The molal depression constant can be written K_f=(RT_f^2M_1)/(1000 Delta_fus)H with M_1 in g mol^-1 and Delta_fusH in J mol^-1. For water take T_f=273 K, M_1=18 g mol^-1, Delta_fusH=6010 J mol^-1 and R=8.314 J mol^-1K^-1. Calculate K_f in K kg mol^-1 to two decimal places.

JEE Main levelNumerical answerDifficulty 2 of 3so183

The molal depression constant can be written $K_{f} = \frac{R T _{f}^{2} M _{1}}{1000 Δ _{fus} H}$ with $M_{1}$ in $g mo l^{- 1}$ and $Δ_{fus} H$ in $J mo l^{- 1}$ . For water take $T_{f} = 273 K$ , $M_{1} = 18 g mo l^{- 1}$ , $Δ_{fus} H = 6010 J mo l^{- 1}$ and $R = 8.314 J mo l^{- 1} K^{- 1}$ . Calculate $K_{f}$ in $K kg mo l^{- 1}$ to two decimal places.

Numerical answer type, answer in K kg/mol, tolerance 0.01.

K_{f} = \frac{8.314 \times 27 3 ^{2} \times 18}{1000 \times 6010} = \frac{8.314 \times 74529 \times 18}{6.010 \times 1 0 ^{6}} = 1.8558 \approx 1.86 K kg mo l^{- 1}