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Grade 12th passPhysical Chemistry

A current of dry air was passed through a solution of 2.5g of a non volatile substance \X\ in 100g of water and then through water alone. The loss of weight of the former was 1.25g and that of the latter was 0.05g . Calculate 1.mole fraction of the solute in solution 2. molecular weight of the solute

Profile image of Ishtiyak Qadir
10 Years agoGrade 12th pass
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1 Answer

Profile image of Mehdi hasan
7 Years ago
Loss in weight of solution --ps = 1.25g
Loss of weight of solvent=p0-ps=0.05g
         P0=(p0-ps)+ps
Therefore,
                  P0=0.05+1.25  i,e., P0=1.30g
Hence p0-ps/p0=0.05/1.30=0.0385
i,e, x2=0.0385
Now,
         P0-ps/p0=n2/n1=w1/M2/w2/M1
0.0385=2.5/M2/100/18
Or         2.5/M2=100/18*0.0385
Or      M2=11.7