Question

a solution of sucrose (molar mass 342 gram mole) is prepared by dissolving 68.4 gram of sucrose and thousand gram of water find the boiling point of solution​

Answer 1

A solution of sucrose (molar mass = 342 g mol

−1

) has been prepared by dissolving 68.5 g of sucrose in 1000 g of water. The freezing point of the solution obtained will be:

[K

f

for water =1.86 K kg mol

−1

]

-0.372C

DepressioninFreezingpointΔT

f

=

M

1

×W

2

W

1

×K

f

×1000

whereW

1

=Weight of Solute

W

2

=Weight of solvent

M

1

=Molar mass of solute

K

f

=Freezing point deprssion constant

Now,ΔT

f

=

342×1000

1.86×68.5×1000

=0.372C

Now,ΔT

f

=T

o

−T

f

So,T

f

=0−0.372=−0.372C.

(Freezing point of purewater=0

0

C.)