Asked by JESSY
5 tonnes of citric acid (C6H8O7) were dissolved in 20L of water. What is the expected vapour pressure of the solution or mixture at 25°C? Note that vapour pressure of H2O at 25°C is 0.0317bars
Answers
Answered by
GPT-4o mini
To calculate the expected vapor pressure of the solution of citric acid (C₆H₈O₇) in water at 25°C, we can use Raoult's Law. Raoult's Law states that the vapor pressure of a solvent in a solution is equal to the vapor pressure of the pure solvent multiplied by the mole fraction of the solvent in the solution.
1. **Determine the moles of citric acid and water:**
- **Molar mass of citric acid (C₆H₈O₇):**
\[
\text{Molar mass} = 6 \times 12.01 \, (\text{C}) + 8 \times 1.01 \, (\text{H}) + 7 \times 16.00 \, (\text{O}) = 192.13 \, \text{g/mol}
\]
- **Mass of citric acid:**
\[
5 \, \text{tonnes} = 5,000 \, \text{kg} = 5,000,000 \, \text{g}
\]
- **Moles of citric acid:**
\[
\text{Moles of citric acid} = \frac{5,000,000 \, \text{g}}{192.13 \, \text{g/mol}} \approx 26040.6 \, \text{mol}
\]
- **Assuming the density of water is approximately 1 g/mL:**
\[
\text{Mass of water} = 20 \, \text{L} \times 1000 \, \text{g/L} = 20,000 \, \text{g}
\]
- **Moles of water:**
\[
\text{Moles of water} = \frac{20,000 \, \text{g}}{18.02 \, \text{g/mol}} \approx 1105.56 \, \text{mol}
\]
2. **Calculate the total moles in the solution:**
\[
\text{Total moles} = \text{Moles of citric acid} + \text{Moles of water} \approx 26040.6 \, \text{mol} + 1105.56 \, \text{mol} \approx 27146.16 \, \text{mol}
\]
3. **Calculate the mole fraction of water (\(X_{\text{water}}\)):**
\[
X_{\text{water}} = \frac{\text{Moles of water}}{\text{Total moles}} = \frac{1105.56 \, \text{mol}}{27146.16 \, \text{mol}} \approx 0.0407
\]
4. **Use Raoult's Law to calculate the vapor pressure of the solution:**
\[
P_{\text{solution}} = X_{\text{water}} \times P^0_{\text{water}}
\]
Given \(P^0_{\text{water}} = 0.0317 \, \text{bars}\), we can calculate:
\[
P_{\text{solution}} = 0.0407 \times 0.0317 \approx 0.00129 \, \text{bars}
\]
Thus, the expected vapor pressure of the solution at 25°C is approximately **0.00129 bars**.
1. **Determine the moles of citric acid and water:**
- **Molar mass of citric acid (C₆H₈O₇):**
\[
\text{Molar mass} = 6 \times 12.01 \, (\text{C}) + 8 \times 1.01 \, (\text{H}) + 7 \times 16.00 \, (\text{O}) = 192.13 \, \text{g/mol}
\]
- **Mass of citric acid:**
\[
5 \, \text{tonnes} = 5,000 \, \text{kg} = 5,000,000 \, \text{g}
\]
- **Moles of citric acid:**
\[
\text{Moles of citric acid} = \frac{5,000,000 \, \text{g}}{192.13 \, \text{g/mol}} \approx 26040.6 \, \text{mol}
\]
- **Assuming the density of water is approximately 1 g/mL:**
\[
\text{Mass of water} = 20 \, \text{L} \times 1000 \, \text{g/L} = 20,000 \, \text{g}
\]
- **Moles of water:**
\[
\text{Moles of water} = \frac{20,000 \, \text{g}}{18.02 \, \text{g/mol}} \approx 1105.56 \, \text{mol}
\]
2. **Calculate the total moles in the solution:**
\[
\text{Total moles} = \text{Moles of citric acid} + \text{Moles of water} \approx 26040.6 \, \text{mol} + 1105.56 \, \text{mol} \approx 27146.16 \, \text{mol}
\]
3. **Calculate the mole fraction of water (\(X_{\text{water}}\)):**
\[
X_{\text{water}} = \frac{\text{Moles of water}}{\text{Total moles}} = \frac{1105.56 \, \text{mol}}{27146.16 \, \text{mol}} \approx 0.0407
\]
4. **Use Raoult's Law to calculate the vapor pressure of the solution:**
\[
P_{\text{solution}} = X_{\text{water}} \times P^0_{\text{water}}
\]
Given \(P^0_{\text{water}} = 0.0317 \, \text{bars}\), we can calculate:
\[
P_{\text{solution}} = 0.0407 \times 0.0317 \approx 0.00129 \, \text{bars}
\]
Thus, the expected vapor pressure of the solution at 25°C is approximately **0.00129 bars**.