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.`

Options

A. `56.46^(@)C`
B. `36.56^(@)C`
C. `56.14^(@)C`
D. `72.52^(@)C`

Answer

A

Explanation

`"Given, "K_(b)=17.2^(@)C,w=0.456,w_(B)=31.4g" and "w_(A)=152`
We know that,
`DeltaT_(b)=(100+K_(b)xxw_(B))/(M_(B)xxw_(A))=(100xx17.2xx0.456)/(31.4xx152)=0.16^(@)C`
`:."Boiling point of solution, "T_(s)=T^(@)+DeltaT_(b)=56.30+0.16=56.46^(@)C`