3 gms of urea is added to 36 gms of boiling water. How much lowering in its vapour pressures is noticed

Options

A. 18.53 mm
B. 38 mm
C. 760 mm
D. 76 mm

Explanation

To solve the problem of how much lowering in vapor pressure is noticed when 3 grams of urea is added to 36 grams of boiling water, we can follow these steps:

### Step 1: Understand the Concept
We need to calculate the lowering of vapor pressure when a non-volatile solute (urea) is added to a solvent (water). The formula for relative lowering of vapor pressure is given by:

\[ \frac{P_0 - P_{\text{solution}}}{P_0} = x_{\text{solute}} \]

Where:
- \(P_0\) = vapor pressure of pure solvent (water)
- \(P_{\text{solution}}\) = vapor pressure of the solution
- \(x_{\text{solute}}\) = mole fraction of the solute

### Step 2: Calculate the Moles of Solvent (Water)
The number of moles of water can be calculated using the formula:

\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]

The molar mass of water (H₂O) is 18 g/mol. Given that we have 36 grams of water:

\[ \text{Moles of water} = \frac{36 \text{ g}}{18 \text{ g/mol}} = 2 \text{ moles} \]

### Step 3: Calculate the Moles of Solute (Urea)
The molar mass of urea (NH₂CO) is calculated as follows:
- Nitrogen (N): 14 g/mol (2 atoms)
- Hydrogen (H): 1 g/mol (4 atoms)
- Carbon (C): 12 g/mol (1 atom)
- Oxygen (O): 16 g/mol (1 atom)

Calculating the total:

\[ \text{Molar mass of urea} = (2 \times 14) + (4 \times 1) + 12 + 16 = 60 \text{ g/mol} \]

Now, we can calculate the moles of urea:

\[ \text{Moles of urea} = \frac{3 \text{ g}}{60 \text{ g/mol}} = 0.05 \text{ moles} \]

### Step 4: Calculate the Mole Fraction of the Solute
The mole fraction of the solute (urea) is given by:

\[ x_{\text{solute}} = \frac{\text{moles of solute}}{\text{moles of solute} + \text{moles of solvent}} \]

Substituting the values:

\[ x_{\text{solute}} = \frac{0.05}{0.05 + 2} = \frac{0.05}{2.05} \approx 0.02439 \]

### Step 5: Calculate the Lowering of Vapor Pressure
The vapor pressure of pure water at boiling point (100°C) is 760 mm Hg. The lowering of vapor pressure can be calculated as:

\[ \Delta P = P_0 \times x_{\text{solute}} = 760 \text{ mm Hg} \times 0.02439 \]

Calculating this gives:

\[ \Delta P \approx 18.53 \text{ mm Hg} \]

### Final Answer
The lowering in vapor pressure when 3 grams of urea is added to 36 grams of boiling water is approximately **18.53 mm Hg**.

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