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 0.0317 bars)
Answers
GPT-4o mini
To calculate the expected vapor pressure of a solution (in this case, citric acid dissolved in water) at a given temperature, we can use Raoult's Law. Raoult's Law states that the vapor pressure of a component in a solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution.
Here's a step-by-step approach for this calculation:
### Step 1: Calculate the moles of citric acid and water
1. **Molecular weight of citric acid (C₆H₈O₇)**:
\[
\text{Molecular weight} = 6 \times 12.01 + 8 \times 1.008 + 7 \times 16.00 \approx 192.13 \, \text{g/mol}
\]
2. **Calculate moles of citric acid**:
\[
5 \, \text{tonnes} = 5,000 \, \text{kg} = 5,000,000 \, \text{g}
\]
\[
\text{Moles of citric acid} = \frac{5,000,000 \, \text{g}}{192.13 \, \text{g/mol}} \approx 26041.56 \, \text{mol}
\]
3. **Calculate moles of water (assuming density of water is about 1 g/mL)**:
\[
\text{Volume of water} = 20 \, \text{L} = 20,000 \, \text{g}
\]
\[
\text{Moles of water} = \frac{20,000 \, \text{g}}{18.015 \, \text{g/mol}} \approx 1105.83 \, \text{mol}
\]
### Step 2: Calculate the total moles in the solution
\[
\text{Total moles} = \text{Moles of citric acid} + \text{Moles of water} = 26041.56 + 1105.83 \approx 27147.39 \, \text{mol}
\]
### Step 3: Calculate the mole fraction of water in the solution
\[
\text{Mole fraction of water} (X_{\text{water}}) = \frac{\text{Moles of water}}{\text{Total moles}} = \frac{1105.83}{27147.39} \approx 0.0407
\]
### Step 4: Calculate the vapor pressure of the solution using Raoult's Law
\[
P_{\text{solution}} = P^{\circ}_{\text{water}} \cdot X_{\text{water}}
\]
where \(P^{\circ}_{\text{water}} = 0.0317 \, \text{bar}\).
\[
P_{\text{solution}} = 0.0317 \, \text{bar} \times 0.0407 \approx 0.001288 \, \text{bar} \text{ (approximately)}
\]
### Conclusion
Therefore, the expected vapor pressure of the citric acid solution at 25°C is approximately **0.00129 bars**.
Here's a step-by-step approach for this calculation:
### Step 1: Calculate the moles of citric acid and water
1. **Molecular weight of citric acid (C₆H₈O₇)**:
\[
\text{Molecular weight} = 6 \times 12.01 + 8 \times 1.008 + 7 \times 16.00 \approx 192.13 \, \text{g/mol}
\]
2. **Calculate moles of citric acid**:
\[
5 \, \text{tonnes} = 5,000 \, \text{kg} = 5,000,000 \, \text{g}
\]
\[
\text{Moles of citric acid} = \frac{5,000,000 \, \text{g}}{192.13 \, \text{g/mol}} \approx 26041.56 \, \text{mol}
\]
3. **Calculate moles of water (assuming density of water is about 1 g/mL)**:
\[
\text{Volume of water} = 20 \, \text{L} = 20,000 \, \text{g}
\]
\[
\text{Moles of water} = \frac{20,000 \, \text{g}}{18.015 \, \text{g/mol}} \approx 1105.83 \, \text{mol}
\]
### Step 2: Calculate the total moles in the solution
\[
\text{Total moles} = \text{Moles of citric acid} + \text{Moles of water} = 26041.56 + 1105.83 \approx 27147.39 \, \text{mol}
\]
### Step 3: Calculate the mole fraction of water in the solution
\[
\text{Mole fraction of water} (X_{\text{water}}) = \frac{\text{Moles of water}}{\text{Total moles}} = \frac{1105.83}{27147.39} \approx 0.0407
\]
### Step 4: Calculate the vapor pressure of the solution using Raoult's Law
\[
P_{\text{solution}} = P^{\circ}_{\text{water}} \cdot X_{\text{water}}
\]
where \(P^{\circ}_{\text{water}} = 0.0317 \, \text{bar}\).
\[
P_{\text{solution}} = 0.0317 \, \text{bar} \times 0.0407 \approx 0.001288 \, \text{bar} \text{ (approximately)}
\]
### Conclusion
Therefore, the expected vapor pressure of the citric acid solution at 25°C is approximately **0.00129 bars**.