Question

5 tonnes of citric acid were dissolved in 20 L of water. What is the expected vapor pressure of the solution or mixture at 25 degree celsius. ( vapor pressure of water at 25 degree celsius is p.0317 bars)

Answer 1

To calculate the expected vapor pressure of a solution containing 5 tonnes (5000 kg) of citric acid dissolved in 20 L of water at 25°C, we will use Raoult's Law. Raoult's Law states that the vapor pressure of a solvent in a solution is proportional to the mole fraction of the solvent.

### Steps to Calculate the Vapor Pressure:

1. **Calculate the number of moles of water:**
- The density of water is approximately 1 kg/L, so 20 L of water has a mass of 20 kg.
- Molar mass of water (H₂O) = 18.015 g/mol = 0.018015 kg/mol.
- Moles of water:
\[
n_{H2O} = \frac{20 \text{ kg}}{0.018015 \text{ kg/mol}} \approx 1104.59 \text{ mol}
\]

2. **Calculate the number of moles of citric acid:**
- The molar mass of citric acid (C₆H₈O₇) = 192.13 g/mol = 0.19213 kg/mol.
- Moles of citric acid:
\[
n_{\text{citric acid}} = \frac{5000 \text{ kg}}{0.19213 \text{ kg/mol}} \approx 26002.17 \text{ mol}
\]

3. **Calculate the total moles in the solution:**
\[
n_{\text{total}} = n_{H2O} + n_{\text{citric acid}} \approx 1104.59 \text{ mol} + 26002.17 \text{ mol} \approx 27106.76 \text{ mol}
\]

4. **Calculate the mole fraction of water:**
\[
X_{H2O} = \frac{n_{H2O}}{n_{total}} = \frac{1104.59}{27106.76} \approx 0.0407
\]

5. **Calculate the vapor pressure of the solution using Raoult's Law:**
- Vapor pressure of pure water at 25°C, \( P_{H2O}^0 = 0.0317 \text{ bar} \).
\[
P_{solution} = X_{H2O} \cdot P_{H2O}^0 = 0.0407 \cdot 0.0317 \text{ bar} \approx 0.00129 \text{ bar}
\]

### Conclusion:
The expected vapor pressure of the citric acid solution at 25°C is approximately **0.00129 bar**.