At room temperature, a dilute solution of urea is prepared by dissolving 0.60 g of urea in 360 g of water. If the vapour pressure of pure water at this temperature is 35 mm Hg, lowering of vapour pressure will be `1.74 xx 10^(-x)` mm Hg. The numerical value of x is _____________. (Molar mass of urea = 60 g `mol^(-1)`).

Explanation

To solve the problem, we need to calculate the lowering of vapor pressure when 0.60 g of urea is dissolved in 360 g of water. We will use Raoult's Law, which states that the lowering of vapor pressure is proportional to the mole fraction of the solute.

### Step-by-Step Solution:

1. **Calculate the number of moles of urea:**
\[ \text{Number of moles of urea} = \frac{\text{mass of urea}}{\text{molar mass of urea}} = \frac{0.60 \, \text{g}}{60 \, \text{g/mol}} = 0.01 \, \text{mol} \]

2. **Calculate the number of moles of water:**
\[ \text{Molar mass of water (H}_2\text{O)} = 2 \times 1 + 16 = 18 \, \text{g/mol} \]
\[ \text{Number of moles of water} = \frac{\text{mass of water}}{\text{molar mass of water}} = \frac{360 \, \text{g}}{18 \, \text{g/mol}} = 20 \, \text{mol} \]

3. **Calculate the total number of moles in the solution:**
\[ \text{Total number of moles} = \text{moles of urea} + \text{moles of water} = 0.01 + 20 = 20.01 \, \text{mol} \]

4. **Calculate the mole fraction of urea (solute):**
\[ \chi_{\text{urea}} = \frac{\text{moles of urea}}{\text{total moles}} = \frac{0.01}{20.01} \approx 0.00049975 \]

5. **Calculate the lowering of vapor pressure (ΔP):**
Using Raoult's Law:
\[ \Delta P = P^0 \cdot \chi_{\text{urea}} \]
where \( P^0 = 35 \, \text{mm Hg} \):
\[ \Delta P = 35 \, \text{mm Hg} \times 0.00049975 \approx 0.01749 \, \text{mm Hg} \]

6. **Express ΔP in the form \( 1.74 \times 10^{-x} \) mm Hg:**
\[ 0.01749 \, \text{mm Hg} = 1.749 \times 10^{-2} \, \text{mm Hg} \approx 1.74 \times 10^{-2} \, \text{mm Hg} \]
Here, we can see that \( x = 2 \).

### Final Answer:
The numerical value of \( x \) is **2**.