Calculate the molal depression constant of a solvent which has freezing point 16.6° C and latent heat of fusion 180.75 Jg -1

Question

Calculate the molal depression constant of a solvent which has freezing point 16.6° C and latent heat of fusion 180.75 Jg -1 (A) 2.68 (B) 4.68 (C) 3.86 (D) 2.86

Tardigrade Admin · Accepted Answer

Kf=(R Tf2/1000 × Lf), R=8.314 JK -1 mol -1 Tf=273+16.6=289.6 K ; Lf=180.75 Jg -1 ∴ Kf=(8.314 × 289.6 × 289.6/1000 × 180.75)=3.86

Q. Calculate the molal depression constant of a solvent which has freezing point $16. 6^{\circ} C$ and latent heat of fusion $180.75$ $J g^{- 1}$

2.68

4.68

3.86

2.86

$K_{f} = \frac{R T _{f}^{2}}{1000 \times L _{f}}$ ,

$R = 8.314 J K^{- 1} m o l^{- 1}$

$T_{f} = 273 + 16.6 = 289.6 K$ ;

$L_{f} = 180.75 J g^{- 1}$

$∴ K_{f} = \frac{8.314 \times 289.6 \times 289.6}{1000 \times 180.75} = 3.86$

Q. Calculate the molal depression constant of a solvent which has freezing point 16.6∘C and latent heat of fusion 180.75 Jg−1