A dry air is passed through a soution which contains 10 g of a solute dissoved in 90 g of water. Then the same air is passed through pure water. The depression in weight of the solution is by 2.5 g and in weight of pure solvent (i.e. water) is 0.05 g. The molar mass of the solute is :

A dry air is passed through a soution which contains 10 g of a solute

1.	A dry air is passed through a soution which contains 10 g of a solute dissoved in 90 g of water. Then the same air is passed through pure water. The depression in weight of the solution is by 2.5 g and in weight of pure solvent (i.e. water) is 0.05 g. The molar mass of the solute is :
Answer» 25 50 100 180 Solution :According to Raoult's Law `(P_(A)^(@)-P_(S))/P_(S)=("Lowring in WEIGHT of SOLVENT")/("Lowring in weight of solution")=((0.05g))/((2.5g))` According to Raoult's Law `(P_(A)^(@)-P_(S))/P_(S)=n_(B)/n_(A)=W_(B)/M_(B)xxM_(A)/W_(A)` `((0.05))/((2.5))=((10g))/((M_(2)))xx((18G mol^(-1)))/((90g))` `M_(2)=((10g)xx(18gmol^(-1))xx2.5)/((0.05)xx(90g))=100 g mol^(-1)`