Question
5 tonnes of citric acid were dissolved 20 L of water.what is the expected vapour pressure of the solution or mixture at 25°c?(note vapour pressure of water at 25°c is 0.0317 bars)
Answers
GPT-4o mini
To calculate the expected vapor pressure of the solution, we will use Raoult's Law, which 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.
1. **Determine Moles of Solute (Citric Acid)**:
- The molar mass of citric acid (C6H8O7) is approximately \( 192.13 \, \text{g/mol} \).
- Given that we have 5 tonnes of citric acid:
- Convert tonnes to grams: \( 5 \, \text{tonnes} = 5,000 \, \text{kg} = 5,000,000 \, \text{g} \).
- Calculate moles of citric acid:
\[
\text{Moles of citric acid} = \frac{5,000,000 \, \text{g}}{192.13 \, \text{g/mol}} \approx 26040.7 \, \text{mol}
\]
2. **Determine Moles of Solvent (Water)**:
- The density of water is approximately \( 1 \, \text{g/mL} \), so for \( 20 \, \text{L} \):
- Convert liters to grams: \( 20 \, \text{L} = 20,000 \, \text{g} \).
- Calculate moles of water:
\[
\text{Moles of water} = \frac{20,000 \, \text{g}}{18.015 \, \text{g/mol}} \approx 1104.1 \, \text{mol}
\]
3. **Calculate the Total Moles in the Solution**:
- Total moles = moles of citric acid + moles of water:
\[
\text{Total moles} = 26040.7 \, \text{mol} + 1104.1 \, \text{mol} \approx 27144.8 \, \text{mol}
\]
4. **Calculate the Mole Fraction of Water**:
- The mole fraction of water (\( X_{\text{water}} \)) is calculated as:
\[
X_{\text{water}} = \frac{\text{Moles of water}}{\text{Total moles}} = \frac{1104.1 \, \text{mol}}{27144.8 \, \text{mol}} \approx 0.0407
\]
5. **Calculate the Vapor Pressure of the Solution**:
- According to Raoult's Law:
\[
P_{\text{solution}} = X_{\text{water}} \times P^0_{\text{water}}
\]
- Where \( P^0_{\text{water}} = 0.0317 \, \text{bars} \):
\[
P_{\text{solution}} = 0.0407 \times 0.0317 \, \text{bars} \approx 0.00129 \, \text{bars} (or 1.29 \, \text{mbar})
\]
Thus, the expected vapor pressure of the solution at 25°C is approximately **0.00129 bars** or **1.29 mbar**.
1. **Determine Moles of Solute (Citric Acid)**:
- The molar mass of citric acid (C6H8O7) is approximately \( 192.13 \, \text{g/mol} \).
- Given that we have 5 tonnes of citric acid:
- Convert tonnes to grams: \( 5 \, \text{tonnes} = 5,000 \, \text{kg} = 5,000,000 \, \text{g} \).
- Calculate moles of citric acid:
\[
\text{Moles of citric acid} = \frac{5,000,000 \, \text{g}}{192.13 \, \text{g/mol}} \approx 26040.7 \, \text{mol}
\]
2. **Determine Moles of Solvent (Water)**:
- The density of water is approximately \( 1 \, \text{g/mL} \), so for \( 20 \, \text{L} \):
- Convert liters to grams: \( 20 \, \text{L} = 20,000 \, \text{g} \).
- Calculate moles of water:
\[
\text{Moles of water} = \frac{20,000 \, \text{g}}{18.015 \, \text{g/mol}} \approx 1104.1 \, \text{mol}
\]
3. **Calculate the Total Moles in the Solution**:
- Total moles = moles of citric acid + moles of water:
\[
\text{Total moles} = 26040.7 \, \text{mol} + 1104.1 \, \text{mol} \approx 27144.8 \, \text{mol}
\]
4. **Calculate the Mole Fraction of Water**:
- The mole fraction of water (\( X_{\text{water}} \)) is calculated as:
\[
X_{\text{water}} = \frac{\text{Moles of water}}{\text{Total moles}} = \frac{1104.1 \, \text{mol}}{27144.8 \, \text{mol}} \approx 0.0407
\]
5. **Calculate the Vapor Pressure of the Solution**:
- According to Raoult's Law:
\[
P_{\text{solution}} = X_{\text{water}} \times P^0_{\text{water}}
\]
- Where \( P^0_{\text{water}} = 0.0317 \, \text{bars} \):
\[
P_{\text{solution}} = 0.0407 \times 0.0317 \, \text{bars} \approx 0.00129 \, \text{bars} (or 1.29 \, \text{mbar})
\]
Thus, the expected vapor pressure of the solution at 25°C is approximately **0.00129 bars** or **1.29 mbar**.