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Calculate the molal depression constant of a solvent which has freezing point `16.6^(@)C` and latent heat of fusion `180.75 Jg^(-1)`
A. `2.68`
B. `3.86`
C. `4.68`
D. `2.86`

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Correct Answer - B
`K_(f)=(RT_(f)^(2))/(1000xx L_(f)), R=8.314 JK^(-1)mol^(-1)`
`T_(f)=273+16.6=289.6K , L_(f)=180.75 Jg^(-1)`
`K_(f)=(8.314xx289.6xx289.6)/(1000xx180.75)=3.86`

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