The freezing point of aqueous solution that contains `5%` by mass urea. `1.0%` by mass `KCl` and `10%` by mass of glucose is:
`(K_(f) H_(2)O = 1.86 K "molality"^(-1))`

Options

A. 290.2 K
B. 285.5 K
C. 269 93 K
D. 250 K

Answer

C

Explanation

`Delta T_(f) = Delta T_(f) ("glucose") + Delta T_(f) (KCl) +Delta T_(f) ("urea")`
But `Delta T_(f)=(K_(f) xx W_(2))/(M_(2) xx W_(2)) xx 1000`
`:. (1000 xx1.86 xx10)/(100 xx 180) +(1000 xx 1.86 xx 1 xx 2)/(74.5 xx 100)+(1000+1.86 xx5)/(100 xx 60)=3.069`
f.pt. =273 - 3.069 = 269.93 K