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

Answer

B

Explanation

`B.Pt (T_(b))=100+Delta T_(b)=100 +K_(b) m `
`F.Pt. (T_(f))=0-Delta T_(f)=-K_(f) m`
`T_(b)-T_(f)=(100+K_(b)m)-(-K_(f)m)`
105=100+0.51m +1.86 m =100 + 2.37 m`
`2.37 m = 105-100=5`
`:. M=5/2.37=2.11
:.` W_(2)` of sucrose to be dissolved in 100 g water
`W_(2)=(m xx M_(2) xx W_(1))/1000=(2.11 xx 342 xx 100)/1000 =72g`