Two liquids A and B form an ideal solution at temperature T K. At T K, the vapour pressures of pure A and B are 55 and 15 kN m^ -2 respectively. What is the mole fraction of A in solution of A and B in equilibrium with

Question

Quantrex Academy · Accepted Answer

Correct answer: C. 0.5217

For an ideal solution with Raoult's law, the partial pressure of A in vapour is

P_A = P_A^0 y_A

where

y_A = 0.8

is the mole fraction in vapour phase. Thus

P_A = 55 \times 0.8 = 44

kNm

^{-2}

.
Using the equilibrium condition

y_A = \frac{P_A^0 \chi_A}{P_{total}}

where

\chi_A

is mole fraction in liquid and

P_{total} = P_A^0 \chi_A + P_B^0(1-\chi_A)

:

0.8 = \frac{55\chi_A}{55\chi_A + 15(1-\chi_A)}

0.8(55\chi_A + 15 - 15\chi_A) = 55\chi_A

44\chi_A + 12 - 12\chi_A = 55\chi_A

12 = 23\chi_A

\chi_A = 0.5217

Two liquids A and B form an ideal solution at temperature T K. At T...

Options

Correct answer

Step-by-step solution

More from Solutions