times 500}{0.52\times 1000}$

= 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 1.50C, how much mass of ascorbic acid is needed to be dissolved in the solution where Kt = 3.9 K kg $mol^{-1}$?

Mass of acetic acid (w1) = 75 g

Molar mass of ascorbic acid (C6H8O6), M2­­ = 6 x 12 + 8 x 1 + 6 x 16 = 176 g $mol^{-1}$

Lowering the melting point $\Delta T_{f}$ = 1.5 K

We know that

$\Delta T_{f} = \frac{K_{f}\times w_{2}\times 1000}{M_{2}\times w_{1}}$

=> $w_{2} = \frac{\Delta T_{f}\times M_{2}\times w_{1}}{K_{f}\times 1000}$

= $\frac{1.5\times 176\times 75}{3.9\times 1000}$

= 5.08 g (approx)

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

Q 2.12) If a solution is 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.

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 = $\frac{1}{185000}$ mol

We know that

Osmotic pressure, $\pi = \frac{n}{V}\; RT$

= $\frac{1}{185000}\;mol\times \frac{1}{0.45\;L}\times 8.314\times 10^{3}\; PaL\; K^{-1}mol^{-1}\times 310\; K$

= 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?

Molar mass of ethane (C2H6) = 2 x 12 + 6 x 1 = 30 g $mol^{-1}$

Therefore, number of moles present in $6.56\times 10^{-2}\;g$ of ethane = $\frac{6.56\times 10^{-2}}{30}$

= $2.187\times 10^{-3}\;mol$

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

$p = K_{H}x$

=> 1 bar = $K_{H}. \frac{2.187\times 10^{-3}}{2.187\times 10^{-3} + x}$

=> 1 bar = $K_{H}. \frac{2.187\times 10^{-3}}{x}$

=> $K_{H} = \frac{x}{2.187\times 10^{-3}}$ bar         (Since x >> $2.187\times 10^{-3}$)

The number of moles present in $5\times 10^{-2}$ g of ethane = $\frac{5\times 10^{-2}}{30}$ mol

= $1.67\times 10^{-3}\;mol$

According to Henry’s law,

$p = K_{H}x$

= $\frac{x}{2.187\times 10^{-3}}\times \frac{1.67\times 10^{-3}}{(1.67\times 10^{-3}) + x}$

= $\frac{x}{2.187\times 10^{-3}}\times \frac{1.67\times 10^{-3}}{x}$           (Since, x>> $1.67\times 10^{-3}$)

= 0.764 bar

Hence, the 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?

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.

$\Delta _{sol}H = 0$

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

$∴ \Delta _{sol}H = Positive$

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

$∴ \Delta _{sol}H = Negative$

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?

Vapour pressure of the solution at normal boiling point, $p_{1}$ = 1.004 bar

Vapour pressure of pure water at normal boiling point, $p^{\circ}_{1}$ = 1.013 bar

Mass of solute, w2 = 2 g

Mass of solvent (water), M1 = 18 g $mol^{-1}$

According to Raoult’s law,

$\frac{p^{\circ}_{1} – p_{1}}{p^{\circ}_{1}} = \frac{w_{2}\times M_{1}}{M_{2}\times w_{1}}$

=> $\frac{1.013 – 1.004}{1.013} = \frac{2\times 18}{M_{2}\times 98}$

=> $\frac{0.009}{1.013} = \frac{2\times 18}{M_{2}\times 98}$

=> $M_{2} = \frac{1.013\times 2\times 18}{0.009\times 98}$

= 41.35 g $mol^{-1}$

Hence, 41.35 g $mol^{-1}$ is the molar mass of the solute.

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?

Vapour pressure of heptanes, $p^{\circ}_{1}$ = 105.2 kPa

Vapour pressure of octane, $p^{\circ}_{2}$ = 46.8 kPa

We know that,

The molar mass of heptanes (C7H16) = 7 x 12 + 16 x 1 = 100 g $mol^{-1}$

Therefore, the number of moles of heptane = $\frac{26}{100}$ = 0.26 mol

The molar mass of octane (C8H18) = 8 x 12 + 18 x 1 = 114 g $mol^{-1}$

Therefore, the number of moles of octane = $\frac{35}{114}$ = 0.31 mol

The mole fraction of heptane, $x_{1} = \frac{0.26}{0.26 + 0.31}$ = 0.456

And, the mole fraction of octane, $x_{2} = 1 – 0.456$ = 0.544

Now, the partial pressure of heptane, $p_{1} = x_{1}p^{\circ}_{1}$

= 0.456 x 105.2

= 47.97 kPa

Partial pressure of octane, $p_{2} = x_{2}p^{\circ}_{2}$

= 0.544 x 46.8

= 25.46 kPa

Hence, vapour pressure of solution, $p_{total} = p_{1} + p_{2}$

= 47.97 + 25.46

= 73.43 kPa

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

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

The molar mass of water = 18 g $mol^{-1}$

Therefore, the number of moles present in 1000 g of water = $\frac{1000}{18}$

= 55.56 mol

Therefore, the mole fraction of the solute in the solution is

$x_{2} = \frac{1}{1 + 55.56}$ = 0.0177

It is given that,

Vapour pressure of water, $p^{\circ}_{1}$ = 12.3 kPa

Applying the relation, $\frac{p^{\circ}_{1} – p_{1}}{p^{\circ}_{1}} = x_{2}$

=> $\frac{12.3 – p_{1}}{12.3}$ = 0.0177

=> 12.3 – p1 = 0.2177

=> p1 = 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%.

Let $p^{\circ}_{1}$ be the vapour pressure of pure octane.

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

$\frac{80}{100}\; p^{\circ}_{1} = 0.8\; p^{\circ}_{1}$

The molar mass of solute, M2 = 40 g $mol^{-1}$

The mass of octane, w1 = 114 g

The molar mass of octane, (C8H18), M1 = 8 x 12 + 18 x 1 = 114 g $mol^{-1}$

Applying the relation,

$\frac{p^{\circ}_{1} – p_{1}}{p^{\circ}_{1}} = \frac{w_{2}\times M_{1}}{M_{2}\times w_{1}}$

=> $\frac{p^{\circ}_{1} – 0.8p^{\circ}_{1}}{p^{\circ}_{1}} = \frac{w_{2}\times 114}{40\times 114}$

=> $\frac{0.2\; p^{\circ}_{1}}{p^{\circ}_{1}} = \frac{w_{2}}{40}$

=> 0.2 = $\frac{w_{2}}{40}$

=> w2 = 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.

(i) Let, the molar mass of the solute be M g $mol^{-1}$

Now, the number of moles of solvent (water), $n_{1} = \frac{90\; g}{18\; g\; mol^{-1}}$ = 5 mol

And, the number of moles of solute, $n_{2} = \frac{30\; g}{M\; mol^{-1}} = \frac{30}{M}\; mol$

p1 = 2.8 kPa

Applying the relation:

$\frac{p^{\circ}_{1} – p_{1}}{p^{\circ}_{1}} = \frac{n_{2}}{n_{1} + n_{2}}$

=> $\frac{p^{\circ}_{1} – 2.8}{p^{\circ}_{1}} = \frac{\frac{30}{M}}{5 + \frac{30}{M}}$

=> $1 – \frac{2.8}{p^{\circ}_{1}} = \frac{\frac{30}{M}}{\frac{5M + 30}{M}}$

=> $1 – \frac{2.8}{p^{\circ}_{1}} = \frac{30}{5M + 30}$

=> $\frac{2.8}{p^{\circ}_{1}} = 1 – \frac{30}{5M + 30}$

=> $\frac{2.8}{p^{\circ}_{1}} = \frac{5M + 30 – 30}{5M + 30}$

=> $\frac{2.8}{p^{\circ}_{1}} = \frac{5M}{5M + 30}$

=> $\frac{p^{\circ}_{1}}{2.8} = \frac{5M + 30}{5M}$           (i)

After the addition of 18 g of water:

$n_{1} = \frac{90 + 18g}{18} = 6\;mol$ $p_{1} = 2.9 \;kPa$

Again applying the relation:

$\frac{p^{\circ}_{1} – p_{1}}{p^{\circ}_{1}} = \frac{n_{2}}{n_{1} + n_{2}}$

=> $\frac{p^{\circ}_{1} – 2.9}{p^{\circ}_{1}} = \frac{\frac{30}{M}}{6 + \frac{30}{M}}$

=> $1 – \frac{2.9}{p^{\circ}_{1}} = \frac{\frac{30}{M}}{\frac{6M + 30}{M}}$

=> $1 – \frac{2.9}{p^{\circ}_{1}} = \frac{30}{6M + 30}$

=> $\frac{2.9}{p^{\circ}_{1}} = 1 – \frac{30}{6M + 30}$

=> $\frac{2.9}{p^{\circ}_{1}} = \frac{6M + 30 – 30}{6M + 30}$

=> $\frac{2.9}{p^{\circ}_{1}} = \frac{6M}{6M + 30}$

=> $\frac{p^{\circ}_{1}}{2.9} = \frac{6M + 30}{6M}$           (ii)

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

$\frac{2.9}{2.8} = \frac{\frac{5M + 30}{5M}}{\frac{6M + 30}{6M}}$

=> $\frac{2.9}{2.8}\times \frac{6M + 30}{6} = \frac{5M + 30}{5}$

=> $2.9\times 5\times (6M + 30) = 2.8\times 6\times (5M + 30)$

=> 87M + 435 = 84M + 504

=> 3M = 69

=> M = 23 g $mol^{-1}$

Therefore, 23 g $mol^{-1}$ is the molar mass of the solute.

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

$\frac{p^{\circ}_{1}}{2.8} = \frac{5\times 23 + 30}{5\times 23}$

=> $\frac{p^{\circ}_{1}}{2.8} = \frac{145}{115}$

=> $p^{\circ}_{1}$ = 3.53 kPa

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 a freezing point of 271K. Calculate the freezing point of 5% glucose in water if the freezing point of pure water is 273.15 K.

$\Delta T_{f}$ = 273.15 – 271 = 2.15 K

The molar mass of sugar (C12H22O11) = 12 x 12 + 22 x 1 + 11 x 16 = 342 g $mol^{-1}$

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, the number of moles of cane sugar = $\frac{5}{342}$ mol = 0.0146 mol

Therefore, the molality of the solution,

$m = \frac{0.0146\; mol}{0.095\; kg} = 0.1537 \;mol\;kg^{-1}$

Applying the relation,

$\Delta T_{f} = K_{f}\times m$

=> $K_{f} = \frac{\Delta T_{f}}{m}$

= $\frac{2.15\; K}{0.1537\;mol\;kg^{-1}}$

= 13.99 K kg $mol^{-1}$

The molar mass of glucose (C6H12O6) = 6 x 12 + 12 x 1 + 6 x 16 = 180 g $mol^{-1}$

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

Therefore, the number of moles of glucose = $\frac{5}{180}$ mol = 0.0278 mol

Therefore, the molality of the solution, m = $\frac{0.0278\;mol}{0.095\;kg}$

= 0.2926 mol $kg^{-1}$

Applying the relation:

$\Delta T_{f} = K_{f}\times m$

= $13.99\;K\;kg\;mol^{-1}\times 0.2926\;mol\;kg^{-1}$

= 4.09 K (approx)

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

Q 2.21) Two elements A and B form compounds having formulas AB2 and AB4. When dissolved in 20 g of benzene (C6H6), 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 kg mol–1. Calculate the atomic masses of A and B.

We know that,

$M_{2} = \frac{1000\times w_{2}\times k_{f}}{\Delta T_{f}\times w_{1}}$

Then, $M_{AB_{2}} = \frac{1000\times 1\times 5.1}{2.3\times 20}$

= 110.87 g $mol^{-1}$

$M_{AB_{4}} = \frac{1000\times 1\times 5.1}{1.3\times 20}$

= 196.15 g $mol^{-1}$

Now, we have the molar masses of AB2 and AB4 as 110.87 g $mol^{-1}$ and 196.15 g $mol^{-1}$ respectively.

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?

Given:

T = 300 K

n = 1.52 bar

R = 0.083 bar L $K^{-1}\; mol^{-1}$

Applying the relation, n = CRT

=> C = $\frac{n}{RT}$

= $\frac{1.52\;bar}{0.083\;bar\;L\;K^{-1}\;mol^{-1}\times 300\;K}$

= 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) I2 and CCl4
(iii) NaClO4 and water
(iv) methanol and acetone
(v) acetonitrile (CH3CN) and acetone (C3H6O)

(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, CH3OH, CH3CN.

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 < CH3CN < CH3OH < KCl

Therefore, the order of increasing solubility is

KCl < CH3OH < CH3CN < 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

(i) Phenol (C6H5OH) has the polar group −OH and non-polar group –C6H5. Thus, phenol is partially soluble in water.

(ii) Toluene (C6H5−CH3) 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 (C5H11OH) has a polar −OH group, but it also contains a very bulky nonpolar −C5H11 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.

The number of moles present in 92 g of Na+ ions = $\frac{92\;g}{23\;g\;mol^{-1}}$

= 4 mol

Therefore, the molality of Na+ ions in the lake = $\frac{4\;mol}{1\;kg}$

= 4 m

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

Solubility product of CuS, $K_{sp} = 6\times 10^{-16}$

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

$CuS \leftrightarrow Cu^{2+} + S^{2-}$

Now,                                                   s                  s

$K_{sp} = [Cu^{2+}] + [S^{2-}]$

= s x s

= s2

Then, we have, $K_{sp} = s^{2} = 6\times 10^{-16}$