What would be the molar mass of a compound if I 6.21 g of it dissolved in 24 .0 g of chloroform form a solution that has a bolling point of `68.04""^(@)C.` The bolling point of pure chloroform is `61.7^(@)C` and the boiling point elevation constant, `K_(b)` for chloroform is `3.63^(@)`C/m.

Explanation

`DeltaT_(b)=K_(b)xx(W_(B))/(M_(B))xx(1000)/(W_(A))`
`T_(b)-T_(b)^(0)=(3.63xx6.21xx1000)/(M_(B)xx24)`
`61.7-68.04 = (3.63xx6210)/(M_(B)xx24)`
`M_(b)=(3.63xx6210)/(24xx6.34)`
`=(22542.3)/(152.16)`
`=148.148g//"mol"^(-1)`