Question

Latent heat of water (ice) is 1436.3 cal per mol. What will be molal freezing point depression constant of water (R = 2 cal/degree/mol)?

Options

• 18.70°C

• 1.870°C

• 0.187°C

• None of these

Answer 1

1.870°C

Explanation:

`K_f = (MRT°_f^(2))/(Δ_(fus) H xx 1000)`

= `(18  g  mol^(-1) xx 2  cal  k^(-1)  mol^(-1) xx (273  K)^2)/(1436.3  cal  mol^(-1) xx 1000)`

= 1.87 K kg mo1^–1 

or 1.87°C kg mol^–1