CPU 4 - Colligative Properties II (Depression In Freezing Point) - Class 12 MCQ

30 mL of CH₃OH (d = 0.7980 g cm^-3) and 70 mL of H₂O(of = 0.9984 g cm^-3) are mixed at 25° C to form a solution of density 0.9575 g cm^-3. Thus, depression in freezing point is (K_f for H₂0 =1.86 K mol^-1 kg)

How much ethyl alcohol (C₂H₅OH) must be added to 1 L of water so that solution will not freeze at - 4 ° F? (K_f of H₂0 = 1.86° m ol^-1 kg)

CH₃OH is 10% by weight of solution of CH₃OH in water (K_f of H₂0 = 1.86° mol^-1 kg). Thus, freezing point of the solution is 

The amount of ice that will separate out on cooling a solution containing 50 g of ethylene glycol in 200 g w ater to -9.3° C is (K_f of H₂0 = 1.86° mol^-1 kg)

A solution of a non-volatile solute in water freezes at - 0.30° C. The vapour pressure o f pure w ater at 298 K is 23.51 torr. For water, K_f = 1.86 K mol^-1 kg. Thus, vapour pressure of the solution (in torr)

A solution is prepared by condensing 4 L o f a gas measured at 27° C and Working Space 748 mm Hg pressure into 58 g benzene (K_f of benzene = 5.12 K mol^-1 kg). Thus, depression in freezing point is

What freezing point depression would be predicted for 0.1 molal solution of benzoic acid in benzene if latent heat of fusion is 30.00 cal g^-1 at 280 K (freezing point) for benzene? (assume no change in molecular state)

Based on the given graph, molar mass of solute X (non-electrolyte) is

Statement Type
This section is based on Statement I and Statement II. Select the correct anser from the codes given below

Q.

Statement I : Freezing point of a solution is smaller than that of the solvent.

Statement II At freezing point, solid and liquid exist in equilibrium and T(fp) = 

One or More than One Options Correct Type
This section contains 2 multiple type questions. Each question has 4 choices (a), (b), (c) and (d), out of which ONE or MORE THAN ONE are correct.

Q.

In the depression of freezing point experiment, it is found that the

A solution containing 3.5 g of solute X is 50.0 g of water has a volume of 52.5 mL and a freezing point of - 0.86° C. Thus, (K_f of H₂0 = 1.86° mol^-1 kg)

Matching List Type
Choices for the correct combination of elements from Column I and Column II are given as options (a), (b), (c) and (d), out of which one is correct

Q. 

Given, K_f of H₂0 =1.86° mol^-1kg.
Match the solutions given in Column I with their depression in freezing points given in Column II and select the answer from the codes given below.

Comprehension Type
This section contains a passage describing theory, experiments, data, etc. Two questions related to the paragraph have been given. Each question has only one correct answer out of the given 4 options (a), (b), (c) and (d)

Passage

1 kg of 1 molal aqueous solution of sucrose is cooled and maintained at - 3.534° C. K_f (H₂0) = 1.86 K mol^-1kg

Q.

 Relative lowering of vapour pressure 

Passage

1 kg of 1 molal aqueous solution of sucrose is cooled and maintained at - 3.534° C. K_f (H₂0) = 1.86 K mol^-1 kg

Q.

How much ice will separate under the given conditions?

One Integer Value Correct Type
This section contains 3 questions, when worked out will result in an integer value from 0 to 9 (both inclusive)

Q.

At 100° C and 1 atm, following equilibrium is set up

Volume occupied by water mole, in 10 L steam is x cm³. What is the value of x?

Molecular formula of white phosphorus in benzene is (P)_x. What is the value of x based on following experiment?

Freezing point of benzene = 5.45° C K_f (benzene) = 5.12 K mol^-1 kg

Freezing point of solution containing 0.675 g of white phosphorus in 50.0 g of benzene = 4.89° C.

One molal aqueous solution of urea freezes at -1.86° C. Aqueous solution of C_nH_2nO_n  36% by mass of water freezes at -3.72° C. What is the value of n?