An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute ?

Explanation

Applying Raoult.s law of dilute solution
`(p^0 - p_s)/(p^0) = (n_2)/(n_1+n_2) = n_2/n_1 = (w_2// M_2)/(w_1//M_1)= w_2/M_2 xx M_1/w_1`
Substituting the values, we have
`((1.013 - 1.004)"bar")/(1.013 "bar") = (2g)/(M_2) xx (18 g "mol"^(-1))/(98 g)`
`M_2 = (2xx 18)/(98) xx (1.013)/(0.009) g "mol"^(-1) =41.g "mol"^(-1)`