Question

If Kb for water is 0.52 Kkgmol^ -1 the boiling point of 0.2 m solution of a non-volatile solute will be
a) 371.96K
b) 373.104K
c) 373.52K
d) 374.OK​

Answer 1

Given: $K_b$ for water is 0.52 K.kgmol/mol and concentration is 0.2 m .

To find: We have to find out the boiling point of the solution.

Solution:

From the law of evaluation of boiling point we know that-

$\Delta T_b=K_b.m$

Where $\Delta T_b$ is the change boiling point.

Given that $K_b=0.52K.kg/mol$

m=0.2m

$\Delta T_b=0.52\times 0.2$

$\Delta T_b=.104$

Boiling point of the solution is 100+.104=100.104 degree celcius.

In kelvin the boiling point of the solution is 273+100.104=373.104K

So, boiling point of the solution is 373.104K.

The correct option is b.