Question

(An aqueous solution of urea freezes at
-0.186°C, K, for water = 1.86 K kg mol-!, K, for
water = 0.512 K kg mol-1. The boiling point of
urea solution will be:
​

Answer 1

Answer:

373.0512K

Explanation:

since,

∆Tb = Kf × m

given that;

Kf = 1.86 Kkg/mol

∆Tf = 0 - ( -0.186)

= 0.186

so,

from above relation;

0.186 = 1.86 × m

m = 0.1

We know that;

∆Tb = Kb × m

∆Tb = 0.512 × 0.1

∆Tb = 0.0512 K

so, boiling point of urea solution is;

Tb = 373K + 0.0512K= 373.0512K