tentuka titik beku larutan 300 gram urea {CO(NH2)}2 dalam 1000 gram air jika diketahui kf = 1,86 derajat C/m dimana Ar H=1 C=12 N=14 dan O=16

diketahui :
W urea = 300 gr
Mr Urea = 60
p air = 1000 Gr
Kf = 1,86

ditanya : Tf ?

jawab :
molalitas (m)
m = W urea/Mr urea . 1000/p
m = 300/60 . 1000/1000
m = 5 . 1
m = 5

∆Tf = Kf . m
∆Tf = 1,86 . 5
∆Tf = 9,3°C

Tf = Tf° - ∆Tf
Tf = 0° - 9,3°
Tf = -9,3°C