Calculate the molal depression constant of a solvent which has
a. Freezing point `16.6^(@)C` and latent heat of fusion `180.75 J g^(-1)` .
b. Freezing point `20.0^(@)C` and latent heat of fusion `200.00 J g^(-1)`.

Explanation

a. `K_(f)=(RT_(f)^(2))/(1000xxl_(f))`
`R = 8.314 JK^(-1)mol^(-1)`
`T_(f) = 16.6^(@)C = 273+16.6 = 289.6 K`
`l_(f) = 180.75 Jg^(-1)`
Substituting values in Eq. (i), we get
`K_(f) = (8.314 xx (289.6))^(2))/(1000 xx 180.75) = 3.86`
b. `K_(f) = (RT_(f)^(2))/(1000 xx l_(f))`
`T_(f) = 273 + 20.0 = 293.0`
`l_(f) = 200.0 J g_(-1) = ((8.314 xx (293.0)^(2)) /(1000 xx 200.0) = 3.56`