Question

Boiling point of water is 373.11K. What is Kb of water, if 0.15 molal aqueous
Solution of substance boils at 373.20K?​

Answer 1

Answer:Kb= 0.6 k kg / mol

Explanation:
As per the given data in the question,
We have to calculate Kb
As we know,
KB is the molal elevation of water .
In the question we have given,
Molal aqueous solution,m = 0.15 m
Boiling point of the solution, Tb = 373.20 K
Boiling point of water, Tb^0 = 373.11 K
To find Kb of water,
We know that the formula to find Kb is
Boiling point of the solution -Boiling point of water, Tb^0 is equal to Kb * m
That is,
Tb - Tb^0 = Kb *m
So, the molal elevation of the water is,
KB= (Tb -Tb^0)/m
= (373.20-373.11)/.15
= 0.6 k kg / mol
So the molal elevation of the water Kb will be equal to 0.6 k kg / mol

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