Question

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

Answer 1

freezing point of water, T = 0°C or 273K

Latent heat of water , L = 1436.3 cal/mol

Universal gas constant, R = 2 cal/K.mol

molar mass of water , M = 18g/mol

now, depression in freezing point, $K_f=\frac{MRT^2}{1000L}$

$=\frac{18g/mol\times2cal/K.mol\times(273K)^2}{1000\times1436.3cal/mol}$

=18 × 2 × 273 × 273/(1436.3 × 1000) gm.K/mol

= 1.868 gm.K/mol

hence, depression in freezing point is 1.868 gm.K/mol

Answer 2

Hey.

For this question, we will use the formula,

     $Kf = \frac{ RT^{2} _0}{1000L_v}$

where, Kf is the freezing depression constant, lv is the latent heat of  fusion per gm of solvent., R is the universal gas Constant or you can also say it here as solution constant.

T₀ is the Freezing point of pure water.

In Question, latent heat of fusion per moles of solvent is given, thus we will use the formula,

  $Kf = \frac{ MRT^{2} _0}{1000H}$

 Kf = (18 × 2 ×  273²)/(1000 × 1436.3)

Kf  = 1.87 KKg/mole.

Hope it helps.