Question

A 0.01 m aqueous solution of AlCl₃ freezes at – 0.068°C. Calculate the percentage of dissociation.

[Given : K_f for Water = 1.86 K kg mol^–1]

Answer 1

We know, \(\Delta T_f = i(K_f)(m)\)

and  i = 1 + (n – 1)\(\alpha\)

where, i = van't Hoff's factor

n = number of ions

\(\alpha\) = degree of dissociation

Now, 0.068 = i (1.86) (0.01)

\(\Rightarrow i \simeq 3.66\)

\(\Rightarrow \) 1 + (4 – 1)\(\alpha\) = 3.66

\(\Rightarrow \alpha = \frac{2.66}{3}= 0.887\) 

\(\therefore\ \% \ \text{dissociation} = 100\ \alpha = 88.7 \%\)

Hence, the percentage of dissociation is 88.7%