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.
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Determine the moles of citric acid and water:
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Molar mass of citric acid (C₆H₈O₇): \[ \text{Molar mass} = 6 \times 12.01 , (\text{C}) + 8 \times 1.01 , (\text{H}) + 7 \times 16.00 , (\text{O}) = 192.13 , \text{g/mol} \]
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Mass of citric acid: \[ 5 , \text{tonnes} = 5,000 , \text{kg} = 5,000,000 , \text{g} \]
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Moles of citric acid: \[ \text{Moles of citric acid} = \frac{5,000,000 , \text{g}}{192.13 , \text{g/mol}} \approx 26040.6 , \text{mol} \]
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Assuming the density of water is approximately 1 g/mL: \[ \text{Mass of water} = 20 , \text{L} \times 1000 , \text{g/L} = 20,000 , \text{g} \]
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Moles of water: \[ \text{Moles of water} = \frac{20,000 , \text{g}}{18.02 , \text{g/mol}} \approx 1105.56 , \text{mol} \]
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Calculate the total moles in the solution: \[ \text{Total moles} = \text{Moles of citric acid} + \text{Moles of water} \approx 26040.6 , \text{mol} + 1105.56 , \text{mol} \approx 27146.16 , \text{mol} \]
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Calculate the mole fraction of water (\(X_{\text{water}}\)): \[ X_{\text{water}} = \frac{\text{Moles of water}}{\text{Total moles}} = \frac{1105.56 , \text{mol}}{27146.16 , \text{mol}} \approx 0.0407 \]
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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.