Calculate the difference between the boiling points of 0.2 m `Na_(2)SO_(4)` and 0.5 m glucose assuming complete dissociation of `Na_(2)SO_(4)`. (`K_(b)` for water ` = 0.52 K m^(-1)` ).

Answer

`0.052`

Explanation

`Delta T_(b)` (glucose) `= 0.52 xx 0.5 = 0.26`
`T_(b) = 100.26`
`Delta T_(b)(Na_(2)SO_(4)) = 3 xx 0.52 xx 0.2 = 0.312`
`T_(b) = 100.312 `
Differene in boiling point = 100.312 - 100.26
` = 0.052^(@)`