A current of dry at was passed through a solution of 2.5 g of a non-volatile substance 'X' in 100 g of w ater and then through water along. The loss of weight of the former was 1.25 g and that of the latter was 0.05 g. Calculate (i) mole fraction of the solute in the solution (ii) molecular weight of the solute.

A current of dry at was passed through a solution of 2.5 g of a non-vo

1.	A current of dry at was passed through a solution of 2.5 g of a non-volatile substance 'X' in 100 g of w ater and then through water along. The loss of weight of the former was 1.25 g and that of the latter was 0.05 g. Calculate (i) mole fraction of the solute in the solution (ii) molecular weight of the solute.
Answer» SOLUTION :Loss in weight of solution `prop p_(s) prop 1.25g`, Loss in weight of solvent (WATER) `prop p^(@)-p_(s) prop 0.05g` `p^(@)=(p_(0)-p_(s))+p_(s) therefore""p^(@) prop 0.05 +1.25, i.e., p^(@) prop 1.30g` Hence, `""(p^(@)-p_(s))/(p^(@))=(0.05)/(1.30)=0.0385, i.e, x_(2)=0.0385` `"Now,"(p^(@)-p_(s))/(p^(@))=(n_(2))/(n_(1))=(w_(1)//M_(2))/(w_(1)//M_(1)) ,` `0.0385=(2.5//M_(2))/(100//18) or (2.5)/(M_(2))=(100)/(18)xx0.0385 or M_(2)=11.7`