In: Chemistry
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?
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