Calculate the molal depression constant of a solvent which has freezing point 16.6° C and latent heat of fusion 180.75 J / g.

Question

Calculate the molal depression constant of a solvent which has freezing point 16.6° C and latent heat of fusion 180.75 J / g. (A) 2.68 (B) 3.86 (C) 4.68 (D) 2.86

Tardigrade Admin · Accepted Answer

Given, T f =273+16.6=289.6 K Lf =180.75 J / g R =8.314 J / K / mol ∴ Kf =(Kf=?/1000 × Lf)=(8.314 ×(289.6)2/1000 × 180.76) ∴ Kf =3.86

Q. Calculate the molal depression constant of a solvent which has freezing point $16. 6^{\circ} C$ and latent heat of fusion $180.75 J / g$ .

2.68

3.86

4.68

2.86

Given,
$T_{f} = 273 + 16.6 = 289.6 K$
$L_{f} = 180.75 J / g$
$R = 8.314 J / K / m o l$
$∴ K_{f} = \frac{K _{f} = ?}{1000 \times L _{f}} = \frac{8.314 \times ( 289.6 ) ^{2}}{1000 \times 180.76}$
$∴ K_{f} = 3.86$

Q. Calculate the molal depression constant of a solvent which has freezing point 16.6∘C and latent heat of fusion 180.75J/g.