The latent heat of vaporisation of water is 9700 cal/mol and if the boiling point is 100∘C,100∘C, the ebullioscopic constant of water is. please give ans with whole solution
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Answered by
15
Use formula ,
Kb = RT²/1000Lv
Where , R is gas constant
T is the boiling point of water in Kelvin
Lv is latent heat of vaporization .
Given, T = 100°C = 373K
R = 2 cal/K/mol
Lv = 9700 cal/mol.
So, Kb = 2 × (373)²/1000 × 9700
= 2 × 373 × 373/(9.7 × 10⁶)
= 0.0286
Hence, ebullioscopic constant is 0.0286 K
Kb = RT²/1000Lv
Where , R is gas constant
T is the boiling point of water in Kelvin
Lv is latent heat of vaporization .
Given, T = 100°C = 373K
R = 2 cal/K/mol
Lv = 9700 cal/mol.
So, Kb = 2 × (373)²/1000 × 9700
= 2 × 373 × 373/(9.7 × 10⁶)
= 0.0286
Hence, ebullioscopic constant is 0.0286 K
Answered by
32
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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