To calculate the expected vapor pressure of a solution, we need to apply Raoult's Law, which states that the vapor pressure of a solvent in a solution (P_solvent) is equal to the vapor pressure of the pure solvent (P^0_solvent) multiplied by the mole fraction of the solvent in the solution (X_solvent).
Given:
- Mass of citric acid = 5 tonnes (5,000 kg)
- Volume of water = 20 L
- Density of water ~ 1 kg/L
- Therefore, mass of water = 20 kg
Next, we need to find the number of moles of citric acid and water:
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Molar mass of citric acid (C₆H₈O₇):
- C: 12.01 g/mol × 6 = 72.06 g/mol
- H: 1.008 g/mol × 8 = 8.064 g/mol
- O: 16.00 g/mol × 7 = 112.00 g/mol
- Total molar mass = 72.06 + 8.064 + 112.00 = 192.124 g/mol
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Convert 5,000 kg of citric acid to grams: \[ 5,000 \text{ kg} = 5,000,000 \text{ g} \]
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Calculate the number of moles of citric acid: \[ n_{\text{citric acid}} = \frac{5,000,000 \text{ g}}{192.124 \text{ g/mol}} \approx 26046.88 \text{ mol} \]
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For water, the molar mass is ~18.015 g/mol: \[ n_{\text{water}} = \frac{20,000 \text{ g}}{18.015 \text{ g/mol}} \approx 1104.23 \text{ mol} \]
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Calculate the mole fractions: \[ X_{\text{water}} = \frac{n_{\text{water}}}{n_{\text{water}} + n_{\text{citric acid}}} = \frac{1104.23}{1104.23 + 26046.88} \approx 0.0406 \] \[ X_{\text{citric acid}} = 1 - X_{\text{water}} \approx 0.9594 \]
Next, we need the vapor pressure of pure water at 25 °C, which is approximately 23.76 mmHg.
Now, apply Raoult's Law: \[ P_{\text{solution}} = P^0_{\text{water}} \cdot X_{\text{water}} = 23.76 \text{ mmHg} \cdot 0.0406 \approx 0.964 \text{ mmHg} \]
Therefore, the expected vapor pressure of the solution at 25 °C would be approximately 0.964 mmHg.
Note: The presence of a large amount of solute (citric acid) compared to the solvent (water) significantly lowers the vapor pressure due to colligative properties.
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