What is the vapour pressure at `100^(@)C` of a solution containing `15.6g` of water and `1.68g` of sucrose `(C_(2)H_(22)O_(11))` ?

Answer

To find the vapor pressure of a solution containing 15.6 g of water and 1.68 g of sucrose (C₁₂H₂₂O₁₁) at 100°C, we will follow these steps:

### Step 1: Calculate the number of moles of water and sucrose.
- **Molar mass of water (H₂O)** = 18 g/mol
- **Molar mass of sucrose (C₁₂H₂₂O₁₁)** = 342 g/mol

**Moles of water:**
\[ \text{Moles of water} = \frac{\text{mass of water}}{\text{molar mass of water}} = \frac{15.6 \, \text{g}}{18 \, \text{g/mol}} = 0.8667 \, \text{mol} \]

**Moles of sucrose:**
\[ \text{Moles of sucrose} = \frac{\text{mass of sucrose}}{\text{molar mass of sucrose}} = \frac{1.68 \, \text{g}}{342 \, \text{g/mol}} = 0.0049 \, \text{mol} \]

### Step 2: Calculate the total number of moles in the solution.
\[ \text{Total moles} = \text{Moles of water} + \text{Moles of sucrose} = 0.8667 \, \text{mol} + 0.0049 \, \text{mol} = 0.8716 \, \text{mol} \]

### Step 3: Calculate the mole fraction of sucrose.
\[ \text{Mole fraction of sucrose (X}_{\text{sucrose}}) = \frac{\text{Moles of sucrose}}{\text{Total moles}} = \frac{0.0049}{0.8716} = 0.0056 \]

### Step 4: Calculate the vapor pressure of pure water at 100°C.
The vapor pressure of pure water (P₀) at 100°C is approximately 760 mmHg.

### Step 5: Use Raoult's Law to find the vapor pressure of the solution (Pₛ).
According to Raoult's Law:
\[ P_s = P_0 \times (1 - X_{\text{sucrose}}) \]
Where:
- \( P_s \) = vapor pressure of the solution
- \( P_0 \) = vapor pressure of pure solvent (760 mmHg)
- \( X_{\text{sucrose}} \) = mole fraction of the solute

Calculating \( P_s \):
\[ P_s = 760 \, \text{mmHg} \times (1 - 0.0056) = 760 \, \text{mmHg} \times 0.9944 = 754.5 \, \text{mmHg} \]

### Step 6: Conclusion
The vapor pressure of the solution at 100°C is approximately **754.5 mmHg**.

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