The molal lowering of vapor pressure for `H_(2)O` at `100^(@)C` is

Options

A. 13.43 mm of Hg
B. 23.43 mm of Hg
C. 1.343 mm of Hg
D. 234. 3 mm of Hg

Answer

To find the molal lowering of vapor pressure for water (H₂O) at 100°C, we can follow these steps:

### Step 1: Understand the Concept
When a non-volatile solute is added to a pure solvent, the vapor pressure of the solution decreases compared to the pure solvent. This decrease in vapor pressure is known as the lowering of vapor pressure.

### Step 2: Define the Formula
The relative lowering of vapor pressure (ΔP) can be expressed as:
\[ \Delta P = \frac{P_0 - P_s}{P_0} = X_{solute} \]
where:
- \(P_0\) = vapor pressure of the pure solvent
- \(P_s\) = vapor pressure of the solution
- \(X_{solute}\) = mole fraction of the solute

### Step 3: Calculate the Mole Fraction of the Solute
Given that we have a 1 molal solution, this means we have 1 mole of solute in 1 kg of solvent (water).

1. **Calculate the moles of solvent (water)**:
- Molar mass of water (H₂O) = 18 g/mol
- Mass of water = 1 kg = 1000 g
\[ \text{Moles of water} = \frac{1000 \text{ g}}{18 \text{ g/mol}} \approx 55.56 \text{ moles} \]

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

3. **Calculate the mole fraction of the solute**:
\[ X_{solute} = \frac{\text{Moles of solute}}{\text{Total moles}} = \frac{1}{56.56} \approx 0.0177 \]

### Step 4: Calculate the Lowering of Vapor Pressure
Using the vapor pressure of pure water at 100°C, which is \(P_0 = 760 \text{ mm Hg}\):
\[ \Delta P = P_0 \times X_{solute} = 760 \text{ mm Hg} \times 0.0177 \approx 13.43 \text{ mm Hg} \]

### Final Answer
The molal lowering of vapor pressure for water at 100°C is approximately **13.43 mm Hg**.

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