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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:\

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:\

Grade:12th Pass

1 Answers

Ashish Jangir
8 Points
5 years ago
Given,
PA = 10 torr {vapour pressure of pure A}
Decrease in vapour pressure = 1torr
so total vapour 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=??
put the values and u will get MB=90u
 

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