Question

The vapour pressure of pure 'A' is 10 Torr and at same temperature when 1 gm of 'B'is dissolved in 20 gm of 'A' it's vapour pressure is reduced to 9 torr . If the molecular weight of 'A' is 200 amu then the molecular mass of 'B' is​

Answer 1

Explanation:

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:

I'll mention vapour pressure of pure component as Po and vapour pressure of solution as P.

Now, here, PoA = 10 torr and Psolution = 9 torr

Let molecular mass of B is x.

Then according to relative lower of vapour pressure,

Po-P/Po = Mole fraction of B

=> 10-9/10 = (1/x)/[(20/200)+(1/x)]

=> 1/10 = 10x/[x(x+10)]

=> x^2+10x = 100x

=> x(x-90) = 0

Either, x=0 or x= 90

Since, molecular weight can't be zero, molecular weight of B is 90 amu