1.

What weight of sucrose will be dissolved in 250 g of water and obtain solution with boiling point at 100.2 degree Celsius

Answer»

Answer:

hope it will help you...

Explanation:

Let the molality of the solution is m.

ΔT

b

=

K

f

m

ΔT

f

= K

f

× m

ΔT

f

+ΔT

b

=

(K

f

+

K

b

m=

(1.86+

0.51)×

m

=

2.37×

m

Hence

m=

5/2.37=

2.11

and

1000

2.11×100

=

0.211

gram of sucrose MUST be DISSOLVED in 100 gram of water.

Required MASS of sucrose

=

0.211mol×

342g/mole=

72.2g



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