If the vapour pressure of pure water at `25^@C` is 23.8 mmHg , then calculate the vapour pressure lowering caused by the addition of 100 g of sucrose (molecular mass = 342 g/mol) to 100 g of water

Options

A. 00.12 mmHg
B. 0.125 mmHg
C. 1.15 mmHg
D. 1.25 mmHg

Answer

To solve the problem of vapor pressure lowering caused by the addition of sucrose to water, we will follow these steps:

### Step 1: Identify the given data
- Vapor pressure of pure water (P₀) at 25°C = 23.8 mmHg
- Mass of sucrose (solute) = 100 g
- Molecular mass of sucrose = 342 g/mol
- Mass of water (solvent) = 100 g
- Molecular mass of water = 18 g/mol

### Step 2: Calculate the number of moles of sucrose (solute)
To find the number of moles of sucrose, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
For sucrose:
\[ \text{Number of moles of sucrose} = \frac{100 \, \text{g}}{342 \, \text{g/mol}} \approx 0.292 \, \text{mol} \]

### Step 3: Calculate the number of moles of water (solvent)
Using the same formula for water:
\[ \text{Number of moles of water} = \frac{100 \, \text{g}}{18 \, \text{g/mol}} \approx 5.55 \, \text{mol} \]

### Step 4: Calculate the relative lowering of vapor pressure
The formula for relative lowering of vapor pressure is given by:
\[ \frac{\Delta P}{P₀} = \frac{\text{Number of moles of solute}}{\text{Number of moles of solvent}} \]
Substituting the values we calculated:
\[ \frac{\Delta P}{23.8} = \frac{0.292}{5.55} \]

### Step 5: Solve for ΔP (the lowering of vapor pressure)
First, calculate the right side:
\[ \frac{0.292}{5.55} \approx 0.0527 \]
Now, multiply both sides by P₀ to find ΔP:
\[ \Delta P = 23.8 \times 0.0527 \approx 1.25 \, \text{mmHg} \]

### Final Answer
The vapor pressure lowering caused by the addition of 100 g of sucrose to 100 g of water is approximately **1.25 mmHg**.

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