Question

Calculate the vapor pressure of a solution of 11.5 grams of urea (a nonvolatile solute, CH4N2O, 124.16 g/mole) in 137.5 grams of water (H2O, 18.02 g/mole) at 45.0oC

The vapor pressure of pure water at 45.0oC is 71.93 mmHg at this temperature

Answer 1

To calculate the vapor pressure of a solution containing urea (a nonvolatile solute) in water, we can use Raoult's Law, which states:

\[
P_{\text{solution}} = X_{\text{solvent}} \times P^0_{\text{solvent}}
\]

where:
- \( P_{\text{solution}} \) is the vapor pressure of the solution,
- \( X_{\text{solvent}} \) is the mole fraction of the solvent (water),
- \( P^0_{\text{solvent}} \) is the vapor pressure of the pure solvent (water).

### Step 1: Calculate the number of moles of urea and water.

1. **Moles of urea:**
\[
\text{Moles of urea} = \frac{11.5 \text{ grams}}{124.16 \text{ g/mole}} \approx 0.0927 \text{ moles}
\]

2. **Moles of water:**
\[
\text{Moles of water} = \frac{137.5 \text{ grams}}{18.02 \text{ g/mole}} \approx 7.634 \text{ moles}
\]

### Step 2: Calculate the total number of moles in the solution.

\[
\text{Total moles} = \text{Moles of urea} + \text{Moles of water} \approx 0.0927 + 7.634 \approx 7.7267 \text{ moles}
\]

### Step 3: Calculate the mole fraction of water.

\[
X_{\text{water}} = \frac{\text{Moles of water}}{\text{Total moles}} = \frac{7.634}{7.7267} \approx 0.988
\]

### Step 4: Calculate the vapor pressure of the solution using Raoult's Law.

Given that \( P^0_{\text{water}} = 71.93 \text{ mmHg} \):

\[
P_{\text{solution}} = X_{\text{water}} \times P^0_{\text{water}} = 0.988 \times 71.93 \text{ mmHg} \approx 71.08 \text{ mmHg}
\]

### Final Result

The vapor pressure of the solution at 45.0°C is approximately **71.08 mmHg**.