Question

Latent heat of fusion of ice is 1436.3 Cal. mol.calculate the molal depression constant of water.​

Answer 1

Answer:

Explanation:

T.f = boiling point of pure water = 373K

R = Universal Gas constant

lfus = latent heat of fusion

Answer 2

Answer:

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}K

f

=

1000L

MRT

2

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

1000×1436.3cal/mol

18g/mol×2cal/K.mol×(273K)

2

=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

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