NCERT Solutions for Class 12 Chemistry Chapter 2 Solutions

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

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NCERT Solutions for Class 12 Chemistry Chapter 2 Solutions

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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).

Answer 2.1:

Mass percentage of Benzene (C₆H₆) =

=

=

= 15.28%

Mass percentage of Carbon Tetrachloride (CCl₄) =

=

=

= 84.72%

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

Answer 2.2:

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

= 78 g

Therefore, the number of moles of C₆H₆ =

= 0.3846 mol

Molar mass of CCl₄ = 1 x 12 + 4 x 355 = 154 g

Therefore, the number of moles of CCl₄ =

= 0.4545 mol

Thus, the mole fraction of C₆H₆ 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. 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 =

(a) Molar mass of Co(NO)₃. 6H₂O = 59 + 2 (14 + 3 x 16) + 6 x 18 = 291 g

Therefore, the moles of Co(NO)₃. 6H₂O =

= 0.103 mol

Therefore, molarity =

= 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 =

= 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 (NH₂CONH₂) 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, the mass of Urea required is 37 g.

Q 2.5) If 1.202 g

(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, the 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 H₂S (a toxic gas with a 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 dissolved in 1000 g of water.

Moles of water =

= 55.56 mol

Therefore, the mole fraction of H₂S, 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:

The 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 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.

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, the mole fraction of liquid A =

=

=

= 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).

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 the vapour pressure of water in the solution. We take vapour pressure as p₁.

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 (C₁₂H₂₂O₁₁),

= 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 1.5⁰C, 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 (w₁) = 75 g

Molar mass of ascorbic acid (C₆H₈O₆), 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 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.

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 (C₂H₆) = 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 =

=>

The number of moles present in

=

According to Henry’s law,

=

=

= 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

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, the absorption of heat takes place.

In the case of solutions showing negative deviations, the 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, w₂ = 2 g

Mass of solvent (water), M₁ = 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,

The molar mass of heptanes (C₇H₁₆) = 7 x 12 + 16 x 1 = 100 g

Therefore, the number of moles of heptane =

The molar mass of octane (C₈H₁₈) = 8 x 12 + 18 x 1 = 114 g

Therefore, the number of moles of octane =

The mole fraction of heptane,

And, the mole fraction of octane,

Now, the 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 the 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).

The molar mass of water = 18 g

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

= 55.56 mol

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

It is given that,

Vapour pressure of water,

Applying the relation,

=>

=> 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

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

The molar mass of solute, M₂ = 40 g

The mass of octane, w₁ = 114 g

The molar mass of octane, (C₈H₁₈), M₁ = 8 x 12 + 18 x 1 = 114 g

Applying the relation,

=>

=>

=> 0.2 =

=> 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

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

And, the number of moles of solute,

p₁ = 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 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.

Answer 2.20:

The molar mass of sugar (C₁₂H₂₂O₁₁) = 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, the number of moles of cane sugar =

Therefore, the molality of the solution,

Applying the relation,

=>

=

= 13.99 K kg

The molar mass of glucose (C₆H₁₂O₆) = 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, the number of moles of glucose =

Therefore, the molality of the solution, m =

= 0.2926 mol

Applying the relation:

=

= 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.

Answer 2.21:

We know that,

Then,

= 110.87 g

= 196.15 g

Now, we have the molar masses of AB₂ and AB₄ 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) 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:

The number of moles present in 92 g of Na⁺ ions =

= 4 mol

Therefore, the 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

= s²

Then, we have,

=>

=

Hence,

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

Therefore, the mass percentage of C₉H₈O₄ =

= 1.424%

Q 2.29) Nalorphene (C₁₉H₂₁NO₃), similar to morphine, is used to combat withdrawal symptoms in narcotic users. The 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 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 the aqueous solution.

Q 2.30) Calculate the amount of benzoic acid (C₆H₅COOH) required for preparing 250
mL of 0.15 M solutions 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 (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 the 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 the least electronegative, while F is the 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 ionises to the largest extent. Now, the more ions produced, the greater 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. K_a = 1.4 × 10^–3, K_f = 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, the number of moles present in 10 g of CH₃CH₂CHCICOOH =

= 0.0816 mol

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

Therefore, the molality of the solution, CH₃CH₂CHCICOOH =

= 0.3264 mol kg^-1

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

CH₃CH₂CHCICOOH 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 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

w_2­ = 19.5 g

K_f = 1.86 K kg

We know that

=

=

Therefore, observed molar mass of CH₂FCOOH,

The calculated molar mass of CH₂FCOOH,

Therefore, van’t Hoff factor,

=

= 1.0753

Let ‘a’ be the degree of dissociation of CH₂FCOOH.

=> i = 1 +

=>

= 1.0753 – 1

= 0.0753

Now, the value of K_a is given as

=

=

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

C =

= 0.5 M

Therefore,

=

=

= 0.00307 (approx)

=

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,

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

Molar mass of water, M₁ = 18 g

Then, the number of moles of glucose,

= 0.139 mol

And, the number of moles of water,

= 25 mol

We know that,

=>

=>

=>

=>

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

According to Henry’s law,

p = k_Hx

=> x =

=

=

=

Hence,

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,

Number of moles of liquid B,

Then, the mole fraction of A,

=

= 0.114

And, the mole fraction of B, x_B = 1 – 0.114 = 0.886

Vapour pressure of pure liquid B,

Therefore, the vapour pressure of liquid B in the solution,

= 500 x 0.886

= 443 torr

Total vapour pressure of the solution,

Therefore, the vapour pressure of liquid A in the solution,

= 475 – 443

= 32 torr

Now,  

=>

=

= 280.7 torr

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

Q 2.37) Vapour pressures of pure acetone and chloroform at 328 K are 741.8 mm Hg and 632.8 mm Hg respectively. Assuming that they form the ideal solution over the entire range of composition, plot p_total, p_chloroform, and p_acetone as a function of x_acetone.The experimental data observed for different compositions of the mixture is:

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

Answer 2.37:

From the question, we have the following data

It can be observed from the graph that the plot for the

Q 2.38) Benzene and toluene form the 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 the 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

Molar mass of toluene (C₆H₅CH₃) = 7 x 12 + 8 x 1 = 92 g

Now, the number of moles present in 80 g of benzene =

And, the number of moles present in 100 g of toluene =

Therefore, the mole fraction of benzene,

And, the mole fraction of toluene,

It is given that the vapour pressure of pure benzene,

And, the vapour pressure of pure toluene,

Therefore, the partial pressure of benzene,

= 0.486 x 50.71

= 24.645 mm Hg

And, the partial vapour pressure of toluene,

= 0.514 x 32.06

= 16.479 mm Hg

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

=

=

= 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 an 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 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,

= 1520 mm Hg

Partial pressure of nitrogen,

= 6004 mm Hg

For oxygen:

=>

=

=

For nitrogen:

=>

=

=

Hence,

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

Answer 2.40:

We know that,

=>

=>

V = 2.5 L

i = 2.47

T = (27 + 273) = 300 K

Here,

R =

M = 1 x 40 + 2 x 35.5

= 111 g

Therefore, w =

= 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,

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 =

M = (2 x 39) + (1 x 32) + (4 x 16) = 174 g

Applying the following relation,

=

=

=

A solution is a homogeneous mixture of two or more than two components. Solved answers on the chapter’s NCERT Solutions for Class 12 are given by our subject experts. These NCERT Solutions will help the students to understand the types of solutions, expressing the concentration of the solution, solubility, ideal and non-ideal solutions, colligative properties and abnormal molar mass. Chemistry Class 12 questions and solutions in Chapter 2 given here are very simple and easy to understand.

Class 12 NCERT Solutions for Chemistry Chapter 2 Solutions

Chapter 2 Solutions of Class 12 Chemistry, is designed as per the CBSE Syllabus for the session 2023-24. This chapter holds approximately 5 marks in the board examination. The Chemistry Class 12 NCERT Solutions Chapter 2 teaches about the types of solutions, the concentration of solutions, the solubility of solids and gases in a liquid, the vapour pressure of liquid solutions, Raoult’s law, ideal and non-ideal solutions, colligative properties and determination of molar masses.

Subtopics for Class 12 Chemistry Chapter 2 – Solutions

1. Types of Solutions
2. Expressing Concentration of Solutions
3. Solubility
   1. The solubility of a Solid in a Liquid
   2. The solubility of a Gas in a Liquid
4. Vapour Pressure of Liquid Solutions
   1. Vapour Pressure of Liquid-Liquid Solutions
   2. Raoult’s Law as a Special Case of Henry’s Law
   3. Vapour Pressure of Solutions of Solids in Liquids
5. Ideal and Nonideal Solutions
   1. Ideal Solutions
   2. Non-ideal Solutions
6. Colligative Properties and Determination of Molar Mass
   1. Relative Lowering of Vapour Pressure
   2. Elevation of Boiling Point
   3. Depression of Freezing Point
   4. Osmosis and Osmotic Pressure
7. Abnormal Molar Masses

NCERT Chemistry book includes a chapter on solutions to introduce various important concepts to the students. In this, students learn to determine the molarity, molality and mole fraction of solutions and know about Henry’s law constant, mass percentage, etc. The topics provided in the NCERT books are not only important for the Class 12 board examination but also for competitive exams like JEE Main and NEET. JEE Main is a national-level engineering entrance examination, and NEET is a national-level examination to take admission to the best medical colleges in India. Also, students must solve NCERT exemplar problems, MCQS, and short and long-answer questions in order to attain a firm grip across subjects.

Solving the NCERT questions using NCERT Solutions will help them to understand the topics in an effective and simple way. Sometimes, the questions are also asked in the JEE Main and NEET examinations.

Students preparing for CBSE examinations are strictly advised to study the chapter thoroughly and should solve the NCERT questions provided at the end of each chapter. The NCERT Solutions are designed in such a way that students can understand each and every chapter of the book. Along with the NCERT questions, students are also encouraged to solve the CBSE Class 12 previous years’ question papers and sample papers. Solving the previous years’ questions and sample papers will help them get acquainted with the different types of questions and their marking scheme. Students are always encouraged to solve the previous years’ questions to boost exam preparations. Solving the question papers in a stipulated time of 3 hours will help them to understand their problem-solving speed and identify the time they take to solve a particular type of question.