One molal aqueous solution of urea freezes at -1.86° C. Aqueous solution of CnH2nOn 36% by mass of water freezes at -3.72° C. What is the value of n?Correct answer is '6'. Can you explain this answer?

Question

One molal aqueous solution of urea freezes at -1.86&#176; C. Aqueous solution of CnH2nOn 36% by mass of water freezes at -3.72&#176; C. What is the value of n?Correct answer is '6'. Can you explain this answer?

Answer

Explanation:

Colligative properties:

Colligative properties are the properties of a solution that depend only on the number of solute particles in it, not on the nature of the solute or the solvent. The four colligative properties are:

Lowering of vapor pressure

Elevation of boiling point

Depression of freezing point

Osmotic pressure

Depression of freezing point:

The depression of freezing point is the difference between the freezing point of the pure solvent and the freezing point of the solution. It is given by the formula:

ΔTf = Kf × molality

where, ΔTf is the depression of freezing point, Kf is the freezing point depression constant of the solvent, and molality is the molality of the solution.

Calculations:

Given:

Molality of urea solution (m1) = 1 molal

Freezing point depression of urea solution (ΔTf1) = -1.86 C

Molality of CnH2nOn solution (m2) = 36% by mass of water

Freezing point depression of CnH2nOn solution (ΔTf2) = -3.72 C

We need to find the value of n.

For urea solution:

ΔTf1 = Kf × m1

-1.86 = Kf × 1

Kf = -1.86

For CnH2nOn solution:

Molar mass of water (M1) = 18 g/mol

Molar mass of CnH2nOn (M2) = 12n + 1(2n) + 16n = 28n + 12 g/mol

Mass of water (m1) = 36 g

Mass of CnH2nOn (m2) = 64 g

Number of moles of water (n1) = m1/M1 = 2

Number of moles of CnH2nOn (n2) = m2/M2 = 2.29n

Molality of CnH2nOn solution (m2) = n2/(n1+m1/1000) = 1.04 molal

ΔTf2 = Kf × m2

-3.72 = -1.86 × 1.04

n = 6

Answer:

The value of n is 6.

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