After adding non-volatile solute freezing point of water decreases to -0.186 ° C. Calculate Δ TB if Kf = 1.86 K kg mol-1 and Kb = 0.521 K kg mol-1

Question

After adding non-volatile solute freezing point of water decreases to -0.186 ° C. Calculate Δ TB if Kf = 1.86 K kg mol-1 and Kb = 0.521 K kg mol-1 (A) 0.521 (B) 0.0521 (C) 1.86 (D) 0.0186

Tardigrade Admin · Accepted Answer

According to Raoult's law, when a non-volatile solid is added to the solvent its vapour pressure decreases and now its vapour pressure would become equal to that of solid solvent at lower temperature. Thus, the freezing point of the solvent decreases.

Δ T_{f} \propto m

(m=molality of solution)

Δ T_{f} = K_{f} m, 0.186 = 1.86 \times m

∴ m = 0.1

Now

Δ T_{b} = K_{b} \times m = 0.521 \times 0.1 = 0.0521

Q. After adding non-volatile solute freezing point of water decreases to $- 0.186^{\circ} C$ . Calculate $Δ T_{B}$ if $K_{f} = 1.86 K k g$ $m o l^{- 1}$ and $K_{b} = 0.521 K k g m o l^{- 1}$

$0.521$

$0.0521$

$1.86$

$0.0186$

Q. After adding non-volatile solute freezing point of water decreases to −0.186∘C. Calculate ΔTB​ if Kf​=1.86Kkg mol−1 and Kb​=0.521Kkgmol−1

0.521

0.0521

1.86

0.0186

Q. After adding non-volatile solute freezing point of water decreases to $- 0.186^{\circ} C$ . Calculate $Δ T_{B}$ if $K_{f} = 1.86 K k g$ $m o l^{- 1}$ and $K_{b} = 0.521 K k g m o l^{- 1}$

$0.521$

$0.0521$

$1.86$

$0.0186$