Question

at 90 degrees a solution of n proponal(X2=0.259 and P2=577.2torr) and water (P1=527.76) has a vapor pressure of 820.3torr, the vapor phase is 39.7% by mass n-proponal

calculate the vapor pressures, activities,activity coefficient for each component

calculate the vapor pressure of this solution using raoults law

Answer 1

It is always true that the sum of mole fractions of all the components in a solution = 1

i.e. X1 + X2 = 1, where X1 and X2 are the mole fractions of water and n-propanol in the liquid phase.

i.e. X1 + 0.259 = 1

i.e. X1 = 1-0.259 = 0.741

I hope that the given values are vapor pressures in their pure state.

i.e. P1^o = 527.76 torr and P2^o = 577.2 torr

The vapor pressure of solution (P) = 820.3 torr

Part 1:

Now, The vapor pressure of water (P1) = Y1*P and the vapor pressure of n-propanol (P2) = Y2*P

Where Y1 and Y2 are the mole fractions of water and n-propanol in the vapor phase.

i.e. P1 = (1-0.397)*820.3 torr = 494.6409 torr

And P2 = 0.397*820.3 torr = 325.6591 torr

Now, activity of water (a1) = P1/P1^o = 494.6409/527.76 = 0.937

And the activity of n-propanol (a2) = P2/P2^o = 325.6591/577.2 = 0.564

Now, the activity coefficient of water (1) = a1/X1 = 0.937/0.741 = 1.26

And the activity coefficient of n-propanol (2) = a2/X2 = 0.564/0.259 = 2.18

Part 2:

According to Raoult's law, P1 = X1*P1^o and P2 = X2*P2^o, where P1 and P2 are the partial pressures of water and n-propanol, respectively in the solution.

Therefore, P1 = 0.741*527.76 torr = 391.07016 torr

And P2 = 0.259*577.2 torr = 149.4948 tor

The vapor pressure of solution (P) = P1 + P2

i.e. P = (391.07016 + 149.4948 ) torr = 540.56496 tor