write the formula to calculate Kb when enthalpy of vapoursation of solvent is given
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Answered by
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
Answered by
0
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:
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