The vapour pressure of an aqueous solution of glucose is `750 mm` of `Hg` at `373 K`. Calculate molality and mole fraction of solute.

Explanation

To solve the problem, we need to calculate the molality and mole fraction of the solute (glucose) in the aqueous solution given the vapor pressure of the solution.

### Step 1: Understand the given data
- Vapor pressure of the solution (Ps) = 750 mmHg
- Vapor pressure of pure water (P0) = 760 mmHg (at 373 K)

### Step 2: Calculate the change in vapor pressure
The change in vapor pressure (ΔP) can be calculated as:
\[ \Delta P = P_0 - P_s = 760 \, \text{mmHg} - 750 \, \text{mmHg} = 10 \, \text{mmHg} \]

### Step 3: Use Raoult's Law to find the mole fraction of the solute
According to Raoult's Law:
\[ \frac{\Delta P}{P_0} = \frac{n_{solute}}{n_{solution}} \]
Where:
- \(n_{solute}\) = moles of solute
- \(n_{solution}\) = moles of solute + moles of solvent

Let's denote the mole fraction of the solute (glucose) as \(X_{solute}\):
\[ X_{solute} = \frac{n_{solute}}{n_{solute} + n_{solvent}} = \frac{\Delta P}{P_0} \]

Substituting the values:
\[ X_{solute} = \frac{10}{760} = 0.01316 \]

### Step 4: Calculate the molality of the solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent. We can rearrange the equation from Raoult's Law:
\[ \Delta P = P_0 \cdot X_{solute} \]
From this, we can express moles of solute in terms of moles of solvent:
\[ n_{solute} = X_{solute} \cdot (n_{solute} + n_{solvent}) \approx X_{solute} \cdot n_{solvent} \quad (\text{for small } X_{solute}) \]

Let \(n_{solvent}\) be the number of moles of water:
\[ n_{solvent} = \frac{W_{solvent}}{M_{solvent}} = \frac{1000 \, \text{g}}{18 \, \text{g/mol}} \approx 55.56 \, \text{mol} \]

Using the mole fraction calculated:
\[ n_{solute} = 0.01316 \cdot 55.56 \approx 0.731 \, \text{mol} \]

Now, we can calculate molality:
\[ m = \frac{n_{solute}}{W_{solvent} \text{ (in kg)}} = \frac{0.731 \, \text{mol}}{1 \, \text{kg}} = 0.731 \, \text{mol/kg} \]

### Final Results
- **Molality (m)**: 0.731 mol/kg
- **Mole Fraction of Solute (X_{solute})**: 0.01316