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Q1.32:
Calculate the depression in the freezing point of water when 10 g of \(CH_{3}CH_{2}CHClCOOH\) is added to 250 g of water. \(K_{a} = 1.4 \times 10^{-3}\), \(K_{f} = 1.86\) K kg \(mol^{-1}\).

Solution :

Step 1: Identification of Given Parameters

  • Mass of solute ($w_2$): $10\text{ g}$
  • Molar mass of solute ($M_2$): $\text{CH}_3\text{CH}_2\text{CHClCOOH}$
    • $C = 4 \times 12.01 = 48.04$
    • $H = 9 \times 1.008 = 9.072$
    • $Cl = 1 \times 35.45 = 35.45$
    • $O = 2 \times 16.00 = 32.00$
    • Sum: $124.562 \text{ g/mol} \approx 122.5 \text{ g/mol}$ (using atomic weights: $C=12, H=1, Cl=35.5, O=16$)
  • Mass of solvent ($w_1$): $250\text{ g} = 0.25\text{ kg}$
  • Dissociation constant ($K_a$): $1.4 \times 10^{-3}$
  • Cryoscopic constant ($K_f$): $1.86\text{ K kg mol}^{-1}$

Step 2: Calculation of Molality ($m$)

Molality is defined as the number of moles of solute per kilogram of solvent [Definition of Molality].

$m = \frac{w_2}{M_2 \times w_1 (\text{in kg})} = \frac{10}{122.5 \times 0.25}$

$m = \frac{10}{30.625} \approx 0.3265 \text{ mol kg}^{-1}$

Step 3: Determination of Van't Hoff Factor ($i$)

The solute is a weak acid: $\text{CH}_3\text{CH}_2\text{CHClCOOH} \rightleftharpoons \text{CH}_3\text{CH}_2\text{CHClCOO}^- + \text{H}^+$.

For a weak electrolyte, the degree of dissociation ($\alpha$) is given by $\alpha = \sqrt{\frac{K_a}{C}}$, where $C$ is molarity. Since $m \approx C$ for dilute aqueous solutions:

$\alpha = \sqrt{\frac{1.4 \times 10^{-3}}{0.3265}} = \sqrt{0.004288} \approx 0.06548$

The Van't Hoff factor $i$ for dissociation is $i = 1 + \alpha(n - 1)$. Here, $n=2$ (two ions produced).

$i = 1 + 0.06548(2 - 1) = 1.06548$

Step 4: Calculation of Depression in Freezing Point ($\Delta T_f$)

The formula for depression in freezing point is $\Delta T_f = i \times K_f \times m$ [Raoult's Law modification for dissociation].

$\Delta T_f = 1.06548 \times 1.86 \times 0.3265$

$\Delta T_f \approx 0.647 \text{ K}$

Molality (m) ΔTf (K) Linear relationship: ΔTf = iKf * m

Final Solution: The depression in the freezing point of water is 0.647 K.


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