Vapour pressure of an aqueous solution of glucose is 750 mm of Hg at 373 K. Calcualte the molality and mole fraction of solution?

Explanation

To solve the problem of finding the molality and mole fraction of an 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

2. **Use Raoult's Law:**
According to Raoult's Law, the relationship between the vapor pressures is given by:
\[ \frac{P_0 - P_s}{P_0} = X_B \]
where \(X_B\) is the mole fraction of the solute (glucose).

3. **Calculate the Mole Fraction of the Solute:**
Substitute the values into the equation:
\[ \frac{760 - 750}{760} = X_B \]
\[ \frac{10}{760} = X_B \]
\[ X_B = 0.01316 \approx 0.013 \]

4. **Relate Mole Fraction to Moles of Solute and Solvent:**
The mole fraction of the solute can also be expressed as:
\[ X_B = \frac{n_B}{n_A + n_B} \]
where \(n_A\) is the moles of solvent (water) and \(n_B\) is the moles of solute (glucose). Since the moles of solute are small compared to the moles of solvent, we can approximate:
\[ X_B \approx \frac{n_B}{n_A} \]

5. **Calculate Moles of Solvent:**
Assume we have 1 liter of solution. The mass of the solvent (water) is approximately 1000 grams (1 kg).
- Molar mass of water = 18 g/mol
\[ n_A = \frac{1000 \text{ g}}{18 \text{ g/mol}} \approx 55.56 \text{ moles} \]

6. **Calculate Moles of Solute:**
From the mole fraction:
\[ 0.013 = \frac{n_B}{55.56} \]
Rearranging gives:
\[ n_B = 0.013 \times 55.56 \approx 0.72 \text{ moles} \]

7. **Calculate Molality:**
Molality (m) is defined as the number of moles of solute per kilogram of solvent:
\[ m = \frac{n_B}{\text{mass of solvent in kg}} = \frac{0.72}{1} = 0.72 \text{ mol/kg} \]

### Final Answers:
- **Mole Fraction of Glucose (X_B):** 0.013
- **Molality (m):** 0.72 mol/kg