NCERT Solutions for Class 12 Chemistry Chapter 2 Solutions

NCERT Solutions for Class 12 Chemistry Chapter 2 Solutions

NCERT 12 Chemistry Solutions for Class 12 Chapter 2 Solutions includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 12 exams must go through NCERT Solutions for Class 12 Chemistry Chapter 2 Solutions. Going through the solutions provided on this page will help you to know how to approach and solve the problems.

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NCERT Solutions for Class 12 Chemistry Chapter 2 Solutions

NCERT EXERCISES

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).

Answer 2.1:

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

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

= $\frac{22}{22 + 122}\times 100$

= 15.28%

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

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

= $\frac{122}{22 + 122}\times 100$

= 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 CCl4 = (100 − 30) g

= 70 g

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

= 78 g $mol^{-1}$

Therefore, 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, 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.

Answer 2.3:

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, 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, 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.

Answer 2.4:

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, 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 $mol^{-1}$

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

= $\frac{\frac{20}{166}}{0.08}\; m$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, 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 rotten egg like smell) in water at STP is 0.195 m

Answer 2.6:

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

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

= 55.56 mol

Therefore, 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.

Answer 2.7:

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, 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:

$P^{\circ}_{A}$ = 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: 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 2.15) 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, $p_{1}$ = 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,

$\frac{p^{\circ}_{1} – p_{1}}{p^{\circ}_{1}} = \frac{w_{2}\times M_{1}}{M_{2}\times w_{1}}$​=M2​×w1​w2​×M1​​

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

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

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

= 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?

Answer 2.16:

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

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

We know that,

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

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

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

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

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

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

Now, 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 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 $mol^{-1}$

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

= 55.56 mol

Therefore, 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, $\frac{p^{\circ}_{1} – p_{1}}{p^{\circ}_{1}} = x_{2}$​=x2​

=> $\frac{12.3 – p_{1}}{12.3}$ = 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%?

Answer 2.18:

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

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

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

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

Mass of octane, w1 = 114 g

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

Applying the relation,

$\frac{p^{\circ}_{1} – p_{1}}{p^{\circ}_{1}} = \frac{w_{2}\times M_{1}}{M_{2}\times w_{1}}$​=M2​×w1​w2​×M1​​

=> $\frac{p^{\circ}_{1} – 0.8p^{\circ}_{1}}{p^{\circ}_{1}} = \frac{w_{2}\times 114}{40\times 114}$​=40×114w2​×114​

=> $\frac{0.2\; p^{\circ}_{1}}{p^{\circ}_{1}} = \frac{w_{2}}{40}$​=40w2​​

=> 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.

Answer 2.19:

(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, $n_{2} = \frac{30\; g}{M\; mol^{-1}} = \frac{30}{M}\; mol$=M30​mol

p1 = 2.8 kPa

Applying the relation:

$\frac{p^{\circ}_{1} – p_{1}}{p^{\circ}_{1}} = \frac{n_{2}}{n_{1} + n_{2}}$​=n1​+n2​n2​​

=> $\frac{p^{\circ}_{1} – 2.8}{p^{\circ}_{1}} = \frac{\frac{30}{M}}{5 + \frac{30}{M}}$​=5+M30​M30​​

=> $1 – \frac{2.8}{p^{\circ}_{1}} = \frac{\frac{30}{M}}{\frac{5M + 30}{M}}$=M5M+30​M30​​

=> $1 – \frac{2.8}{p^{\circ}_{1}} = \frac{30}{5M + 30}$=5M+3030​

=> $\frac{2.8}{p^{\circ}_{1}} = 1 – \frac{30}{5M + 30}$=1–5M+3030​

=> $\frac{2.8}{p^{\circ}_{1}} = \frac{5M + 30 – 30}{5M + 30}$=5M+305M+30–30​

=> $\frac{2.8}{p^{\circ}_{1}} = \frac{5M}{5M + 30}$=5M+305M​

=> $\frac{p^{\circ}_{1}}{2.8} = \frac{5M + 30}{5M}$=5M5M+30​           (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:

$\frac{p^{\circ}_{1} – p_{1}}{p^{\circ}_{1}} = \frac{n_{2}}{n_{1} + n_{2}}$​=n1​+n2​n2​​

=> $\frac{p^{\circ}_{1} – 2.9}{p^{\circ}_{1}} = \frac{\frac{30}{M}}{6 + \frac{30}{M}}$​=6+M30​M30​​

=> $1 – \frac{2.9}{p^{\circ}_{1}} = \frac{\frac{30}{M}}{\frac{6M + 30}{M}}$=M6M+30​M30​​

=> $1 – \frac{2.9}{p^{\circ}_{1}} = \frac{30}{6M + 30}$=6M+3030​

=> $\frac{2.9}{p^{\circ}_{1}} = 1 – \frac{30}{6M + 30}$=1–6M+3030​

=> $\frac{2.9}{p^{\circ}_{1}} = \frac{6M + 30 – 30}{6M + 30}$=6M+306M+30–30​

=> $\frac{2.9}{p^{\circ}_{1}} = \frac{6M}{6M + 30}$=6M+306M​

=> $\frac{p^{\circ}_{1}}{2.9} = \frac{6M + 30}{6M}$=6M6M+30​           (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:

$\frac{p^{\circ}_{1}}{2.8} = \frac{5\times 23 + 30}{5\times 23}$=5×235×23+30​

=> $\frac{p^{\circ}_{1}}{2.8} = \frac{145}{115}$=115145​

=> $p^{\circ}_{1}$ = 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 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:

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

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

Therefore, 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 $mol^{-1}$

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

Therefore, 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 5 % glucose solution is (273.15 – 4.09) K = 269.06 K.

Q 2.21) Two elements A and B form compounds having formula 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 atomic masses of A and B.

Answer 2.21:

We know that,

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

Then, $M_{AB_{2}} = \frac{1000\times 1\times 5.1}{2.3\times 20}$=2.3×201000×1×5.1​

= 110.87 g $mol^{-1}$

 

$M_{AB_{4}} = \frac{1000\times 1\times 5.1}{1.3\times 20}$=1.3×201000×1×5.1​

= 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?

Answer 2.22:

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).

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, CH3OH, CH3CN.

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 < 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.

Answer 2.25:

(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

Answer 2.26:

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

= 4 mol

Therefore, 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.

Answer 2.27:                                                                   

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, $K_{sp} = s^{2} = 6\times 10^{-16}$

=> $s = \sqrt{6\times 10^{-16}}$

= $2.45\times 10^{-8}\;mol\;L^{-1}$

Hence, $2.45\times 10^{-8}\;mol\;L^{-1}$is the maximum molarity of CuS in an aqueous solution.

Q 2.28) Calculate the mass percentage of aspirin (C9H8O4) in acetonitrile (CH3CN) when
6.5 g of C9H8O4 is dissolved in 450 g of CH3CN.

Answer 2.28:

6.5 g of C9H8O4 is dissolved in 450 g of CH3CN.

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

Therefore, mass percentage of C9H8O4 = $\frac{6.5}{456.5}\times 100$

= 1.424%

Q 2.29) Nalorphene (C19H21NO3), 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 (C19H21NO3) = 19 x 12 + 21 x 1 + 1 x 14 + 3 x 16 = 311 g mol-1

In $1.5\times 10^{-3}\;m$ aqueous solution of nalorphene,

1 kg (1000 g) of water contains $1.5\times 10^{-3}\;mol$ = $1.5\times 10^{-3}\times 311$ g

= 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:

$\frac{1000.4665\times 1.5\times 10^{-3}}{0.4665}$ g

= 3.22 g

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

Q 2.30) Calculate the amount of benzoic acid (C6H5COOH) 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 $\frac{0.15\times 250}{1000}$ mol of benzoic acid

= 0.0375 mol of benzoic acid

Molar mass of benzoic acid (C6H5COOH) = 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 CH3CH2CHClCOOH is added to 250 g of water. Ka = 1.4 × 10–3, Kf = 1.86 K kg mol–1

Answer 2.32:

Molar mass of CH3CH2CHCICOOH = 15 + 14 + 13 + 35.5 + 12 + 16 + 16 + 1

= 122.5 g mol-1

Therefore, No. of moles present in 10 g of CH3CH2CHCICOOH = $\frac{10\;g}{122.5\;g\;mol^{-1}}$

= 0.0816 mol

It is given that 10 g of CH3CH2CHCICOOH is added to 250 g of water.

Therefore, Molality of the solution, CH3CH2CHCICOOH = $\frac{0.0186}{250}\times 1000$

= 0.3264 mol kg-1

Let ’a’ be the degree of dissociation of CH3CH2CHCICOOH.

CH3CH2CHCICOOH undergoes dissociation according to the following equation:

$∴ K_{a} = \frac{C\alpha .C\alpha}{C (1 – \alpha)}$

= $\frac{C\alpha^{2}}{1 – \alpha}$

Since a is very small with respect to 1, $1 – a \approx 1$ $K_{a} = \frac{C \alpha^{2}}{1}$

Now,

=> $K_{a} = C \alpha^{2}$

=> $\alpha = \sqrt{\frac{K_{a}}{C}}$

= $\sqrt{\frac{1.4\times 10^{-3}}{0.3264}}\;\;\;\;(∵ K_{a}=1.4\times 10^{-3})$(∵Ka​=1.4×10−3)

= 0.0655

Again,

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

$∴ i = \frac{1 + \alpha}{1}$

= $1 + \alpha$

= 1 + 0.0655

= 1.0655

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

$\Delta T_{f} = i.K_{f}m$

= $1.0655\times 1.86\;K\;kg\;mol^{-1}\times 0.3264\;mol\;kg^{-1}$

= 0.65 K

Q 2.33) 19.5 g of CH2 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:

w1 = 500 g

w2­ = 19.5 g

Kf = 1.86 K kg $mol^{-1}$ $\Delta T_{f}$ = 1 K

We know that:

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

= $\frac{1.86\;K\;kg\;mol^{-1}\times 19.5\;g\times 1000\;g\;kg^{-1}}{500\;g\times 1\;K}$

= $72.54\;g\;mol^{-1}$

Therefore, observed molar mass of CH2FCOOH, $(M_{2})_{obs} = 72.54\;g\;mol^{-1}$

The calculated molar mass of CH2FCOOH,

$(M_{2})_{cal}$ = 14 + 19 + 12 + 16 + 16 + 1 = 78 g $mol^{-1}$

Therefore, van’t Hoff factor, $i = \frac{(M_{2})_{cal}}{(M_{2})_{obs}}$​ is:

= $\frac{78\;g\;mol^{-1}}{72.54\;g\;mol^{-1}}$​

= 1.0753

Let ‘a’ be the degree of dissociation of CH2FCOOH

$∴ i = \frac{C(1 + \alpha)}{C}$

=> i = 1 + $\alpha$

=> $\alpha$ = i – 1

= 1.0753 – 1

= 0.0753

Now, the value of Ka is given as:

$K_{a} = \frac{[CH_{2}FCOO^{-}][H^{+}]}{[CH_{2}FCOOH]}$​

= $\frac{C\alpha.\; C\alpha}{C(1 – \alpha)}$

= $\frac{C\alpha^{2}}{1 – \alpha}$

Taking the volume of the solution as 500 mL, we have the concentration:

C = $\frac{\frac{19.5}{78}}{500}\times 1000\;M$×1000M

= 0.5 M

Therefore, $K_{a} = \frac{C\alpha^{2}}{1 – \alpha}$

= $\frac{0.5\times (0.0753)^{2}}{1 – 0.0753}$

= $\frac{0.5\times 0.00567}{0.9247}$

= 0.00307 (approx)

= $3.07\times 10^{-3}$

Q 2.34) Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.

Answer 2.34:

Vapour pressure of water, $p^{\circ}_{1}$ = 17.535 mm of Hg

Mass of glucose, w2 = 25 g

Mass of water, w1 = 450 g

We know that,

Molar mass of glucose (C6H12O6), M2 = 6 x 12 + 12 x 1 + 6 x 16 = 180 g $mol^{-1}$

Molar mass of water, M1 = 18 g $mol^{-1}$

Then, number of moles of glucose, $n_{2} = \frac{25}{180\;g\;mol^{-1}}$

= 0.139 mol

And, number of moles of water, $n_{1} = \frac{450\;g}{18\;g\;mol^{-1}}$

= 25 mol

We know that,

$\frac{p^{\circ}_{1} – p_{1}}{p^{\circ}_{1}} = \frac{n_{1}}{n_{2} + n_{1}}$​=n2​+n1​n1​​

=> $\frac{17.535 – p_{1}}{17.535} = \frac{0.139}{0.139 + 25}$=0.139+250.139​

=> $17.535 – p_{1} = \frac{0.139\times 17.535}{25.139}$

=> $17.535 – p_{1} = 0.097$

=> $p_{1} = 17.44$ mm of Hg

Hence, 17.44 mm of Hg is the vapour pressure of water.

Q 2.35) Henry’s law constant for the molality of methane in benzene at 298 K is 4.27 × 105 mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg

Answer 2.35:

Given:

p = 760 mm Hg

$k_{H} = 4.27\times 10^{5}$ mm Hg

According to Henry’s law,

p = kHx

=> x = $\frac{p}{k_{H}}$

= $\frac{760\;mm\;Hg}{4.27\times 10^{5}\;mm\;Hg}$

= $177.99\times 10^{-5}$

= $178\times 10^{-5}$ (approx)

Hence, $178\times 10^{-5}$ is the mole fraction of methane in benzene.

Q 2.36) 100 g of liquid A (molar mass 140 g mol–1) was dissolved in 1000 g of liquid B (molar mass 180 g mol–1). The vapour pressure of pure liquid B was found to be 500 torrs. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.

Answer 2.36:

Number of moles of liquid A, $n_{A} = \frac{100}{140}$ = 0.714 mol

Number of moles of liquid B, $n_{B} = \frac{1000}{180}$ = 5.556 mol

Then, mole fraction of A, $x_{A} = \frac{n_{A}}{n_{A} + n_{B}}$

= $\frac{0.714}{0.714 + 5.556}$

= 0.114

And, mole fraction of B, xB = 1 – 0.114 = 0.886

Vapour pressure of pure liquid B, $p^{\circ}_{B}$ = 500 torr

Therefore, vapour pressure of liquid B in the solution,

$p_{B} = p^{\circ}_{B}\;x_{B}$

= 500 x 0.886

= 443 torr

Total vapour pressure of the solution, $p_{total}$ = 475 torr

Therefore, Vapour pressure of liquid A in the solution,

$p_{A} = p_{total} – p_{B}$

= 475 – 443

= 32 torr

Now,  $p_{A} = p^{\circ}_{A}\;x_{A}$

=> $p^{\circ}_{A} = \frac{p_{A}}{x_{A}}$

= $\frac{32}{0.114}$

= 280.7 torr

Hence, 280.7 torr is the vapour pressure of pure liquid A

Q 2.38) 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 mm Hg and 32.06 mm Hg, respectively. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of toluene.

Answer 2.38:

Molar mass of benzene (C6H6) = 6 x 12 + 6 x 1 = 78 g $mol^{-1}$

Molar mass of toluene (C6H5CH3) = 7 x 12 + 8 x 1 = 92 g $mol^{-1}$

Now, number of moles present in 80 g of benzene = $\frac{80}{78}$ = 1.026 mol

And, number of moles present in 100 g of toluene = $\frac{100}{92}$ = 1.087 mol

Therefore, Mole fraction of benzene, $x_{b} = \frac{1.026}{1.026 + 1.087}$ = 0.486

And, mole fraction of toluene, $x_{t}$ = 1 – 0.486 = 0.514

It is given that vapour pressure of pure benzene, $p^{\circ}_{b}$ = 50.71 mm Hg

And, vapour pressure of pure toluene, $p^{\circ}_{t}$ = 32.06 mm Hg

Therefore, partial pressure of benzene, $p_{b} = x_{b}\times p^{\circ}_{b}$

= 0.486 x 50.71

= 24.645 mm Hg

And, partial vapour pressure of toluene, $p_{t} = x_{t}\times p^{\circ}_{t}$

= 0.514 x 32.06

= 16.479 mm Hg

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

$\frac{p_{b}}{p_{b} + p_{t}}$

= $\frac{24.645}{24.645 + 16.479}$

= $\frac{24.645}{41.124}$

= 0.599

= 0.6 (approx)

Q 2.39) 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 atm. At 298 K if the Henry’s law constants for oxygen and nitrogen at 298 K are 3.30 × 107 mm and 6.51 × 107 mm respectively, calculate the composition of these gases in water.

Answer 2.39:

Percentage of oxygen in air = 20 %

Percentage of nitrogen in air = 79 %

Also, it is given that water is in equilibrium with air at a total pressure of 10 atm, that is (10 x 760) mm Hg = 7600 mm Hg

Therefore,

Partial pressure of oxygen, $p_{O_{2}} = \frac{20}{100}\times 7600$=10020​×7600 mm Hg

= 1520 mm Hg

Partial pressure of nitrogen, $p_{N_{2}} = \frac{79}{100}\times 7600$=10079​×7600

= 6004 mm Hg

For oxygen:

$p_{O_{2}} = K_{H}. x_{O_{2}}$=KH​.xO2​​

=> $x_{O_{2}} = \frac{p_{O_{2}}}{K_{H}}$=KH​pO2​​​

= $\frac{1520\;mm\;Hg}{3.30\times 10^{7}\;mm\;Hg}\;\;\;\;\;\;\;\;\;(Given\;K_{H} = 3.30\times 10^{7}\;mm\;Hg)$(GivenKH​=3.30×107mmHg)

= $4.61\times 10^{-5}$

For nitrogen:

$p_{N_{2}} = K_{H}. x_{N_{2}}$=KH​.xN2​​

=> $x_{N_{2}} = \frac{p_{N_{2}}}{K_{H}}$=KH​pN2​​​

= $\frac{6004\;mm\;Hg}{6.51\times 10^{7}\;mm\;Hg}$

= $9.22\times 10^{-5}$

Hence, $4.61\times 10^{-5}$ and $9.22\times 10^{-5}$ are the mole fractions of oxygen and nitrogen in water.

Q 2.40) Determine the amount of CaCl2 (i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27° C.

Answer 2.40:

We know that,

$\pi = i\; \frac{n}{V}\; RT$

=> $\pi = i\; \frac{w}{MV}\; RT$

=> $w = \frac{\pi MV}{iRT}$ $\pi$ = 0.75 atm

V = 2.5 L

i = 2.47

T = (27 + 273) = 300 K

Here,

R = $0.0821\;L\;atm\;K^{-1}\;mol^{-1}$

M = 1 x 40 + 2 x 35.5

= 111 g $mol^{-1}$

Therefore, w = $\frac{0.75\times 111\times 2.5}{2.47\times 0.0821\times 300}$

= 3.42 g

Hence, 3.42 g is the required amount of CaCl2.

Q 2.41) Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K2SO4
in 2 litres of water at 25° C, assuming that it is completely dissociated.

Answer 2.41:

When K2SO4 is dissolved in water, $K^{+}\; and\; SO_{4}^{2-}$ ions are produced.

$K_{2}SO_{4} \rightarrow 2K^{+} + SO_{4}^{2-}$

Total number of ions produced = 3

Therefore, i = 3

Given:

w = 25 mg = 0.025 g

V = 2 L

T = 250C = (25 + 273) = 298 K

Also, we know that:

R = $0.0821\;L\;atm\;K^{-1}\;mol^{-1}$

M = (2 x 39) + (1 x 32) + (4 x 16) = 174 g $mol^{-1}$

Applying the following relation,

$\pi = i\; \frac{n}{V}\; RT$

= $i \frac{w}{M} \frac{1}{v} RT$

= $3\times \frac{0.025}{174}\times \frac{1}{2}\times 0.0821\times 298$

= $5.27\times 10^{-3}\; atm$

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