Calculate the bolling point of one mole aqueous solution (density 1.06 g `cm^(-3)`) of KBr.
(Given : `K_(b)`, for `H_(2)O =0.52`K kg `mol^(-1)`, Atomic mass : K =39 , Br = 80 ]

Explanation

One molar aqueous solution Boiling point = 7 d = 1.06 g/ml assuming volume 1 L of KBr. Weight of KBr = 1.06 g
`DeltaK_(b)=0.52` K kg /mol .
`DeltaT_(b)=iK_(b)` m weight of solvent 1 kg
`DeltaT_(b) = T_(2)-T_(1)`
where `T_(1)= 373 K`
`DeltaT_(b) = iK_(b) xx(W_(B)xx1000)/(M_(B)W_(A))`
`DeltaT_(b)=2xx0.52 xx(1.06xx1000)/(119xx1)`
`DeltaT_(b)=(1.1024xx1000)/(119)`
=9.26
`T_(2)-373=9.26`
`=382.26 K` .