If the vapor pressure of a dilute aqueous solution of glucose is `750 mm` of Hg at `373 K`, then molality of solute is

Options

A. 0.26
B. 0.73
C. 0.74
D. 0.039

Answer

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

### Step-by-Step Solution:

1. **Identify Given Values:**
- Vapor pressure of the solution (P_s) = 750 mm Hg
- Vapor pressure of pure water (P^0) = 760 mm Hg (at 373 K)

2. **Calculate the Relative Lowering of Vapor Pressure:**
\[ \text{Relative lowering of vapor pressure} = \frac{P^0 - P_s}{P^0} = \frac{760 - 750}{760} \]
\[ = \frac{10}{760} = \frac{1}{76} \approx 0.0132 \]

3. **Determine the Mole Fraction of the Solute (Glucose):**
- Let \( x_B \) be the mole fraction of the solute (glucose).
- From the previous calculation, \( x_B \approx 0.0132 \).

4. **Calculate the Mole Fraction of the Solvent (Water):**
\[ x_A = 1 - x_B = 1 - 0.0132 = 0.9868 \]

5. **Set Up the Relationship Between Mole Fractions and Moles:**
- The mole fraction of solute and solvent can be expressed as:
\[ x_B = \frac{n_B}{n_B + n_A} \quad \text{and} \quad x_A = \frac{n_A}{n_B + n_A} \]
- Therefore, the ratio of mole fractions can be expressed as:
\[ \frac{x_B}{x_A} = \frac{n_B}{n_A} \]
\[ \Rightarrow \frac{0.0132}{0.9868} = \frac{n_B}{n_A} \]

6. **Calculate the Number of Moles of Solute (n_B) in Terms of n_A:**
- Let \( n_A \) be the number of moles of solvent (water).
\[ n_B = n_A \cdot \frac{0.0132}{0.9868} \]

7. **Convert Moles of Solvent to Mass:**
- The molar mass of water (H₂O) is approximately 18 g/mol.
- Therefore, the mass of the solvent in grams is:
\[ \text{Mass of solvent} = n_A \cdot 18 \text{ g} \]

8. **Calculate Molality (m):**
- Molality (m) is defined as:
\[ m = \frac{n_B}{\text{mass of solvent in kg}} = \frac{n_B}{\frac{n_A \cdot 18}{1000}} \]
- Substituting \( n_B \):
\[ m = \frac{n_A \cdot \frac{0.0132}{0.9868}}{\frac{n_A \cdot 18}{1000}} \]
- Simplifying:
\[ m = \frac{0.0132 \cdot 1000}{0.9868 \cdot 18} \]

9. **Perform the Calculation:**
\[ m = \frac{13.2}{17.76} \approx 0.743 \]

10. **Final Answer:**
- The molality of the solution is approximately **0.743 mol/kg**.