Question

The vapour pressures of two volatile liquids A and B at 25°C are 50 Torr and 100 Torr, respectively. If the liquid mixture, contains 0.3 mole fraction of A, then the mole fraction of liquid B in the vapour phase is \(\frac {x}{17}\). The value of x is _____.

Answer 1

Correct Answer: 14

To solve the problem of determining the value of \( x \), which is the numerator of the mole fraction of liquid B in the vapour phase, we use Raoult's law and Dalton's law of partial pressures. According to Raoult's law, the partial pressure of a component in a mixture is its mole fraction in the liquid phase times its pure component vapour pressure. For components A and B, the partial pressures are:
P_A = x_A × P_{pure A} = 0.3 × 50 Torr = 15 Torr
P_B = x_B × P_{pure B} = (1 - 0.3) × 100 Torr = 0.7 × 100 Torr = 70 Torr
The total pressure P_total = P_A + P_B = 15 Torr + 70 Torr = 85 Torr.
Now, using Dalton's law, the mole fraction of B in the vapour phase (y_B) can be found as:
y_B = P_B / P_total = 70 Torr / 85 Torr = 14/17.
Given y_B = x/17, this implies x = 14, which falls within the expected range of 14 to 14, confirming our computation. Thus, the value of \( x \) is 14.