Question

2. 18 g of glucose is dissolved in 1 kg of water. At what temperature will solution boil. Kb for

Water is 0.52Kkgmol-1?​

Answer 1

Answer:

Refer to the above attachment ✌️✌️

Since water boils at 373.15 K at 1.013 bar pressure, therefore, the boiling point of solution will be 373.15 + 0.052 = 373.202 K.

I hope it helps u

Answer 2

Answer: 30gm of water is evaporated when the temperature is heated to 100.375k.

Explanation:

Total water containing sucrose solution = 50gm

The boiling temperature of water = 100.15-degree Celsius

Raise in the boiling temperature = 100.375 degree Celsius

Given Kb = 0.52kKgmol^-1​

We have 0.15K=Kb(n20.05kg) and 0.375K=Kb

We have 0.15K=Kb(n20.05kg) and 0.375K=Kb

Consequently, m1= (0.15K0.375K) (0.05Kg) =0.02kg

or

Consequently, m1= (0.15K0.375K) (0.05Kg) =0.02kg

Hence, the amount of water evaporated is 30gm.

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