How many grams of sucrose (molecular weight 342) should be dissolved in `100 g` water in order to produce a solution with `105^(@)C` difference between the freezing point and the boiling point ? `(K_(b) =0.51^(@)C m^(-1) , (K_(f) =1.86^(@)C m^(-1))`

Options

A. `34.2g`
B. `72g`
C. `342g`
D. `460g`

Answer

B

Explanation

`"Boiling point "(T_(b))=100+DeltaT_(b)=100+k_(b)m`
`"Freezing point "(T_(f))=0-DeltaT_(f)=-k_(f)m`
`T_(b)-T_(f)=(100+k_(b)m)-(-k_(f)m)`
`105=100+0.51m+1.86m`
`2.37m=5" or m"=(5)/(2.37)=2.11`
`:."Weight of sucrose to be dissolved in 100g water"`
`=(2.11xx342)/(1000)xx100=72g`