Relative decrease in vapour pressure of an aqueous `NaCl` is `0.167`. Number of moles of `NaCl` present in `180g` of `H_(2)O` is:

Options

A. 2 mol
B. 1 mol
C. 3 mol
D. 4 mol

Explanation

To solve the problem, we need to find the number of moles of NaCl present in 180 g of water given that the relative decrease in vapor pressure is 0.167. We will use the concepts of vapor pressure lowering and the van 't Hoff factor.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We know that the relative decrease in vapor pressure (ΔP/P₀) is given as 0.167.
- We need to find the number of moles of NaCl in 180 g of water.

2. **Calculate the Number of Moles of Solvent (Water)**:
- The molar mass of water (H₂O) is approximately 18 g/mol.
- The number of moles of water (n) can be calculated using the formula:
\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{180 \text{ g}}{18 \text{ g/mol}} = 10 \text{ moles} \]

3. **Identify the van 't Hoff Factor (i)**:
- For NaCl, which dissociates into Na⁺ and Cl⁻ ions, the van 't Hoff factor (i) is 2.

4. **Using the Formula for Relative Decrease in Vapor Pressure**:
- The formula for the relative decrease in vapor pressure is:
\[ \frac{\Delta P}{P_0} = \frac{i \cdot n}{n + n_s} \]
- Where:
- \( \Delta P/P_0 \) = relative decrease in vapor pressure = 0.167
- \( i \) = van 't Hoff factor = 2
- \( n \) = number of moles of solute (NaCl)
- \( n_s \) = number of moles of solvent (water) = 10 moles

5. **Setting Up the Equation**:
- Plugging in the known values:
\[ 0.167 = \frac{2n}{2n + 10} \]

6. **Cross-Multiplying to Solve for n**:
- Cross-multiplying gives:
\[ 0.167(2n + 10) = 2n \]
- Expanding the left side:
\[ 0.334n + 1.67 = 2n \]

7. **Rearranging the Equation**:
- Rearranging gives:
\[ 2n - 0.334n = 1.67 \]
- Simplifying:
\[ 1.666n = 1.67 \]

8. **Calculating n**:
- Finally, solving for n:
\[ n = \frac{1.67}{1.666} \approx 1 \text{ mole} \]

### Final Answer:
The number of moles of NaCl present in 180 g of water is **1 mole**.