Question

The molal elevation constant for water is 0.52 K. `"molality"^(-)`. The elevation caused in the boiling point of water by dissolving 0.25 mole of a non volatile solute in 250 g of water will be :
A. `52^(@)`
B. `5.2 ^(@)C`
C. `0.52^(@)C`
D. `0.052 ^(@)C`

Answer 1

Correct Answer - C
`(Delta T_(f)=1000 xx K_(f))/W_(1) xx W_(2)/M_(2)=(1000 xx K_(f))/W_(1)xx n`
`Delta T_(f) = (1000 xx 0.52)/250 xx 0.25 =0.52^(@)C`