Two liquids 'A' and 'B' form an ideal solution. At 300 K, the vapour pressure of a solution containing 1 mole of 'A' and 3 moles of 'B' is 550 mm Hg. At the same temperature, if one more mole of 'B' is added to the solution, the vapour pressure of solution increases to 560 mm Hg. Then the ratio of vapour pressures of A and B in their pure state is
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For problems involving ideal solutions and Raoult's Law with two different compositions, you will always get a system of two linear equations. Set up the equations $P_{T1} = X_{A1}P_A^\circ + X_{B1}P_B^\circ$ and $P_{T2} = X_{A2}P_A^\circ + X_{B2}P_B^\circ$ and solve for the pure component vapour pressures.
Let the vapour pressure of pure liquid A be $P_A^\circ$ and that of pure liquid B be $P_B^\circ$.
According to Raoult's law for an ideal solution, the total vapour pressure $P_T$ is given by:
$P_T = X_A P_A^\circ + X_B P_B^\circ$, where $X_A$ and $X_B$ are the mole fractions.
Case 1: Solution with 1 mole of A and 3 moles of B.
Total moles = $1+3=4$.
Mole fraction of A, $X_A = 1/4$.
Mole fraction of B, $X_B = 3/4$.
Total pressure $P_{T1} = 550$ mm Hg.
So, $550 = \frac{1}{4}P_A^\circ + \frac{3}{4}P_B^\circ \implies 2200 = P_A^\circ + 3P_B^\circ$. (Eq. 1)
Case 2: One more mole of B is added. The solution now has 1 mole of A and 4 moles of B.
Total moles = $1+4=5$.
Mole fraction of A, $X_A = 1/5$.
Mole fraction of B, $X_B = 4/5$.
Total pressure $P_{T2} = 560$ mm Hg.
So, $560 = \frac{1}{5}P_A^\circ + \frac{4}{5}P_B^\circ \implies 2800 = P_A^\circ + 4P_B^\circ$. (Eq. 2)
Now we solve the system of two linear equations for $P_A^\circ$ and $P_B^\circ$.
Subtract Eq. 1 from Eq. 2:
$(P_A^\circ + 4P_B^\circ) - (P_A^\circ + 3P_B^\circ) = 2800 - 2200$.
$P_B^\circ = 600$ mm Hg.
Substitute $P_B^\circ = 600$ into Eq. 1:
$2200 = P_A^\circ + 3(600) \implies 2200 = P_A^\circ + 1800 \implies P_A^\circ = 400$ mm Hg.
The question asks for the ratio of vapour pressures of A and B, which is $P_A^\circ : P_B^\circ$.
Ratio = $400 : 600 = 4:6 = 2:3$.
