Question 1

Identify the colligative property from the followings?

Accepted Answer

Colligative properties depend only on the number of solute particles in a solution and not on their nature. The four colligative properties are: 1. Relative lowering of vapour pressure 2. Elevation in boiling point 3. Depression in freezing point 4. Osmotic pressure Vapour pressure, boiling point, and freezing point of a solution are not colligative properties themselves. Osmotic pressure of a solution is a colligative property. Answer: Osmotic pressure of solution.

Question 2

Find the osmotic pressure of 0.01 M solution at 25 °C (R = 0.0821\ L atm mol ^ -1 K ^ -1 )

Accepted Answer

The osmotic pressure is given by the formula: = CRT Here, C = 0.01 M, T = 25^ C = 298 K, and R = 0.0821 L atm mol^ -1 K^ -1 . = 0.01 0.0821 298 = 0.244658 atm 0.245 atm Answer: 0.245 atm

Question 3

When the bottle of soft drink is opened, what would take place from the following? (A) Solubility of dissolved gas decreases. (B) bottle released internal pressure. (C) effervences evolved from the bo

Accepted Answer

Soft drinks are bottled under high pressure to increase the solubility of CO_2 gas in the liquid, according to Henry's law. When the bottle is opened, the high internal pressure is released to match the atmospheric pressure. Thus, statement (B) is correct. As the pressure decreases, the solubility of the dissolved gas decreases. Thus, statement (A) is correct. The excess dissolved gas escapes rapidly in the form of bubbles, causing effervescence. Thus, statement (C) is correct. The external atmo

Question 4

Which from following solutions in water on complete dissociation exhibits minimum freezing point depression?

Accepted Answer

The depression in freezing point is given by T_f = i K_f m. For complete dissociation, the van't Hoff factor i is equal to the number of ions produced per formula unit. For 0.2 m Potassium chloride (KCl): i = 2, i m = 2 0.2 = 0.4 For 0.1 m Sodium chloride (NaCl): i = 2, i m = 2 0.1 = 0.2 For 0.05 m Aluminium phosphate (AlPO_4): i = 2, i m = 2 0.05 = 0.1 For 0.15 m Magnesium sulphate (MgSO_4): i = 2, i m = 2 0.15 = 0.3 Since 0.05 m Aluminium phosphate has the minimum value of i m, it exhibits the

Question 5

Calculate the mass of nonvolatile solute dissolved in 0.3 dm^3 water having osmotic pressure 0.1 atm at 300K. [Molar mass of solute = 328 g mol^ -1 , R = 0.082 dm^3atm K^ -1 mol^ -1 ]

Accepted Answer

The osmotic pressure is given by the formula: = W M V R T Rearranging for the mass of the solute W: W = M V R T Substituting the given values: W = 0.1 328 0.3 0.082 300 W = 9.84 24.6 W = 0.4 g Answer: 0.4 g

Question 6

Calculate T_b of a 0.45 m solution of a nonvolatile solute in solvent if molal elevation constant of solvent is 3.0 K kg mol^ -1

Accepted Answer

The elevation in boiling point is given by the formula: T_b = K_b m Given K_b = 3.0 K kg mol^ -1 and m = 0.45 m. Substituting the values: T_b = 3.0 0.45 = 1.35 K Answer: 1.35 K

Question 7

Calculate the solubility of certain gas in solvent with pressure 3 atm at 25^ C (Henry's law constant is 3.0 10^ -2 \, mol dm ^ -3 \, atm ^ -1 )

Accepted Answer

According to Henry's law, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. S = K_H P Given: K_H = 3.0 10^ -2 mol dm ^ -3 atm ^ -1 P = 3 atm Substituting the values: S = (3.0 10^ -2 mol dm ^ -3 atm ^ -1 ) 3 atm S = 9.0 10^ -2 mol dm ^ -3 Since 1 mol dm ^ -3 = 1 M , the solubility is 0.09 M . Answer: 0.09 M

Question 8

Calculate the molality of aqueous solution of electrolyte that freezes at -0.93\,^ C if K_f for water and van't Hoff factor respectively are 1.86\, K kg mol ^ -1 and 1.25. (Freezing point of water = 0

Accepted Answer

The depression in freezing point is given by the formula: T_f = i K_f m Given: Freezing point of solution, T_f = -0.93\,^ C Freezing point of water, T_f^ = 0\,^ C T_f = T_f^ - T_f = 0 - (-0.93) = 0.93\,^ C van't Hoff factor, i = 1.25 Cryoscopic constant, K_f = 1.86\, K kg mol ^ -1 Substituting the values into the formula: 0.93 = 1.25 1.86 m m = 0.93 1.25 1.86 m = 0.5 1.25 = 0.4\, m Answer: 0.4 m

Question 9

van't Hoff factor for BaCl _2 is 2.47, calculate the percentage dissociation of in its aqueous solution.

Accepted Answer

The dissociation reaction of BaCl _2 is: BaCl _2 Ba ^ 2+ + 2 Cl ^ - The number of ions produced per formula unit, n = 3. The relation between van't Hoff factor i and degree of dissociation is: i = 1 + (n - 1) Substituting the given values: 2.47 = 1 + (3 - 1) 2.47 = 1 + 2 2 = 1.47 = 0.735 Percentage dissociation = 100 = 73.5\%

Question 10

A solution of non volatile solute has boiling point elevation 0.70 K if K_b for the solvent is 2.44\ K kg mol ^ -1 . What is the molality of the solution?

Accepted Answer

The elevation in boiling point is given by the formula: T_b = K_b m where T_b is the boiling point elevation, K_b is the ebullioscopic constant, and m is the molality of the solution. Given: T_b = 0.70 K K_b = 2.44 K kg mol^ -1 Substituting the given values into the formula: 0.70 = 2.44 m m = 0.70 2.44 m 0.286 m Answer: 0.286 m

Question 11

Which among the following pairs of solutions is isotonic in nature?

Accepted Answer

Isotonic solutions are those which have the same osmotic pressure at a given temperature. Osmotic pressure is given by = iCRT, where i is the van't Hoff factor, C is the molar concentration, R is the gas constant, and T is the temperature. For two solutions to be isotonic, their effective concentrations must be equal: i_1C_1 = i_2C_2. For 0.1 M urea and 0.1 M sucrose: Urea is a non-electrolyte, so i = 1. Effective concentration = 1 0.1 = 0.1 M. Sucrose is a non-electrolyte, so i = 1. Effective c

Question 12

Calculate the molar mass of nonelectrolyte solute when 6 gram of it is dissolved in 1 dm ^3 water has osmotic pressure 2.4\ atm at 300 K (R = 0.0821\ atm dm ^3\ K ^ -1 \ mol ^ -1 )

Accepted Answer

The osmotic pressure is given by the formula: = C R T = W R T M V Rearranging the formula to solve for the molar mass M: M = W R T V Substituting the given values W = 6\ g , R = 0.0821\ atm dm ^3\ K ^ -1 \ mol ^ -1 , T = 300\ K , = 2.4\ atm , and V = 1\ dm ^3: M = 6 0.0821 300 2.4 1 M = 147.78 2.4 M = 61.575\ g mol ^ -1 Rounding to two decimal places, the molar mass is 61.58\ g mol ^ -1 . Answer: 61.58\ g mol ^ -1

Question 13

Calculate the mole fraction of solute in solution if vapour pressure of a solution and pure solvent are 27 mm Hg and 30 mm Hg respectively at 298 K.

Accepted Answer

According to Raoult's law, the relative lowering of vapour pressure is equal to the mole fraction of the solute. P^0 - P_s P^0 = x_ solute Given P^0 = 30 mm Hg and P_s = 27 mm Hg . x_ solute = 30 - 27 30 x_ solute = 3 30 = 0.1 Answer: 0.1

Question 14

Identify correct statements from following regarding 0.2\ M urea and 0.2\ M sucrose solutions.

Accepted Answer

Osmotic pressure is given by = iCRT. For urea, i = 1 and C = 0.2\ M . For sucrose, i = 1 and C = 0.2\ M . Since both solutions have the same concentration and the same van 't Hoff factor, their osmotic pressures are equal at a given temperature. Therefore, these solutions exhibit the same osmotic pressure and are isotonic to each other. Answer: These solutions exhibit same osmotic pressure.

Question 15

Calculate the molality of solution of nonvolatile solute if boiling point of solution, molal elevation constant for solvent are 319.8\ K and 2.5\ K kg mol ^ -1 respectively [Boiling point of solvent =

Accepted Answer

Elevation in boiling point is given by T_b = T_b - T_b^ Substituting the given values: T_b = 319.8 - 319.5 = 0.3\ K The relation between elevation in boiling point and molality is T_b = K_b m Rearranging for molality m: m = T_b K_b m = 0.3 2.5 = 0.12\ mol kg ^ -1 Answer: 0.12\ mol kg ^ -1

Question 16

40 gram nonelectrolyte solute having molar mass 180\ g mol ^ -1 dissolved in water has osmotic pressure 2\ atm at 300\ K . Calculate the volume of solution. ( R = 0.0821\ atm mol ^ -1 K ^ -1 )

Accepted Answer

Using the formula for osmotic pressure: = n V RT Rearranging for volume V: V = nRT Number of moles of solute n = 40 180 = 2 9 \ mol Substituting the given values: V = 2 9 0.0821 300 2 V = 24.63 9 V = 2.736\ L Since 1\ L = 1\ dm ^3, the volume of the solution is 2.74\ dm ^3. Answer: 2.74\ dm ^3

Question 17

Identify the correct statement from following.

Accepted Answer

When a non-volatile solute is added to a solvent, the vapour pressure of the solvent decreases. This is known as the relative lowering of vapour pressure. Due to the lowering of vapour pressure, the boiling point of the solution increases. This is known as the elevation in boiling point. Osmotic pressure is a colligative property, meaning it depends on the number of solute particles in the solution. Electrolytes dissociate into ions in solution, resulting in a van't Hoff factor i > 1. Non-electr

Question 18

Calculate the cryoscopic constant of a solvent if depression in freezing point of 0 3 m solution of nonelectrolyte is 0 48 K

Accepted Answer

The depression in freezing point is given by the formula: T_f = i K_f m For a nonelectrolyte, the van't Hoff factor i = 1. Substituting the given values: 0 48 = 1 K_f 0 3 K_f = 0 48 0 3 = 1 6 K kg mol ^ -1 Answer: 1 6 K kg mol ^ -1

Question 19

Calculate the molar mass of a nonvolatile solute if 6.4 g of it dissolved in 100 g water produces a relative lowering in vapour pressure of 0 016 at 300 K.

Accepted Answer

The relative lowering in vapour pressure is given by Raoult's law: P P^0 = n_2 n_1 + n_2 For a dilute solution, n_2 n_1, so n_1 + n_2 n_1. The equation simplifies to: P P^0 n_2 n_1 = w_2 M_1 M_2 w_1 Given: Relative lowering in vapour pressure, P P^0 = 0.016 Mass of solute, w_2 = 6.4 g Mass of solvent (water), w_1 = 100 g Molar mass of water, M_1 = 18 g mol ^ -1 Substituting the values into the formula: 0.016 = 6.4 18 M_2 100 M_2 = 6.4 18 0.016 100 M_2 = 115.2 1.6 = 72 g mol ^ -1 Answer: 72 gmol

Question 20

Calculate the molar mass of nonvolatile solute when 4 g of it is dissolved in 100 g solvent that boils at 319.4 K [K_b = 2.4 K kg mol ^ -1 ; B.P of pure solvent = 319 K]

Accepted Answer

Elevation in boiling point is given by T_b = T_b - T_b^0 T_b = 319.4 - 319 = 0.4 K Using the formula for elevation in boiling point: T_b = K_b m = K_b w_2 1000 M_2 w_1 Substituting the given values: 0.4 = 2.4 4 1000 M_2 100 0.4 = 96 M_2 M_2 = 96 0.4 = 240 g mol^ -1 Answer: 240\ gmol ^ -1

Solutions — Previous Year Questions & Solutions