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Two liquids A and B form ideal solution....

Two liquids A and B form ideal solution. At `300 K`, the vapour pressure of a solution containing 1 mole of A and 3 moles of B is `550 mm` of Hg. At the same temperature, if one more mole of B is added to this solution, the vapour pressure of the solution increases by `10 mm` of Hg. Determine the vapour pressure of a and B in their pure states.

A

`p_A^@`=600 mm Hg and `p_B^@`=400 mm Hg

B

`p_A^@`=550 mm Hg and `p_B^@`=560 mm Hg

C

`p_A^@`=450 mm Hg and `p_B^@`=650 mm Hg

D

`p_A^@`=400 mm Hg and `p_B^@`=600 mm Hg

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To solve the problem, we will use Raoult's Law, which states that the vapor pressure of a solution is equal to the sum of the partial pressures of each component in the solution. The partial pressure of each component can be calculated using the formula: \[ P_i = P_{i}^{0} \cdot x_i \] where: - \( P_i \) is the partial pressure of component \( i \), - \( P_{i}^{0} \) is the vapor pressure of the pure component \( i \), - \( x_i \) is the mole fraction of component \( i \) in the solution. ### Step 1: Determine the initial conditions We have: - Moles of A = 1 - Moles of B = 3 - Total moles = 1 + 3 = 4 - Vapor pressure of the solution = 550 mm Hg ### Step 2: Calculate mole fractions - Mole fraction of A, \( x_A = \frac{1}{4} \) - Mole fraction of B, \( x_B = \frac{3}{4} \) ### Step 3: Apply Raoult's Law for the initial solution Using Raoult's Law, we can express the total vapor pressure as: \[ P_{total} = P_{A}^{0} \cdot x_A + P_{B}^{0} \cdot x_B \] Substituting the known values: \[ 550 = P_{A}^{0} \cdot \frac{1}{4} + P_{B}^{0} \cdot \frac{3}{4} \] ### Step 4: Rearranging the equation Multiplying the entire equation by 4 to eliminate the fractions: \[ 2200 = P_{A}^{0} + 3P_{B}^{0} \tag{1} \] ### Step 5: Determine the conditions after adding one mole of B When one more mole of B is added: - New moles of A = 1 - New moles of B = 4 - New total moles = 1 + 4 = 5 - New vapor pressure = 550 mm Hg + 10 mm Hg = 560 mm Hg ### Step 6: Calculate new mole fractions - New mole fraction of A, \( x_A = \frac{1}{5} \) - New mole fraction of B, \( x_B = \frac{4}{5} \) ### Step 7: Apply Raoult's Law for the new solution Using Raoult's Law again: \[ 560 = P_{A}^{0} \cdot \frac{1}{5} + P_{B}^{0} \cdot \frac{4}{5} \] ### Step 8: Rearranging the equation Multiplying the entire equation by 5: \[ 2800 = P_{A}^{0} + 4P_{B}^{0} \tag{2} \] ### Step 9: Solve the system of equations Now we have two equations: 1. \( P_{A}^{0} + 3P_{B}^{0} = 2200 \) 2. \( P_{A}^{0} + 4P_{B}^{0} = 2800 \) Subtract equation (1) from equation (2): \[ (P_{A}^{0} + 4P_{B}^{0}) - (P_{A}^{0} + 3P_{B}^{0}) = 2800 - 2200 \] This simplifies to: \[ P_{B}^{0} = 600 \text{ mm Hg} \] ### Step 10: Substitute back to find \( P_{A}^{0} \) Substituting \( P_{B}^{0} \) back into equation (1): \[ P_{A}^{0} + 3(600) = 2200 \] \[ P_{A}^{0} + 1800 = 2200 \] \[ P_{A}^{0} = 400 \text{ mm Hg} \] ### Final Answer The vapor pressures of the pure components are: - \( P_{A}^{0} = 400 \) mm Hg - \( P_{B}^{0} = 600 \) mm Hg
To solve the problem, we will use Raoult's Law, which states that the vapor pressure of a solution is equal to the sum of the partial pressures of each component in the solution. The partial pressure of each component can be calculated using the formula: \[ P_i = P_{i}^{0} \cdot x_i \] where: - \( P_i \) is the partial pressure of component \( i \), - \( P_{i}^{0} \) is the vapor pressure of the pure component \( i \), - \( x_i \) is the mole fraction of component \( i \) in the solution. ...
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