The degree of ionisation of HF in 0.100 M aqueous solution is (freezing point of the solution =-0.197° C and Kf for water =186°C)

Question

The degree of ionisation of HF in 0.100 M aqueous solution is (freezing point of the solution =-0.197° C and Kf for water =186°C)  (A) 6% (B) 12% (C) 3% (D) 9%

Tardigrade Admin · Accepted Answer

Δ Tf=i k1 m 0.197=i × 1.86 × 0.1 i=(0.197/1.86 × 0.1)=1.059 ≈ 1.06 In case of dissociation, α=(8-1/n-1) =(1.06-1/2-1)=0.06=6 %

Q. The degree of ionisation of $H F$ in $0.100 M$ aqueous solution is (freezing point of the solution $= - 0.19 7^{\circ} C$ and $K_{f}$ for water $= 18 6^{\circ} C)$

6%

12%

3%

9%

$Δ T_{f} = i k_{1} m$
$0.197 = i \times 1.86 \times 0.1$
$i = \frac{0.197}{1.86 \times 0.1} = 1.059 \approx 1.06$
In case of dissociation, $α = \frac{8 - 1}{n - 1}$
$= \frac{1.06 - 1}{2 - 1} = 0.06 = 6%$

Q. The degree of ionisation of HF in 0.100M aqueous solution is (freezing point of the solution =−0.197∘C and Kf​ for water =186∘C)

6%

12%

3%

9%

ΔTf​=ik1​m 0.197=i×1.86×0.1 i=1.86×0.10.197​=1.059≈1.06 In case of dissociation, α=n−18−1​ =2−11.06−1​=0.06=6%

Q. The degree of ionisation of $H F$ in $0.100 M$ aqueous solution is (freezing point of the solution $= - 0.19 7^{\circ} C$ and $K_{f}$ for water $= 18 6^{\circ} C)$

$Δ T_{f} = i k_{1} m$
$0.197 = i \times 1.86 \times 0.1$
$i = \frac{0.197}{1.86 \times 0.1} = 1.059 \approx 1.06$
In case of dissociation, $α = \frac{8 - 1}{n - 1}$
$= \frac{1.06 - 1}{2 - 1} = 0.06 = 6%$