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 of an aqueous NaCl solution is 0.167.

### Step-by-Step Solution:

1. **Understand the Concept of Relative Decrease in Vapor Pressure**:
The relative decrease in vapor pressure (RLBP) can be expressed using Raoult's Law:
\[ \text{RLBP} = i \cdot \frac{n}{n + N} \]
where:
- \( i \) is the van 't Hoff factor (number of particles the solute dissociates into),
- \( n \) is the number of moles of solute (NaCl),
- \( N \) is the number of moles of solvent (water).

2. **Identify the van 't Hoff Factor for NaCl**:
Sodium chloride (NaCl) dissociates into two ions: Na\(^+\) and Cl\(^-\). Therefore, the van 't Hoff factor \( i \) for NaCl is:
\[ i = 2 \]

3. **Calculate the Number of Moles of Water**:
The molar mass of water (H\(_2\)O) is approximately 18 g/mol. To find the number of moles of water in 180 g:
\[ N = \frac{180 \text{ g}}{18 \text{ g/mol}} = 10 \text{ moles} \]

4. **Substitute Known Values into the RLBP Equation**:
Now we can substitute the known values into the RLBP equation:
\[ 0.167 = 2 \cdot \frac{n}{n + 10} \]

5. **Solve for \( n \)**:
Rearranging the equation gives:
\[ 0.167 = \frac{2n}{n + 10} \]
Cross-multiplying:
\[ 0.167(n + 10) = 2n \]
Expanding the left side:
\[ 0.167n + 1.67 = 2n \]
Rearranging gives:
\[ 2n - 0.167n = 1.67 \]
\[ 1.833n = 1.67 \]
Dividing both sides by 1.833:
\[ n = \frac{1.67}{1.833} \approx 0.911 \text{ moles} \]

### Final Answer:
The number of moles of NaCl present in 180 g of water is approximately **0.911 moles**.