Question

What is the boiling point of a 0.195 aqueous solution k2s?

Answer 1

Answer:- Boiling point of the given solution is 100.3 degree C.

Solution:- The equation for solving this type of boiling point elevation problems is, delta Tb = i x m x kb

Where, delta Tb is the elevation in boiling point

i is Van't hoff factor and it's theoretical value is the number of ions an ionic compound gives. For example, NaCl gives total two ions(one sodium and one chloride ion).

m is molality and kb is the molal elevation constant. For water, kb is 0.512.

$K_2S \rightarrow 2K^+ + S^2^-$

It gives two potassium ions and one sulfide ion. So, i = 2+1 = 3

let's plug in the values in the equation.

delta Tb = 3 x 0.195 x 0.512

delta Tb = 0.3

pure water boils at 100 degree C. So, the boiling point of the solution = 100 + 0.3 = 100.3 degree C