Calculate the boiling point of solution containing `0.456g` of camphor (molar mass =152 ) dissolved in `31.4g` of acetone (boiling point `=56.30^(@)C` ), if the molar elevation constant per 100g of acetone is `17.2^(@)C.`

Explanation

Here, we have `w_(2)=0.456 g, M_(2)=152, w_(1)=31.4 g, T_(0)=56.30^(@)C, K_(b)=17.2^(@)C//100 g`
`Delta T_(b)=(100 K_(b).w_(2))/(w_(1)M_(2))=(100xx17.2xx0.456)/(31.4xx152)=0.16^(@)C`
`therefore` Boiling point of solution `(T_(b))=T_(b)^(@)+Delta T_(b)=56.30+0.16 = 56.46^(@)C`