Question
5 tonnes of citric acid were dissolved in 20 L of water.what is the vapour pressure of the solution at 25°c.? note vapour pressure of water at 25°c is 0.0317 bars
Answers
GPT-4o mini
To find the vapor pressure of the solution containing citric acid in water, you can use Raoult's law, which 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.
### Step 1: Calculate moles of citric acid and water.
1. **Molar mass of citric acid (C₆H₈O₇):**
- Carbon (C): 12.01 g/mol
- Hydrogen (H): 1.008 g/mol
- Oxygen (O): 16.00 g/mol
Molar mass of citric acid = \(6 \times 12.01 + 8 \times 1.008 + 7 \times 16.00 = 192.13 \, \text{g/mol}\)
2. **Moles of citric acid:**
\[
\text{Mass of citric acid} = 5000 \, \text{g}
\]
\[
\text{Moles of citric acid} = \frac{5000 \, \text{g}}{192.13 \, \text{g/mol}} \approx 26.0 \, \text{mol}
\]
3. **Moles of water:**
- Density of water ≈ 1 g/mL, thus 20 L of water is approximately \(20000 \, \text{g}\).
- Molar mass of water (H₂O) = \(2 \times 1.008 + 16.00 = 18.016 \, \text{g/mol}\)
\[
\text{Moles of water} = \frac{20000 \, \text{g}}{18.016 \, \text{g/mol}} \approx 1105.6 \, \text{mol}
\]
### Step 2: Calculate the mole fractions.
1. **Total moles in solution:**
\[
\text{Total moles} = \text{Moles of water} + \text{Moles of citric acid} \approx 1105.6 + 26.0 \approx 1131.6 \, \text{mol}
\]
2. **Mole fraction of water (\(X_{\text{water}}\)):**
\[
X_{\text{water}} = \frac{\text{Moles of water}}{\text{Total moles}} \approx \frac{1105.6}{1131.6} \approx 0.975
\]
### Step 3: Calculate the vapor pressure of the solution.
Using Raoult's law, the vapor pressure of the solution (\(P_{\text{solution}}\)) can be found as follows:
\[
P_{\text{solution}} = X_{\text{water}} \times P_{\text{water}}
\]
Where:
- \(P_{\text{water}} = 0.0317 \, \text{bars}\)
Substituting the values:
\[
P_{\text{solution}} = 0.975 \times 0.0317 \, \text{bars} \approx 0.0309 \, \text{bars}
\]
### Conclusion:
The vapor pressure of the citric acid solution at 25°C is approximately **0.0309 bars**.
### Step 1: Calculate moles of citric acid and water.
1. **Molar mass of citric acid (C₆H₈O₇):**
- Carbon (C): 12.01 g/mol
- Hydrogen (H): 1.008 g/mol
- Oxygen (O): 16.00 g/mol
Molar mass of citric acid = \(6 \times 12.01 + 8 \times 1.008 + 7 \times 16.00 = 192.13 \, \text{g/mol}\)
2. **Moles of citric acid:**
\[
\text{Mass of citric acid} = 5000 \, \text{g}
\]
\[
\text{Moles of citric acid} = \frac{5000 \, \text{g}}{192.13 \, \text{g/mol}} \approx 26.0 \, \text{mol}
\]
3. **Moles of water:**
- Density of water ≈ 1 g/mL, thus 20 L of water is approximately \(20000 \, \text{g}\).
- Molar mass of water (H₂O) = \(2 \times 1.008 + 16.00 = 18.016 \, \text{g/mol}\)
\[
\text{Moles of water} = \frac{20000 \, \text{g}}{18.016 \, \text{g/mol}} \approx 1105.6 \, \text{mol}
\]
### Step 2: Calculate the mole fractions.
1. **Total moles in solution:**
\[
\text{Total moles} = \text{Moles of water} + \text{Moles of citric acid} \approx 1105.6 + 26.0 \approx 1131.6 \, \text{mol}
\]
2. **Mole fraction of water (\(X_{\text{water}}\)):**
\[
X_{\text{water}} = \frac{\text{Moles of water}}{\text{Total moles}} \approx \frac{1105.6}{1131.6} \approx 0.975
\]
### Step 3: Calculate the vapor pressure of the solution.
Using Raoult's law, the vapor pressure of the solution (\(P_{\text{solution}}\)) can be found as follows:
\[
P_{\text{solution}} = X_{\text{water}} \times P_{\text{water}}
\]
Where:
- \(P_{\text{water}} = 0.0317 \, \text{bars}\)
Substituting the values:
\[
P_{\text{solution}} = 0.975 \times 0.0317 \, \text{bars} \approx 0.0309 \, \text{bars}
\]
### Conclusion:
The vapor pressure of the citric acid solution at 25°C is approximately **0.0309 bars**.