Question

the entropy of vaporization of a liquids the molar mass 70 gram mole is 64 joule Kelvin mole if the boiling point of the liquid is 400 Kelvin then the amount of heat required to vaporize 3.5 kg of liquid is X.then find the value ofX/40​

Answer 1

Explenation

Solution:- (A) 26cal/mol−K

As  we know that,

ΔS=  

T

ΔH

​  

 

ΔH=540cal/g

Mol. wt. of water =18g

ΔH  

(per mole)

​  

=ΔH  

(per gm)

​  

×Mol. wt.

ΔH  

(per mole)

​  

=540×18=9720cal/mol

Now, as we know that,

ΔS=  

T

ΔH

​  

 

Given T=100℃=(273+100)=373K

∴ΔS=  

373

9720

​  

=26cal/(mol.K)

I hope this answer will help u.........