An aqueous solution freezes at 272.4 K while pure water freezes at 273 K. Given `K_(f)=1.86 K kg "mol"^(-1)`,`K_(b)=0.512 K kg "mol"^(-1)` and vapour pressure of water at 298 K = 23.756 mm Hg. Determine the following.
Lowering in vapour pressure at 298 K is

Options

A. `0.13`
B. `0.15`
C. `0.16`
D. `0.1378`

Explanation

To determine the lowering in vapor pressure of an aqueous solution at 298 K, we will follow these steps:

### Step 1: Calculate the Depression in Freezing Point (ΔTf)
The depression in freezing point (ΔTf) can be calculated using the formula:
\[ \Delta T_f = T_f^0 - T_f \]
Where:
- \(T_f^0\) = Freezing point of pure water = 273 K
- \(T_f\) = Freezing point of the solution = 272.4 K

Substituting the values:
\[ \Delta T_f = 273 K - 272.4 K = 0.6 K \]

### Step 2: Calculate the Molality (m)
Using the formula for molality:
\[ m = \frac{\Delta T_f}{K_f} \]
Where:
- \(K_f\) = Freezing point depression constant for water = 1.86 K kg/mol

Substituting the values:
\[ m = \frac{0.6 K}{1.86 K \cdot kg/mol} \approx 0.322 \, mol/kg \]

### Step 3: Calculate the Mole Fraction of the Solute (X_solute)
The mole fraction of the solute can be calculated using the relationship between molality and mole fraction:
\[ X_{solute} = \frac{m \cdot M_{solvent}}{m \cdot M_{solvent} + 1} \]
Where:
- \(M_{solvent}\) = Molar mass of water = 18 g/mol = 0.018 kg/mol

Substituting the values:
\[ X_{solute} = \frac{0.322 \cdot 0.018}{0.322 \cdot 0.018 + 1} \approx \frac{0.005796}{1.005796} \approx 0.0058 \]

### Step 4: Calculate the Lowering of Vapor Pressure (ΔP)
According to Raoult's Law, the lowering of vapor pressure can be calculated as:
\[ \Delta P = P^0 \cdot X_{solute} \]
Where:
- \(P^0\) = Vapor pressure of pure water at 298 K = 23.756 mm Hg

Substituting the values:
\[ \Delta P = 23.756 \, mm \, Hg \cdot 0.0058 \approx 0.1378 \, mm \, Hg \]

### Final Answer
The lowering in vapor pressure at 298 K is approximately **0.1378 mm Hg**.

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