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: \[\Delta T_f = i \cdot K_f \cdot m.\] \(i\) represents the van ’t Hoff factor, \(K_f\) is the cryoscopic constant (\(1.86 \, \text{K kg mol}^{-1}\)), and \(m\) is the molality of the solution. Step 1: Calculate the molality of the solution. The mass of acetic acid is calculated as: \[\text{Mass of acetic acid} = \text{Volume} \times \text{Density} = 5 \, \text{mL} \times 1.2 \, \text{g/mL} = 6 \, \text{g}.\] 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 determined 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 yields: \[\alpha = \sqrt{\frac{6.25 \times 10^{-5}}{0.1}} = \sqrt{6.25 \times 10^{-4}} = 0.025.\] 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: \(x = 19\).