Question

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))`

Answer 1

The given values are
`W_(solvent)=100 g`
`Mw_(solute)=342 g mol^(-1)`
`T_(b)-T_(f)=105.0^(@)C`
Using the formula, `DeltaT_(b) =K_(b)m` and `DeltaT_(f) =K_(f)m`
Boiling point of solution` (T_(b))=100+DeltaT_(b)`
=`100+K_(b)m`
Freezing point of solution` (T_(f))=0-DeltaT_(f)`
=`0-K_(f)m`
Subtracting Eq. (ii) from Eq.(i),
`T_(b)-T_(f) =(100+K_(b)m)-(-K_(f)m)`
`105=100+0.51m+1.86m`
`m=2.11`
Weight of sucrose to be dissolved in `100 g` =`(2.11xx342xx100)/(1000)=72.16 g`