At $ 80^{\circ} \mathrm{C}, $ the vapour pressure of pure liquid $ A $ is 520 mmHg and that of pure liquid $ B $ is 1000 mmHg. If a mixture solution of $ A $ and $ B $ boils at $ 80^{\circ} $ and 1 atm pressure, the amount of $ A $ (mole percent) in the mixture is:

Question

Accepted Answer

$ 32 \% $. At the boiling point, the solution’s total vapour pressure equals the external pressure, 760 mmHg. For an ideal binary solution, Raoult’s law gives P = x_A P_A^0 + x_B P_B^0, so you can solve for x_A using x_B = 1 - x_A. The result is x_A = 0.32, i.e. 32% A.

At \( 80^{\circ} \mathrm{C}, \) the vapour pressure of pure liquid \( A \) is 520 mmHg and that of pure liquid \( B \) is 1000 mmHg. If a mixture solution of \( A \) and \( B \) boils at \( 80^{\circ} \) and 1 atm pressure, the amount of \( A \) (mole percent) in the mixture is:

Solution

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