0.6ml of CH3COOH of density 1.06g/ml is dissolved is 1L water. The depression of freezing point observed was 0.02050C. The dissociation constant of acid is (kf=1.86km−1)

A
$1.60 \times 10^{- 5}$
B
$1.86 \times 10^{- 5}$
C
$1.08 \times 10^{- 6}$
D
$1.06 \times 10^{- 5}$

Solution:

$\begin{array}{l} n_{{CH}_{3} COOH} = \frac{V_{A}}{M_{B}} = \frac{0 .6 \times 1 .06}{60} = 0 .0106 \\ mola r i ty = 0 .0106 mol / ks \\ △ T_{f} = k_{f} m = 1 .86 \times 0 .0106 = 0 .0197 \\ i = \frac{(△ T_{f})}{(△ T_{f})} = \frac{0 .0205}{0 .0197} = 1 .041 \\ α = \frac{i - 1}{n - 1} = \frac{1 .014 - 1}{2 - 1} = 0 .041 \\ k = \frac{{cα}^{2}}{1 - α} = \frac{0 .0106 \times {(0 .041)}^{2}}{1 - 0 .041} = 1 .85 \times 10^{- 5} \end{array}$

0.6ml of CH₃COOH of density 1.06g/ml is dissolved is 1L water. The depression of freezing point observed was 0.0205⁰C. The dissociation constant of acid is ( $k_{f} = 1 .86 {km}^{- 1}$ )

Solution:

Related content