Question

A two-phase ($\alpha + \beta$) mixture of an A–B binary system has the following properties:

[(i)] Phase $\alpha$ has equal weight percentages of A and B.
[(ii)] Phase $\beta$ has twice the mole fraction of A compared to B.
[(iii)] The two-phase mixture has equal amounts of $\alpha$ and $\beta$.
[(iv)] Atomic mass of A is twice that of B.
The mole fraction of A in the resultant two-phase mixture is _________. (Round off to one decimal)

Hint:

For equal phase fractions, the overall mole fraction is the average of the mole fractions of the phases.

Answer 1

Correct Answer: 0.5

Let the atomic mass of B = 1 unit. Then the atomic mass of A = 2 units. Phase $\alpha$: Equal weights of A and B. Take 1 g of A and 1 g of B. \[ \text{moles of A} = \frac{1}{2} = 0.5,\qquad \text{moles of B} = 1 \] Total moles = 1.5 \[ x_A^\alpha = \frac{0.5}{1.5} = 0.333,\qquad x_B^\alpha = 0.667 \] Phase $\beta$: Mole fraction of A is twice that of B. Let: \[ x_B^\beta = p,\qquad x_A^\beta = 2p \] Since: \[ 2p + p = 1 \] \[ p = \frac{1}{3} \] Thus: \[ x_A^\beta = \frac{2}{3},\qquad x_B^\beta = \frac{1}{3} \] Given: Equal amounts of $\alpha$ and $\beta$ \[ x_A = \frac{x_A^\alpha + x_A^\beta}{2} \] \[ x_A = \frac{0.333 + 0.667}{2} = 0.5 \] \[ \boxed{0.5} \]