The lowering of vapour pressure due to a solute in a `1 m` aqueous solution at `100^(@)C` is

Options

A. 13.44 torr
B. 14.12 torr
C. 312 torr
D. 352 torr

Explanation

To solve the problem of finding the lowering of vapor pressure due to a solute in a 1 m aqueous solution at 100°C, we can follow these steps:

### Step-by-Step Solution:

**Step 1: Understand the relationship between molality and mole fraction.**
- The molality (m) of a solution is defined as the number of moles of solute per kilogram of solvent. In this case, we have a 1 m aqueous solution, meaning there is 1 mole of solute in 1 kg of water.

**Step 2: Calculate the mole fraction of the solute.**
- The mole fraction (χ) of the solute can be calculated using the formula:
\[ \chi_{solute} = \frac{n_{solute}}{n_{solute} + n_{solvent}} \]
- Here, \(n_{solute} = 1 \, \text{mol}\) (since it's a 1 m solution) and \(n_{solvent} = \frac{1000 \, \text{g}}{18 \, \text{g/mol}} = 55.56 \, \text{mol}\) (for water, where the molar mass of water is approximately 18 g/mol).

**Step 3: Calculate the total number of moles.**
- Total moles = \(n_{solute} + n_{solvent} = 1 + 55.56 = 56.56 \, \text{mol}\).

**Step 4: Calculate the mole fraction of the solute.**
- Now, substituting the values into the mole fraction formula:
\[ \chi_{solute} = \frac{1}{56.56} \approx 0.0177 \]

**Step 5: Use Raoult's Law to find the lowering of vapor pressure.**
- According to Raoult's Law, the relative lowering of vapor pressure is given by:
\[ \Delta P = P_0 - P_s = P_0 \cdot \chi_{solute} \]
- Where \(P_0\) is the vapor pressure of pure solvent (water at 100°C is approximately 760 torr).

**Step 6: Calculate the lowering of vapor pressure.**
- Thus, substituting the values:
\[ \Delta P = 760 \, \text{torr} \cdot 0.0177 \approx 13.44 \, \text{torr} \]

### Final Answer:
The lowering of vapor pressure due to a solute in a 1 m aqueous solution at 100°C is approximately **13.44 torr**.