Class 12 Chemistry Revision Questions Chapter 2 - Solutions

 

Question 1: The vapour pressure of a 2 molal aqueous solution of a solute is 11.872 kPa at 27°C.

At 27°C, calculate the vapour pressure of pure water.

Solution: 2 molal solution means 2 moles of solution is present in 1 kg of the solvent (water since solution is aqueous)

Therefore, number of moles of solute = 2 mol

And, number of moles of solvent =

Then, mole fraction of the solute,

According to Raoult’s law,

 

Given, p₁ = 11.872 kPa

Then,

Therefore, the vapour pressure of pure water at 27°C is 12.3 kPa.

 

Question 2: 1 g of an organic compound is dissolved in 36 g of water. The freezing point of this solution is found to be 272.25 K. The freezing point of water is 273 K and K_f for water is 1.86 K kg mol⁻¹. Calculate the molar mass of the compound.

Solution: Here, depression in freezing point,  

Mass of solute, w₂ = 1 g

Mass of solvent, w₁ = 36 g

Freezing point depression constant, K_f = 1.86 K kg mol⁻¹

It is known that,

Question 3: Under what conditions do non-ideal solutions exhibit negative deviations from Raoult’s law?

Solution: Non-ideal solutions exhibit negative deviations from Raoult’s law if the vapour pressure of the solution is lower than the value predicted by Raoult’s law. This happens when the forces of interaction between the compounds are greater than those in the pure components.

Question 4: Between 0.1 molal solutions of glucose and sodium chloride, which will have a higher boiling point?

Solution: 0.1 m solution of NaCl will have a higher boiling point than 0.1 m solution of glucose. This is because sodium chloride undergoes dissociation in the solution. Thus, an elevation is observed in the boiling point.

 

Question 5: The mole fraction of the solute of an X molal solution of a compound in benzene is 0.2. What is the value of X?

Solution: Let the mole fraction of solute X be x_2.

Let the mole fraction of the solvent (benzene) be x_1.

Hence,

And,

As Molecular mass of C₆H₆ = 78, number of moles of benzene

 

Hence, the molality of the solutions 3.20.

 

Question 6: The vapour pressure of a pure liquid X is 0.05 atm at 300 K. The vapour pressure of this liquid in solution with liquid Y is 0.04 atm. Find the mole fraction of the liquid Y in the solution.

Solution: Applying Raoult’s law,

 Question 7: What happens to molarity when the temperature of a solution is increased?

Solution: When the temperature of a solution is increased, the molarity decreases. This is because the volume of the solution increases with an increase in temperature but the number of moles of the solute remains the same.

Hence, molarity is a function of temperature.

 

Question 8:

a. The freezing point of a solution containing 0.3 g of acetic acid in 30.0 g of benzene is lowered by 0.45 K. Calculate Van’t Hoff factor. (K_f for benzene = 5.12 Kkgmol⁻¹)

b. The osmotic pressure of a 0.0103 molar solution of an electrolyte is found to be 0.70 atm at 27°C. Calculate Van’t Hoff factor. (R = 0.082 Latmmol⁻¹K⁻¹)

Solution:

a. Depression in freezing point can be calculated as follows:

ΔT_{f(calculated)} =  

Van’t Hoff factor can be calculated as:

b. Van’t Hoff equation for a dilute solution, π = i CRT

∴ 0.70 atm = i × 0.0103 molL⁻¹ × 0.082 Latmmol⁻¹K⁻¹ × 300 K

Question 9:

a. The outer shell of two eggs is removed and kept in dilute HCl. Then, one shell is placed in distilled water, while the other is placed in a saturated solution of NaCl. What will be observed?

b. Explain why a bottle containing liquid ammonia is kept in ice before it is opened.

Solution:

a. The egg shell kept in distilled water will get swollen. On the other hand, the egg shell kept in NaCl solution will shrink. This happens because of osmosis, in which the net flow of solvent from the less concentrated to the more concentrated solution takes place. Here, the membrane beneath the outer shell of the egg acts as a semi-permeable membrane.

b. The vapour pressure of liquid ammonia at room temperature is very high. On cooling, the vapour pressure inside the bottle containing liquid ammonia decreases. As a result, on opening the bottle, liquid ammonia does not splash out.

Question 10:

a. Name and explain the factor introduced in 1880 to account for the extent of association or dissociation.

b. Arrange the following solutions in the increasing order of their osmotic pressure. Give reasons for your answer.

i. 34.2 g/L Sucrose ii. 60 g/L Urea            iii. 58.5 g/L Sodium chloride

Solution:

a. In 1880, Van’t Hoff introduced the Van’t Hoff factor i.e., ‘i’ to account for the extent of association or dissociation. The Van’t Hoff factor i.e., 'i' is defined as the ratio of experimental value of colligative property to its calculated value i.e.,

Or,

b. Molar mass of sucrose (C₁₂H₂₂O₁₁) = 342 gmol⁻¹

Molar mass of urea (NH₂CONH₂) = 60 gmol⁻¹

Molar mass of sodium chloride (NaCl) = 58.5 gmol⁻¹

Molar concentration of 34.2 g/L solution of sucrose =

Molar concentration of 60 g/L solution of urea 

Molar concentration of 58.5 g/L solution of sodium chloride =

However, NaCl is a strong electrolyte and one formula unit of NaCl dissociates to give two ions i.e., Na⁺ and Cl⁻. Thus, the molar concentration of particles in the solution is 2 M.

Therefore, the order of the increasing concentration is as follows:

Sucrose < Urea < NaCl

The osmotic pressure is directly proportional to the number of particles in the solution. Hence, the increasing order of the osmotic pressure will be:

Sucrose < Urea < NaCl

Question 11:

a. Why is molality preferred over molarity while expressing the concentration of a solution?

b. Why does the boiling point of water increase when sodium chloride is added to it?

c. Why is phenol partially soluble in water?

Solution:

a. While molarity decreases with an increase in temperature, molality is independent of temperature. This happens because molality involves mass, which does not change with a change in temperature, while molarity involves volume, which is temperature dependent. Hence, molality is preferred over molarity while expressing the concentration of a solution.

b. When a non-volatile solute such as sodium chloride is dissolved in water, the vapour pressure of water decreases. This happens because on addition of NaCl, some of the solvent molecules on the surface are replaced by the non-volatile solute molecules. Hence, the solution has to be heated at a higher temperature to make the vapour pressure equal to the external pressure. Hence, the boiling point of the solution increases.

c. As a general rule, like dissolves like. Phenol has a polar −OH group but an aromatic phenyl C₆H₅­ group. Hence, it is partially soluble in water.

Question 12: 6 g of ethanoic acid (CH₃COOH) is dissolved in 80 g of benzene (C₆H₆). Calculate the respective mass percentages of benzene and ethanoic acid.

Solution: Mass of ethanoic acid = 6 g [Given]

Mass of benzene = 80 g [Given]

Therefore, mass of the solution = Mass of ethanoic acid + Mass of benzene

= (6 + 80) g       = 86 g

Then, mass percentage of benzene =

  =

And, mass percentage of ethanoic acid =

  =

Question 13: Find the mass of sodium hydroxide (NaOH) required in making 4 kg of 0.5 molal aqueous solution.

Solution: 0.5 molal aqueous solution of NaOH means 0.5 mol of NaOH is dissolved in 1 kg i.e., 1000 g of water (solvent).

Molar mass of NaOH = (23 + 16 + 1) g mol⁻¹  = 40 g mol⁻¹

Therefore, 0.5 mol of NaOH = 0.5 × 40 g  = 20 g

That is, 20 g of NaOH is dissolved in (1000 + 20) g of solution = 1020 g of solution

Mass of NaOH required to make 1020 g of 0.5 molal aqueous solution = 20g

Therefore, mass of NaOH required to make 4 kg, i.e. 4000 g, of 0.5 molal aqueous solution

=  = 78.43 g

Question 14: The vapour pressure of a liquid mixture containing liquids X and Y is 500 mm Hg at 75°C. At 75°C, the vapour pressure of pure liquids X and Y are 800 mm Hg and 300 mm Hg respectively.

In vapour phase, calculate the respective mole fractions of X and Y.

Solution: Given,

Applying Raoult’s law, we obtain

Then,

Therefore,

And,

Hence, in vapour phase,

Mole fraction of X =

Mole fraction of Y =

Question 15: A non-volatile solute of molar mass 101.2 g mol⁻¹ is dissolved in 120 g of benzene. Then, it is found that the boiling point of benzene is raised by 0.75 K. Also, K_b for benzene is 2.53 K kg mol⁻¹. What is the dissolved mass of the non-volatile solute?

Solution:

Given,

K_b = 2.53 K kg mol⁻¹

Molar mass of solute, M₂ = 101.2 g mol⁻¹

Elevation of boiling point, ΔT_b= 0.75 K

Mass of solvent, w₁ = 120 g

Let the dissolved mass of the non-volatile solute be w₂.

It is known that,

Hence, the dissolved mass of the non-volatile solute is 3.6 g.

Question 16: At 25 °C, 30 g of urea (NH₂CONH₂) is dissolved in 720 g of water. At 25°C, the vapour pressure of pure water is 3173.06 Pa. After dissolving urea, the vapour pressure of water for this solution will be

A) 3029 Pa                 B) 3135 Pa                 C) 3173 Pa                 D) 3219 Pa

Solution: Applying Raoult’s law,

 

 

Given, w₂ = 30 g

      w₁ = 720 g

            Pa

Again, molar mass of water, M₁ = 18 g mol⁻¹

Molar mass of urea, M₂ = 60 g mol⁻¹

Then,

Therefore, the vapour pressure of water for this solution will be 3135 Pa. The correct answer is B.

Question 17: Calculate the osmotic pressure exerted by the resulting solution if at 300 K, 2.5 g of a protein of molar mass 102000 g mol⁻¹ is dissolved in 500 mL of water.

Solution: Osmotic pressure is given by,

Given, V = 500 mL    = 0.5 L

      T = 300 K

It is known that, R = 8.314 × 10³ Pa L K⁻¹ mol⁻¹

Here, number of moles of solute,

   

Hence, the osmotic pressure exerted by the resulting solution is 122.2 Pa.

Question 18: A solution of K₂SO₄ ( vant Hoff factor, i = 3 ) is prepared by dissolving certain amount of it in 3 L of water at 300 K. The osmotic pressure of the solution is found to be 0.9 atm.

Find the amount of K₂SO₄ dissolved.

Solution: Osmotic pressure is given by,

Given, π = 0.9 atm,      V = 3 L,       i = 3, R = 0.0821 L atm K⁻¹ mol⁻¹,       T = 300 K

 

Then,

Molar mass of K₂SO₄ =  

Therefore, amount of K₂SO₄ dissolved = 3.65 ×10−2 × 174 g = 6.351 g

 

Question 19: Which of the following statements regarding solubility is correct?

·        A) Solubility of gases increases with increase in temperature while that of solids decreases with decrease in temperature.

·        B) Solubility of gases decreases while that of solids increases with increase in temperature.

·        C) Solubility of gases as well as solids decreases with decrease in temperature.

·        D) Solubility of gases is independent of change in temperature while that of solids decreases with decrease in temperature.

Solution: Solubility of gases in liquids decreases with increase in temperature. This is because dissolution of gas in liquid is an exothermic process.

Gas  +  Solvent  ↔  Solution  +  Heat

As the temperature is increased, heat is supplied to the system and as a result, equilibrium shifts backwards.

Solubility of solids in liquids increases with the increase in temperature and decreases with the decrease in temperature. This is because dissolution of solids in liquids is an endothermic process.

The correct answer is B.

 

Question 20: An aqueous solution of urea ( NH₂CONH₂ ) has boiling point 100.13 °C. For water, K_f = 1.86 °C and K_b = 0.512 °C. Which of the following temperatures represents the freezing point of the given solution?

A) − 0.37 °C              B) − 0.47 °C               C) − 0.59 °C                 D) − 0.69 °C

Solution:

It is known that,   

Here,

 

 

 

Then,

 

 

Given, K_f = 1.86 °C

It is known that, ΔT_f = K_f m = 1.86 × 0.254 °C = 0.472 °C

It is known that, freezing point of pure water,  

Hence, freezing point of the solution

 

The correct answer is B.

 

Question 21: An aqueous solution contains 10% of urea by weight and 15% of glucose by weight. Freezing point depression constant (K_f) for water = 1.86 °C. The freezing point of the solution is

A) −1.86 °C   B) −3.38 °C                C) −6.21 °C           D) −7.63 °C

Solution: Let mass of the solution be 100 g

Now,

Mass of urea = 10 g

Mass of glucose = 15 g

Therefore, mass of water = 100 − (10 + 15) = 75 g

Molar mass of urea (NH₂CONH₂) = 60 g mol⁻¹

Therefore, number of moles present in 10 g urea =  mol = 0.17 mol

Then, molality of urea,

Molar mass of glucose ( C₆H₁₂O₆ ) = 180 g mol^–1

Therefore, number of moles present in 15 g glucose = mol = 0.08 mol

And, molality of glucose,  

For the solution,  

Therefore, freezing point of the solution = (0 − 6.21) °C = − 6.21 °C

The correct answer is C.

Question 22: The osmotic pressure of a sugar solution is 2.5 × 10⁵ Pa at 25°C.

The concentration of the given sugar solution is

A) 3.42 g L⁻¹ B) 6.84 g L⁻¹              C) 34.20 g L⁻¹        D) 68.40 g L⁻¹

Solution: Osmotic pressure (π) is given by,

π = CRT

Where, C is the concentration of the solution

Given, π = 2.5 × 10⁵ Pa,      T = (25 + 273) K = 298 K

R = 8.314 × 10³ Pa L K⁻¹ mol⁻¹

π = CRT

Molar mass of sugar ( C₁₂H₂₂O₁₁ ) = 342 g mol^–1

Therefore, concentration of the sugar solution = 0.1 × 342 g L⁻¹ = 34.2 g L⁻¹

The correct answer is C.

Question 23: Henry’s law constant for the solubility of methane in benzene at 25°C is 4.27 × 10⁵ mm Hg.

The mole fraction of methane in benzene is 1.5 × 10⁻³ when the pressure is

A) 42.7 mm Hg B) 64.05 mm Hg        C) 427.0 mm Hg         D) 640.5 mm Hg

Solution: According to Henry’s law,

Therefore, the required pressure is 640.5 mm Hg. The correct answer is D.

 

Question 24: Two liquids A (molar mass 80 g mol⁻¹) and B (molar mass 120 g mol⁻¹) form ideal solution over the entire range of composition. 60 g of A is mixed with 180 g of B. The vapour pressures of the liquids A and B are 70.5 mm Hg and 62.5 mm Hg respectively at 298 K.

The respective mole fractions of the liquids A and B in the vapour phase are

A) 0.219 and 0.781    B) 0.357 and 0.643     C) 0.643 and 0.357        D) 0.781 and 0.219

Solution: Number of moles present in 60 g of liquid A =  = 0.75 mol

Number of moles present in 180 g of liquid B =  = 1.5 mol

Then, mole fraction of liquid A,

And, mole fraction of liquid B, 

Vapour pressure of pure liquid A,  [Given]

Vapour pressure of pure liquid B,  [Given]

According to Raoult’s law,

 

 

Therefore, in vapour phase,

Mole fraction of liquid

And, mole fraction of liquid B = 1 − 0.357 = 0.643

Hence, in the vapour phase, the mole fractions of the liquids A and B are 0.357 and 0.643 respectively. The correct answer is B.

Question 25: A non-volatile solute of molar mass 55g mol⁻¹ is dissolved in 117 g of benzene. As a result, the vapour pressure of benzene is reduced to 75%. The dissolved amount of non-volatile solute is

A) 12.5 g               B) 17.5 g           C) 22.5 g       D) 27.5 g

Solution:

It is known that,

Here, w₁ and w₂ are the masses and M₁ and M₂ are the molar masses of the solvent and solute respectively.   is the vapour pressure of the pure solvent.

Given,

      w₁ = 117 g,       M₂  =  55 g mol^–1

Molar mass of benzene, M₁= 78 g mol⁻¹. Then,

Hence, the dissolved amount of non-volatile solute is 27.5 g. The correct answer is D.

 

Question 26: The correct order of increasing solubility of n-octane, NaCl, and CH₃OH in benzene is

A) NaCl < CH₃OH < n-octane         B) n-octane < CH₃OH < NaCl

C) CH₃OH < n-octane < NaCl         D) CH₃OH < NaCl < n-octane

Solution: Both n-octane and benzene are non-polar. Therefore, they mix completely in all proportions.

NaCl is an ionic compound and therefore, it is insoluble in non-polar benzene.

CH₃OH is less polar than NaCl but more polar than n-octane. Hence, the solubility of CH₃OH is intermediate between those of NaCl and n-octane.

Therefore, the correct order of increasing solubility of the given compounds is

NaCl < CH₃OH < n-octane

The correct answer is A.

Question 27: What will happen to the boiling point of a solution if the weight of the solute dissolved in it is doubled and the weight of the solvent taken is reduced by half?

Solution: If the weight of the solute dissolved is doubled and the weight of the solvent is reduced by half, then an elevation takes place in the boiling point of the solution. The boiling point of the solution becomes four times the original value.

Question 28: What is meant by ‘10% aqueous solution of sodium carbonate’?

Solution: ‘10% aqueous solution of sodium carbonate’ means that 10 g of sodium carbonate (solute) is present in 100 g of solution containing water and sodium carbonate.

Question 29: When is the value of Van’t Hoff factor more than unity?

Solution: The value of Van’t Hoff factor is more than unity when a solute undergoes dissociation in the solution.

Question 30: If the density of a 25% (mass/mass) aqueous solution of a compound (molar mass = 75 g mol⁻¹) is 1.45 g mL⁻¹, then calculate its molarity.

Solution: 25% (mass/mass) aqueous solution means 25 g of the compound dissolves in 100 g of the solution.

Density of solution = 1.45 g mL⁻¹ [Given]

Therefore, volume of the solution =  

Molar mass of the compound = 75 g mol⁻¹ [Given]

Thus, number of moles in 25 g of the compound =  

Therefore, molarity of the solution =

Question 31: The density of KCl solution labelled as 9% w/w is 1.4 g mL⁻¹.

The molarity of the solution is

A) 2.17 M              B) 2.5 M            C) 1.69 M     D) 1.5 M

Solution: In the given solution,

9 g of KCl is dissolved in 100 g of solution.

Molar mass of KCl = 1 × 39 + 1 × 35.5

= 74.5 g mol⁻¹

Thus, number of moles present in 9 g of KCl

Given, density of the solution = 1.4 g mL⁻¹

Thus, volume of 100 g solution  

Therefore, molarity of the solution

The correct answer is C.

 

Question 32: 350 g of a 20% w/w solution is mixed with 300 g of 35% w/w solution.

Calculate the mass percentage of solute in the resulting solution.

Solution: Mass of solute in 350 g of 20 % w/w solution

Mass of solute in 300 g of 35 % w/w solution

Therefore, total mass of solute in the resulting solution = (70 + 105) g = 175 g

Total mass of the resulting solution = (350 + 300) g = 650 g

Therefore, mass percentage of solute in the resulting solution

 

 

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