For a solvent of molar mass M (in g mol^-1), normal freezing point T_f (in K) and molar enthalpy of fusion Delta H_fus, which expression gives K_f in K kg mol^-1?

K_f = (M R T_f^2)/(1000 Delta H_fus)

K_f = (M R T_f)/(1000 Delta H_fus)

K_f = (M R T_f^2)/(1000 Delta H_fus)

K_f = dfrac1000 Delta H_fusM R T_f^2

K_f = (M R T_f^2)/(Delta H_fus)

JEE Main levelSingle correct optionDifficulty 2 of 3so064

For a solvent of molar mass $M$ (in $g mo l^{- 1}$ ), normal freezing point $T_{f}$ (in $K$ ) and molar enthalpy of fusion $Δ H_{f u s}$ , which expression gives $K_{f}$ in $K kg mo l^{- 1}$ ?

From the equilibrium of solid solvent and solution,

Δ T_{f} = \frac{R T _{f}^{2}}{Δ H _{f u s}} x_{2}

, and for a dilute solution

x_{2} \approx m M /1000

. Hence

K_{f} = \frac{MR T _{f}^{2}}{1000 Δ H _{f u s}}

, which has

T_{f}^{2}

in the numerator,

Δ H_{f u s}

in the denominator and the factor 1000 for conversion to molality.