A 5 percent aqueous solution by mass of a non-volatile solute boils at 100.`15^@C`. Calculate the molar mass of the solute. `K_b = 0.52 K kg mol^(-1)`.

Explanation

The normal boiling point of water is `100^@C`.
`:.` The elevation in boiling point is `0.15^@C`.
A 5 percent solution means that in a 100 g solution 5 g of the solute is present. This implies that the solvent is 95 g.
If the molar mass of the solute is M, then the molality of the solute in the solution is,
`(5//M)/95xx1000,DeltaT_b=K_bm`
`0.15=0.52xx(5//m)/95xx1000`
`:.` M = 182.4 g/mol