Question

write the formula to calculate Kb when enthalpy of vapoursation of solvent is given


​

Answer 1

Answer:

Explanation:

The expression for the molal elevation constant is k  

b

​  

=  

L  

v

​  

 

0.002(T  

b

o

​  

)  

2

 

​  

 

Here, T  

b

o

​  

 is the boiling point of pure solvent and L  

v

​  

 is the latent heat of vaporization in cal/g of the pure solvent. It is equal to 9.72 kcal mol  

−1

 or 540 cal/g.

∴k  

b

​  

=  

540

0.002×(373.15)  

2

 

​  

=0.515 K kg mol  

−1

 

Th depression in freezing point is ΔT  

b

​  

=k  

b

​  

×m=0.515×0.1=0.0515

The boiling point is T  

b

​  

=373.15+0.0515≈373.20 K

Answer 2

Answer:Kb = 2 × 373 ×373 ×18 /1000 ×9700

Molecular mass is also multiplied

Ebullioscopic constant value comes out to be = 0.516 ℃

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Explanation: