A liquid mixture had 30 moles of A and 70 moles of B. The vapor pressure...

A liquid mixture had 30 moles of A and 70 moles of B. The vapor pressure of pure A is 790 torr and pure B is 630 torr. What is the pressure at which the mixture boils?

Solutions

Expert Solution

A = moles of A/(moles of A+moles of B) = 30/(30+70) = 0.3

B = moles of B/(moles of A+moles of B) = 70/(30+70) = 0.7

vapor pressure of A in solution = A X pressure

= 0.3 X 790 = 237 torr

vapor pressure of B in solution = B X pressure

= 0.7 X 630 = 441 torr

Pressure of the mixture = 237 + 441 = 678 torr

queen_honey_blossom answered 1 year ago

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