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
-
Molecular weight of citric acid (C₆H₈O₇): \[ \text{Molecular weight} = 6 \times 12.01 + 8 \times 1.008 + 7 \times 16.00 \approx 192.13 , \text{g/mol} \]
-
Calculate moles of citric acid: \[ 5 , \text{tonnes} = 5,000 , \text{kg} = 5,000,000 , \text{g} \] \[ \text{Moles of citric acid} = \frac{5,000,000 , \text{g}}{192.13 , \text{g/mol}} \approx 26041.56 , \text{mol} \]
-
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.