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Step-by-Step Solution
Step 1: Identify the Given Data
(i) Vapour pressure of pure component A at 25 ^\circ C, P_A^0 = 90 \text{ mm Hg} .
(ii) Vapour pressure of pure component B at 25 ^\circ C, P_B^0 = 15 \text{ mm Hg} .
(iii) Mole fraction of A in the liquid mixture, x_A = 0.6 .
(iv) Mole fraction of B in the liquid mixture, x_B = 1 - x_A = 0.4 .
We need to find the mole fraction of B in the vapour phase, y_B , given in the form
y_B = x \times 10^{-1} , and determine the value of x .
Step 2: Calculate the Total Vapour Pressure using Raoult's Law
Raoult’s law states that for an ideal solution, the total vapour pressure P_T is given by:
P_T = x_A \, P_A^0 + x_B \, P_B^0
Substituting the given values:
P_T = (0.6 \times 90) + (0.4 \times 15)
P_T = 54 + 6 = 60 \text{ mm Hg}.
Step 3: Relate Liquid Phase Mole Fraction with Vapour Phase Mole Fraction
The partial pressure of each component is also given by:
P_A = x_A \, P_A^0, \quad P_B = x_B \, P_B^0.
In the vapour phase, the mole fraction of A, y_A , and that of B, y_B , satisfy:
P_A = y_A \, P_T, \quad P_B = y_B \, P_T.
For A:
x_A \, P_A^0 = y_A \, P_T.
Substituting the known values:
0.6 \times 90 = y_A \times 60.
54 = 60 \, y_A \quad \Rightarrow \quad y_A = \frac{54}{60} = 0.9.
Hence, the mole fraction of B in the vapour phase is:
y_B = 1 - y_A = 1 - 0.9 = 0.1.
Step 4: Express y_B in the Requested Form and Find x
We want y_B in the form x \times 10^{-1} . We have:
y_B = 0.1 = 1 \times 10^{-1}.
So, the value of x is 1.
Final Answer
The nearest integer value of x is 1.
