The vapoure pressure of an aqueous solution of glucose is 750 mm of Hg at 373 K. the molality is

Options

A. `0.64`
B. `0.741`
C. `0.68`
D. `0.94`

Answer

To find the molality of the aqueous solution of glucose given its vapor pressure, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Vapor pressure of pure water (P₀) at 373 K = 760 mmHg
- Vapor pressure of the glucose solution (Pₛ) = 750 mmHg

2. **Calculate the Lowering of Vapor Pressure**:
\[ \Delta P = P₀ - Pₛ = 760 \, \text{mmHg} - 750 \, \text{mmHg} = 10 \, \text{mmHg} \]

3. **Calculate the Relative Lowering of Vapor Pressure**:
\[ \text{Relative Lowering} = \frac{\Delta P}{P₀} = \frac{10 \, \text{mmHg}}{760 \, \text{mmHg}} = 0.01316 \]

4. **Relate Relative Lowering to Mole Fraction**:
The relative lowering of vapor pressure is equal to the mole fraction of solute (glucose):
\[ \frac{n_b}{n_a + n_b} = 0.01316 \]
Here, \( n_b \) is the number of moles of solute (glucose) and \( n_a \) is the number of moles of solvent (water).

5. **Assume a Basis for Calculation**:
Assume 1 kg of water (solvent). The number of moles of water (n_a) can be calculated as:
\[ n_a = \frac{1000 \, \text{g}}{18 \, \text{g/mol}} = 55.56 \, \text{mol} \]

6. **Set Up the Equation**:
From the mole fraction equation:
\[ n_b = 0.01316 \times (n_a + n_b) \]
Rearranging gives:
\[ n_b = 0.01316 \times (55.56 + n_b) \]

7. **Solve for n_b**:
\[ n_b - 0.01316 n_b = 0.01316 \times 55.56 \]
\[ n_b (1 - 0.01316) = 0.731 \]
\[ n_b = \frac{0.731}{0.98684} \approx 0.740 \, \text{mol} \]

8. **Calculate Molality**:
Molality (m) is defined as:
\[ m = \frac{n_b}{\text{mass of solvent in kg}} = \frac{0.740 \, \text{mol}}{1 \, \text{kg}} = 0.740 \, \text{mol/kg} \]

### Final Answer:
The molality of the glucose solution is approximately **0.740 mol/kg**.

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