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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))

Answer»

72 gm
34.2 gm
342 gm
460 gm

Solution :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.


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