Question

Question No. 10
10 molal aqueous solution of an electrolyte XY is 100% tonised The boiling point of the solution is (K for HO=0.52kg/ mol
375 5 K
374 04 K
377 12 K
373 25 K​

Answer 1

I think there is correction in the question. I guess the original question says 1 molal solution instead of 10 molal, so here is the answer.

For this first we calculate the vant Hoff factor first.

We know for dissociation

α = i-1/n-1

Here n = 2

α = 1

So

1 = i-1/2-1

Therefore i = 2

We know that,

ΔT = iKm

= 2x0.52x1

= 1.04

So increase in boiling point is

373+1.04 = 374.04K

I hope this helps you.

Incase you still have doubt feel free to ask.