An aqueous solution of `2` per cent `(wt.//wt)` non-volatile solute exerts a pressure of `1.004` bar at the boiling point of the solvent. What is the molecular mass of the solute?

Explanation

According to Raoult's Law.
`(P_(A)^(@)-P_(S))/P_(S)=n_(B)/n_(A)=W_(B)/M_(B)xxM_(A)/W_(A)`
`P_(A)^(@)"(for water)"=1=1.013 "bar" , P_(S)=1.004"bar", W_(B)=100-2=98g, M_(A)=18 g mol^(-1)`
`((1.013-1.004)"bar")/((1.004"bar"))=((2g)xx(18g mol^(-01)))/(M_(B)xx98g)`
`M_(B)=((2g)xx(18g mol^(-1))xx(1.004 "bar"))/((0.009 "bar")xx(98g))=41.0 g mol^(-1)`