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 \(0.041\)