When $0.0106$ mole of acetic acid was dissolved in $1$ kg of water, the freezing point depression for this strength of acid was $0.0205$ K. If the calculated freezing point depression is $0.0197$ K, Van't Hoff factor (i) and degree of dissociation of acetic acid respectively are

The Van't Hoff factor $i$ is given by the ratio of the observed freezing point depression to the calculated freezing point depression:

$i = \dfrac{\Delta T_{f(\text{observed})}}{\Delta T_{f(\text{calculated})}}$

Substituting the given values:

$i = \dfrac{0.0205}{0.0197} = 1.0406 \approx 1.041$

Acetic acid dissociates in water as follows:

$CH_3COOH \rightleftharpoons CH_3COO^- + H^+$

The number of particles produced per molecule of acetic acid is $n = 2$. The relationship between the Van't Hoff factor $i$ and the degree of dissociation $\alpha$ is:

$i = 1 + (n - 1)\alpha$

$1.041 = 1 + (2 - 1)\alpha$

$\alpha = 1.041 - 1 = 0.041$

Thus, the Van't Hoff factor is $1.041$ and the degree of dissociation is $0.041$.

Answer: $1.041$ and $0.041$