What would be the molar mas of a compound if 6.21 g of it dissolved in 24.0 g of chloroform a solution that has a boiling point of `68.04^(@)C`? Given that the boiling point of pure chloroform is `61.7^(@)C` and `K_(b) for chloroform= 3.63^(@)C/m`.

Explanation

Elevation in boiling point ,
`Delta T_(b) = 68.04 - 61.7 = 6.34^(@)C`
Mass of substance `w_(B) = 6.21 g`
Mass of chloroform `w_(A) = 24.0 g`
`k_(b) = 3.63^(@)`C/m
`k_(b) = (Delta T_(b) xx M_(B) xx w_(A))/(w_(B) xx 1000)`
or `" "M_(B) = (k_(b) xx w_(B) xx 1000)/(Delta T_(b) xx w_(A))`
` = (3.63 xx 6.21 xx 1000)/(6.34 xx 24)`
` = 148.15 g mol^(-1)`
`Delta T_(b)` in `C^(@) = Delta T_(b)` in K , because it is difference of two temperatuers.