Freezing point of urea solution is `-0.06^(@)C`. How much urea (M.W. = 60 g/mole) is required to dissolve in 3 kg water `(K_(f) = 1.5^(@)Ckg mol^(-1))`

Options

A. 3.6 g
B. 2.4 g
C. 7.2 g
D. 6.0 g

Answer

C

Explanation

`Delta T_(f)=K_(f)xx(1000xxW_(A))/(M_(A)xxW_(B))`
[where `W_(A)` = weight of solute, `W_(B)` = weight of solvent, `M_(A)` = molecular weight of solute]
or, `0-(-0.6)=(1.5xx1000xxW_(A))/(60xx3000)`
or, `W_(A)=(60xx3000xx0.6)/(1.5xx1000)=72g`
If we consider the mass of water given to 300 gram `0.6^(@)C=(1000g kg^(-1)xx1.5^(@)Ckg mol^(-1)xx " weight of urea")/(60g mol^(-1)xx300 g)`
`therefore` Weight of urea = 7.2 g.