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Q.

 A Current of dry air was first passed through the bulb containing solution of 'A' in water and then through the bulb containing pure water. The loss in mass of a solution bulb is 1.92g gm. Where as that in pure water bulb is 0.08g, then mole fraction of 'A' is

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a

0.96

b

0.04

c

0.2

d

0.86

answer is D.

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Detailed Solution

We have to find XA.
loss in mass of solution bulb (P)        = 1.92  loss in mass of solvent bulb(P'-P ) = 0.08

P'=P'-P+P=0.08+1.92=2
(P'-P)/P'=XA XA= 0.082=0.04.

Hence, the correct option (D).0.04 

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