Question

Considering acetic acid dissociates in water, its dissociation constant is $6.25 \times 10^{-5}$. If 5 mL of acetic acid is dissolved in 1 litre water, the solution will freeze at $-x \times 10^{-1}$ °C, provided pure water freezes at 0 °C. $x = $ _____ (Nearest integer)
Given: $(K_f)_{water} = 1.86$ K kg mol$^{-1}$
density of acetic acid is 1.2 g mol$^{-1}$
molar mass of water = 18 g mol$^{-1}$
molar mass of acetic acid = 60 g mol$^{-1}$
density of water = 1 g cm$^{-3}$
Acetic acid dissociates as CH$_3$COOH $\rightleftharpoons$ CH$_3$COO$^-$ + H$^+$

Answer 1

Correct Answer: 19

The depression in freezing point \(\Delta T_f\) is calculated using the formula:
\[\Delta T_f = i \cdot K_f \cdot m,\]
where:
\(i\) is the van ’t Hoff factor,
\(K_f\) is the cryoscopic constant (\(1.86 \, \text{K kg mol}^{-1}\)),
\(m\) is the molality of the solution.
Step 1: Calculate the molality of the solution
The mass of acetic acid dissolved is:
\[\text{Mass of acetic acid} = \text{Volume} \times \text{Density} = 5 \, \text{mL} \times 1.2 \, \text{g/mL} = 6 \, \text{g}.\]
The number of moles of acetic acid is:
\[\text{Moles of acetic acid} = \frac{\text{Mass of acetic acid}}{\text{Molar mass of acetic acid}} = \frac{6}{60} = 0.1 \, \text{mol}.\]
The molality of the solution is:
\[m = \frac{\text{Moles of solute}}{\text{Mass of solvent (kg)}} = \frac{0.1}{1} = 0.1 \, \text{mol/kg}.\]
Step 2: Calculate the van ’t Hoff factor (\(i\))
The dissociation constant (\(K_a\)) of acetic acid is:
\[K_a = 6.25 \times 10^{-5}.\]
The degree of dissociation (\(\alpha\)) is given by:
\[\alpha = \sqrt{\frac{K_a}{C}},\]
where \(C\) is the molarity of the solution.
The molarity is:
\[C = \frac{\text{Moles of solute}}{\text{Volume of solution (L)}} = \frac{0.1}{1} = 0.1 \, \text{mol/L}.\]
Substituting the values:
\[\alpha = \sqrt{\frac{6.25 \times 10^{-5}}{0.1}} = \sqrt{6.25 \times 10^{-4}} = 0.025.\]
The van ’t Hoff factor is:
\[i = 1 + \alpha = 1 + 0.025 = 1.025.\]
Step 3: Calculate \(\Delta T_f\)
\[\Delta T_f = i \cdot K_f \cdot m = 1.025 \cdot 1.86 \cdot 0.1 = 0.19065 \, \text{K}.\]
Converting to \(-x \times 10^{-2}\):
\[x = 19.\]
Final Answer: \(x = 19\).