1.

A solution of urea (mol. Mass 56 g mol ^(-1)) boils at 100.18^(@)C at the atmospheric pressure. If K_(f) and K_(b) for water are 1.86 and 0.512 K kg mol ^(-1) respectively, the above solution will freeze at

Answer»

`0.654^(@)C`
`-0.654^(@)C`
`6.54^(@)C`
`-6.54^(@)C`

SOLUTION :As`Delta T_(f) =K_(f)` m
`Delta T_(b) =K_(b).m`
Hence, we have `m =(Delta T_(f))/(K _(f)) =(Delta T_(b))/(K_(b))`
or `Delta T _(f) = Delta T_(b) = (K_(f))/(K _(b))`
`implies [Delta T _(b) =100. 18-100= 0.18^(@)C]`
`= 0.18xx (1.86)/(0.512) =0.654^(@)C`
As the Frecezing POINT of pure WATER is `0^(@)C, Delta T _(f) =0-T_(f)`
`0.654 =-0T_(f)`
`therefore T _(f) =-0.654`
Thus the FREEZING point of solution will be `- 0.654 ^(@)C.`


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