To produce difference between freezing point and boiling point of a solution by `105.0^(@)C`, how much sucrose should be dissolved in 100 gm of water ? `(K_(f)=1.86^(@)C, kg mol^(-1) " and " K_(b)=0.151^(@)Ckg mol^(-1))`

Options

A. 72 gm
B. 34.2 gm
C. 342 gm
D. 460 gm

Answer

A

Explanation

Boiling point `(T_(b))=100+Delta T_(b)`
`= 100+K_(b).m`
Freezing point `(T_(f))=0-Delta T_(f)=K_(f).m`
`therefore T_(b)-T_(f)=(100+K_(b).m)-(-K_(f).m)`
`105=100+0.151 m+1.86 m`
`2.37 m =5`
`m = (5)/(2.37)=2.11`
So, mass of sucrose to dissolve in 100 gm water is,
`= (2.11xx342xx100)/(1000)=72` gm.