Question

Molal enthalpy of fusion of water at 273 k is 6.0246 kg mol^–1 calculated molal depression constant.

Options

• 1.85 km^–1

• 2.00 km

• 2.25 km^–1

• 4.35 km

Answer 1

1.85 km^–1 

Explanation:

`Kf = (MR (T_f^circ)^2)/((ΔH  "fusion"))`

`ΔH_(fusion) = 6.0246  kJ  mol^-1 = 6024.6  J  mol^-1`

M = `18  g  mol^-1  = 0.018  kg  mol^-1`

R = `8.314  JK^-1  mol^-1 = 6024.6  J  mol^-1`

M = `18  g  mol^-1  = 0.018  kg  mol^-1`

R = `8.314  KJ^-1  mol^-1, T_f^circ = 273  K`

`k_f = ((0.018  kg  mol^-1) xx (8.314  JK^-1  mol^-1) xx (273  K)^2)/((6024.6  J  mol^-1))`

= `1.85  K  kg  mol^-1` = 1.85 km^–1