Question

The vapour pressure of a pure liquid A is 10 torr at the same temperature when,1 g of B is dissolved in 20 g of A,the vapour pressure is reduced by 1 torr. If the mol. wt.of A is 200 u then the molecular weight of B is:\

Answer 1

Given,

PA = 10 torr {vapour pressure of pure A}

Decrease in vapour pressure = 1torr

Thus total vapor pressure after adding B is 9 torr and B is non volatile substance therefore PB=0

then according to Dalton’s law of partial pressure-

PT = XA.PA + XB.PB

9 = {(mA/MA) / (mA/MA + mB/MB)}.10 + 0 [0 because PB = 0]

mA(mass of A)=20gm, MA=200, mB=1gm , MB=??

Adding the values we get

MB=90u

Answer 2

Answer:

Explanation:

P1°=10 torr

Ps= 9torr