What is the boiling point of an aqueous solution of LiBr that is 0.527*mol*kg^-1 with respect to the salt?

Aug 9, 2017

This site lists ${K}_{B}$, the ebullioscopic constant of water, as 0.512*""^@C*kg*mol^-1.....

Explanation:

We know that in aqueous solution the lithium bromide salt speciates to its constituent ions.........

$L i B r \left(s\right) \stackrel{{H}_{2} O}{\rightarrow} L {i}^{+} + B {r}^{-}$

And $\Delta {T}_{\text{boiling point elevation"=ixx"molal conc.}} \times {K}_{B}$

And $i = 2$ because of the speciation of the ionic solute......

DeltaT=2xx0.527*mol*kg^-1xx0.512*kg*mol^-1*""^@C

$= 0.540$ ""^@C

And thus the observed boiling point of the solution should be $100.540$ ""^@C.