At  1000C   the vapour pressure of a solution of  6.5g of a solute in 100g water is 732mm. If kb=0.52 the boiling point of this solution will be

Given :WB=6.5 g,WA=100 g,ps=732 mm,Kb=0.52Tb°=100°C,p°=760 mmp°-psp°=n2n1⇒760-732760=n2100/18⇒n2=28×100760×18=0.2046 moles ΔTb=Kb×mTb-Tb°=Kb×n2×1000WA(g)Tb-100°C=0.52×0.2046×1000100=1.06Tb=100+1.06=101.06°C