© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the Relevant Concept
We are dealing with depression in freezing point, a colligative property.
The formula for depression in freezing point is
$ \Delta T_f = i \, K_f \, m $,
where $ \Delta T_f $ is the decrease in freezing point,
$i$ is the van’t Hoff factor,
$K_f$ is the cryoscopic constant (here, 1.85),
and $m$ is the molality of the solution.
Step 2: Determine the Van’t Hoff Factor ($i$)
The weak acid HX partially dissociates as:
$ HX \leftrightharpoons H^+ + X^- $
Given the degree of ionization (α) is 0.3,
initially let 1 mole of HX be present.
Then, $ (1-\alpha) $ moles of HX remain undissociated, and
$ \alpha $ moles each of $ H^+ $ and $ X^- $ are produced.
Thus, total particles for every 1 mole of HX become
$ (1-\alpha) + \alpha + \alpha = 1 + \alpha $.
Hence:
$ i = 1 + \alpha = 1 + 0.3 = 1.3 $.
Step 3: Apply the Depression in Freezing Point Formula
Molality $m = 0.2$ (given),
$K_f = 1.85$,
and $ i = 1.3 $.
Substituting into the formula:
$ \Delta T_f = i \times K_f \times m = 1.3 \times 1.85 \times 0.2 $.
Calculating this:
$ \Delta T_f = 1.3 \times 1.85 \times 0.2 \approx 0.48^\circ C $.
Step 4: Determine the Final Freezing Point
Since the normal freezing point of water is $0^\circ C$,
the solution’s freezing point is:
$ T_f = 0^\circ C - 0.48^\circ C = -0.48^\circ C $.
Therefore, the freezing point of the solution is approximately
$ -0.480^\circ C $.
