The freezing point of a 1.00 m aqueous solution of HF is found to be -1.91 °C. The freezing point constant of water, Kf is 1.86 K kg mol-1. The percentage dissociation of HF at this concentration is

Question

The freezing point of a 1.00 m aqueous solution of HF is found to be -1.91 °C. The freezing point constant of water, Kf is 1.86 K kg mol-1. The percentage dissociation of HF at this concentration is (A) 30% (B) 10% (C) 5.2% (D) 2.7%

Tardigrade Admin · Accepted Answer

Δ Tf = Kf × m × i i = (Δ Tf/Kf × m) = (1.91/1.86 ×1) = 1.02 For underset(1-α)HI leftharpoons undersetαH+ + undersetαI- 1-α +α+α = i = 1.027 1+α = 1.027 α = 0.027 or 2.7 %

Q. The freezing point of a $1.00 m$ aqueous solution of HF is found to be $- 1.9 1^{\circ}$ C. The freezing point constant of water, $K_{f}$ is $1.86 K k g m o l^{- 1}$ . The percentage dissociation of HF at this concentration is

30%

10%

5.2%

2.7%

$Δ T_{f} = K_{f} \times m \times i$
$i = \frac{Δ T _{f}}{K _{f} \times m} = \frac{1.91}{1.86 \times 1} = 1.02$
For $(1 - α) H I ⇌ α H^{+} + α I^{-}$
$1 - α + α + α = i = 1.027$
$1 + α = 1.027$
$α = 0.027$ or $2.7%$

Q. The freezing point of a 1.00m aqueous solution of HF is found to be −1.91∘C. The freezing point constant of water, Kf​ is 1.86Kkgmol−1. The percentage dissociation of HF at this concentration is

30%

10%

5.2%

2.7%

ΔTf​=Kf​×m×i i=Kf​×mΔTf​​=1.86×11.91​=1.02 For (1−α)HI​⇌αH+​+αI−​ 1−α+α+α=i=1.027 1+α=1.027 α=0.027 or 2.7%

Q. The freezing point of a $1.00 m$ aqueous solution of HF is found to be $- 1.9 1^{\circ}$ C. The freezing point constant of water, $K_{f}$ is $1.86 K k g m o l^{- 1}$ . The percentage dissociation of HF at this concentration is

$Δ T_{f} = K_{f} \times m \times i$
$i = \frac{Δ T _{f}}{K _{f} \times m} = \frac{1.91}{1.86 \times 1} = 1.02$
For $(1 - α) H I ⇌ α H^{+} + α I^{-}$
$1 - α + α + α = i = 1.027$
$1 + α = 1.027$
$α = 0.027$ or $2.7%$