Lowering of vapour pressure of an aqueous solution of a non-volatile non-electrolyte 1 m aqueous solution at `100^(@)C` is,

Options

A. 14.12 torr
B. 312 torr
C. 13.45 torr
D. 352 torr

Answer

To find the lowering of vapor pressure of a 1 molal aqueous solution of a non-volatile non-electrolyte at 100°C, we can follow these steps:

### Step 1: Understand Raoult's Law
According to Raoult's law, the relative lowering of vapor pressure of a solution is equal to the mole fraction of the solute:
\[ \frac{\Delta P}{P^0} = X_{solute} \]
where:
- \(\Delta P\) = lowering of vapor pressure
- \(P^0\) = vapor pressure of the pure solvent
- \(X_{solute}\) = mole fraction of the solute

### Step 2: Determine the Mole Fraction of the Solute
For a 1 molal solution, we have 1 mole of solute in 1000 grams of solvent (water).

1. **Calculate the number of moles of solute**:
- Moles of solute = 1 mole (given)

2. **Calculate the number of moles of solvent (water)**:
- Mass of solvent = 1000 g
- Molar mass of water (H₂O) = 18 g/mol
- Moles of water = \(\frac{1000 \text{ g}}{18 \text{ g/mol}} \approx 55.56 \text{ moles}\)

3. **Calculate the total moles**:
- Total moles = Moles of solute + Moles of solvent
- Total moles = \(1 + 55.56 = 56.56\)

4. **Calculate the mole fraction of the solute**:
\[ X_{solute} = \frac{\text{Moles of solute}}{\text{Total moles}} = \frac{1}{56.56} \]

### Step 3: Determine the Vapor Pressure of Pure Water
The vapor pressure of pure water at 100°C is given as:
\[ P^0 = 760 \text{ Torr} \]

### Step 4: Calculate the Lowering of Vapor Pressure
Using Raoult's law, we can now calculate the lowering of vapor pressure:
\[ \Delta P = P^0 \times X_{solute} \]
Substituting the values:
\[ \Delta P = 760 \text{ Torr} \times \frac{1}{56.56} \]

### Step 5: Perform the Calculation
Calculating the above expression:
\[ \Delta P \approx 760 \div 56.56 \approx 13.45 \text{ Torr} \]

### Conclusion
The lowering of vapor pressure of the 1 molal aqueous solution at 100°C is approximately **13.45 Torr**.