Question

Calculate the freezing point of the solution containing 0.52gm of glucose dissolved in 80.20gm of water. (C=12, H=1 , 0=16) KF value of water = 1.86k kg/mole​

Answer 1

Answer:

According to the problem:

Let us consider the molecular mass of glucose

C

6

H

12

O

6

=72+12+96

Implies that

=180gmol

−1

Hence the freezing point,

ΔT

f

=

M

B

×W

A

K

f

×W

B

×1000

=>

=

180×80.20

1.86×0.520×1000

=0.067