Question

21.
Calculate the molal elevation constant if water evaporates at 100°C with the absorption of 536 calories de
gm (R = 2 cals).
(1)0.519K-kg/mol (2) 0.0519K-kg/mol (3)1.519K-kg/mol (4)2.519Kkg/mol​

Answer 1

0.529

Explanation:

T²b=373x373

lv=536

• kb=RT²b/lv×1000
• =2×373x373/536x1000=0.519

Answer 2

Answer:

0.519K-kg/mol

Explanation:

We know that the formulae of boiling temperature is given as (Tb)^2=373x373.

From the question we also know that the latent heat of vaporisation is given as lv=536 cal.

So, we know that the elevation in boiling point is directly proportional to the boiling temperature and inversely to the latent heat of vaporisation hence, the elevation in boiling point will be :-(kb)=R(Tb)^2/lv×1000

=2×373x373/536x1000.

Which on solving will be equal to the 0.519K-kg/mol.