Question

Liquids A and B form an ideal solution. The vapor pressure of pure A is 0.700 atm at the normal boiling point of a solution prepared from 0.250 mole of B and 0.650 mole of A. What is the vapor pressure of pure B at this temperature

Answer 1

The vapor pressure of pure B is 1.77atm

Given:-

Vapor pressure of Pure A = 0.700atm

Mole of A = 0.650mole

Mole of B = 0.250mole

Dalton's law of partial pressures states that the total pressure exerted by a gas mixture is equal to the sum of the partial pressures of all of the constituent gases.

So, Ptotal =  $P_A + P_B$

According to Raoult's law, a solvent's partial vapor pressure in a solution (or mixture) is equal to or the same as the pure solvent's vapor pressure times the mole fraction present in the solution.

So, Ptotal =  $X_AP_A + X_BP_B$

Ptotal = 1atm

1atm = (0.650/0.650+0.250 X 0.700atm) + 0.250/0.650+0.250 X $P_B$

1atm = 0.722 X 0.700atm + 0.278 X $P_B$

1atm = 0.506atm + 0.278 X $P_B$

1atm - 0.506atm =  0.278 X $P_B$

0.494atm = 0.278 X $P_B$

$P_B$ =  0.494atm / 0.278

$P_B$ = 1.77atm

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