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))`

Explanation

`DeltaT_(b)=K_(b)xxm`
`"B. pt of solution "(T_(b))=100+DeltaT_(b)=100+K_(b)m`
`"F. pt of solution "(T_(b))=0-DeltaT_(f)=0-K_(f)m, T_(b)-T_(f)=(100+K_(b)m)-(-K_(f)m)`
`"F. pt. of solution "(T_(f))= 0-DeltaT_(f)=0-K_(f)m," "T_(b)-T_(f)=(100+K_(b)m)-(K_(f)m)`
`" "105=100+0.51xxm+1.86m.`
This gives m = 2.11, i.e., 1000 g of water contain sucrose = 2.11 mole `=2.11xx342 g`
`therefore" Mass of sucrose to be dissolved in 100 g of water "=(2.1xx342)/(1000)xx100=72g.`