Vapour pressure of pure water at 298 k is 23.8 mm Hg. Calculate the lowering of vapour pressure caused by adding 5g of sucrose in 50 g of water.

Answer

To solve the problem of calculating the lowering of vapor pressure caused by adding 5 g of sucrose to 50 g of water, we can follow these steps:

### Step 1: Determine the vapor pressure of pure water
The vapor pressure of pure water at 298 K is given as 23.8 mm Hg.

### Step 2: Calculate the number of moles of sucrose
The molecular formula of sucrose is C₁₂H₂₂O₁₁. To find its molar mass:
- Carbon (C): 12 × 12.01 g/mol = 144.12 g/mol
- Hydrogen (H): 22 × 1.008 g/mol = 22.176 g/mol
- Oxygen (O): 11 × 16.00 g/mol = 176.00 g/mol

Adding these together:
\[ \text{Molar mass of sucrose} = 144.12 + 22.176 + 176.00 = 342.30 \text{ g/mol} \]

Now, calculate the number of moles of sucrose:
\[ \text{Number of moles of sucrose} = \frac{\text{mass}}{\text{molar mass}} = \frac{5 \text{ g}}{342.30 \text{ g/mol}} \approx 0.0146 \text{ moles} \]

### Step 3: Calculate the number of moles of water
The molar mass of water (H₂O) is:
- Hydrogen (H): 2 × 1.008 g/mol = 2.016 g/mol
- Oxygen (O): 16.00 g/mol = 16.00 g/mol

Adding these together:
\[ \text{Molar mass of water} = 2.016 + 16.00 = 18.016 \text{ g/mol} \]

Now, calculate the number of moles of water:
\[ \text{Number of moles of water} = \frac{50 \text{ g}}{18.016 \text{ g/mol}} \approx 2.77 \text{ moles} \]

### Step 4: Calculate the total number of moles in the solution
\[ \text{Total number of moles} = \text{moles of sucrose} + \text{moles of water} = 0.0146 + 2.77 \approx 2.7846 \text{ moles} \]

### Step 5: Calculate the mole fraction of sucrose
\[ \text{Mole fraction of sucrose} (X_{\text{sucrose}}) = \frac{\text{moles of sucrose}}{\text{total moles}} = \frac{0.0146}{2.7846} \approx 0.00525 \]

### Step 6: Calculate the lowering of vapor pressure
Using the formula for relative lowering of vapor pressure:
\[ \Delta P = P_0 \times X_{\text{sucrose}} \]
Where \(P_0\) is the vapor pressure of pure water (23.8 mm Hg):
\[ \Delta P = 23.8 \text{ mm Hg} \times 0.00525 \approx 0.124 \text{ mm Hg} \]

### Final Answer
The lowering of vapor pressure caused by adding 5 g of sucrose in 50 g of water is approximately **0.124 mm Hg**.

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