Q.Explain with diagram the boiling point elevation in terms of vapour pressure lowering.

Boiling point elevation is the rise in a solution's boiling point above the pure solvent's boiling point, caused by dissolving a nonvolatile solute: $\Delta T_b=T_b-T_b^0$. It follows directly from vapour pressure lowering -- since the solution's vapour pressure is lower than the pure solvent's at every temperature, the solution's vapour-pressure-vs-temperature curve sits below the solvent's curve, so a HIGHER temperature is needed for the solution's curve to reach the 760 mm (1 atm) line where boiling occurs. Experimentally, $\Delta T_b$ is directly proportional to molality: $\Delta T_b=K_bm$, where $K_b$ (the ebullioscopic or molal elevation constant) is a property of the solvent alone, numerically equal to the elevation a 1-molal solution would produce. This becomes a molar-mass-determination formula: substituting molality's definition ($m=1000W_2/(M_2W_1)$) gives $M_2=\dfrac{1000K_bW_2}{\Delta T_bW_1}$ …