Calculate the relative lowering in vapour pressure if 100 g of a nonvolatile solute.(mol.wt.100 ) are dissolved in 432 g watere.

Explanation

To calculate the relative lowering in vapor pressure when a non-volatile solute is dissolved in a solvent, we can follow these steps:

### Step 1: Calculate the number of moles of the solute (A)
The formula to calculate the number of moles is:

\[ \text{Number of moles of solute (A)} = \frac{\text{Weight of solute (g)}}{\text{Molar mass of solute (g/mol)}} \]

Given:
- Weight of solute (A) = 100 g
- Molar mass of solute (A) = 100 g/mol

Calculating the number of moles of solute:

\[ \text{Number of moles of A} = \frac{100 \, \text{g}}{100 \, \text{g/mol}} = 1 \, \text{mol} \]

### Step 2: Calculate the number of moles of the solvent (B)
Using the same formula for the solvent:

\[ \text{Number of moles of solvent (B)} = \frac{\text{Weight of solvent (g)}}{\text{Molar mass of solvent (g/mol)}} \]

Given:
- Weight of solvent (B) = 432 g
- Molar mass of water (H2O) = 18 g/mol (calculated as \(2 \times 1 + 16 = 18\))

Calculating the number of moles of solvent:

\[ \text{Number of moles of B} = \frac{432 \, \text{g}}{18 \, \text{g/mol}} = 24 \, \text{mol} \]

### Step 3: Calculate the relative lowering in vapor pressure
The relative lowering in vapor pressure (\( \Delta P \)) can be calculated using the formula:

\[ \Delta P = \frac{\text{Number of moles of solute}}{\text{Number of moles of solvent}} \]

Substituting the values we calculated:

\[ \Delta P = \frac{1 \, \text{mol}}{24 \, \text{mol}} = \frac{1}{24} \approx 0.04167 \]

### Final Answer
The relative lowering in vapor pressure is approximately \(0.04167\).

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