The difference between the boiling point and freezing point of an aqueous solution containing sucrose (molecular mass `= 342 g mol^(-1))` in 100 g of water is `105.04`. If `K_(f)` and `K_(b)` of water are 1.86 and `0.51 Kg mol^(-1)` respectively, the weight of sucrose in the solution is about

Options

A. 34.2g
B. 342 g
C. 7.2 g
D. 72g

Answer

D

Explanation

Let w be the mass of sucrose dissolved in 100 g of water .
Molality ,`m = (w)/(342) xx (1000)/(100)`
` = 0.0292 w`
`Delta T _(f) = k_(f) xx m`
` = 1.86 xx 0.0292 w`
`Delta T_(b) = k_(b) xx m`
` = 0.51 xx 0.0292w`
`T_(f) = 0-1.86 xx 0.0292 w`
`T_(b) = 100+0.51 xx 0.0292 w`
`T_(b) - T_(f) = 100 +0.51 xx 0.0292 w`
` - 0+1.86 xx 0.0292 w`
` = 100 + 0.0692 w`
` or " " 105 = 100 +0.0692 w`
`0.062 w = 5`
` w = (5)/(0.0692)`
`= 72.2 ~~ 72 g`