Question

\(20 \%\) of acetic acid is dissociated when its \(5\, g\) is added to \(500 \, mL\) of water The depression in freezing point of such water is ____ \(\times 10^{-3}{ }^{\circ} C\)Atomic mass of \(C , H\) and \(O\) are 12,1 and 16 amu respectively [Given : Molal depression constant and density of water are \(186 \,K \,kg\, mol ^{-1} and \  1 \,g \,cm ^{-3}\)respectively

Hint:

Remember the formula for freezing point depression and how to calculate the van’t Hoff factor for weak electrolytes. Pay attention to units and ensure they are consistent throughout the calculation.

Answer 1

Correct Answer: 372

Step 1: Moles of Acetic Acid
Molar mass of acetic acid (\(\text{CH}_3\text{COOH}\)) is:
\[2 \times 12 + 4 \times 1 + 2 \times 16 = 60 \, \text{g/mol}.\]
\[\text{Moles of acetic acid} = \frac{5 \, \text{g}}{60 \, \text{g/mol}} = \frac{1}{12} \, \text{mol}.\]
Step 2: Molality of Acetic Acid
Mass of water:
\[\text{Volume} \times \text{Density} = 500 \, \text{mL} \times 1 \, \text{g/mL} = 500 \, \text{g} = 0.5 \, \text{kg}.\]
\[\text{Molality} (m) = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}} = \frac{\frac{1}{12} \, \text{mol}}{0.5 \, \text{kg}} = \frac{1}{6} \, \text{mol/kg}.\]
Step 3: van't Hoff Factor (\(i\))
Acetic acid dissociates as follows:
\[\text{CH}_3\text{COOH} \leftrightharpoons \text{CH}_3\text{COO}^- + \text{H}^+.\]
Since 20% of acetic acid dissociates, the degree of dissociation (\(\alpha\)) is \(0.2\). For dissociation:
\[i = 1 + \alpha(n - 1),\]
where \(n\) is the number of particles formed after dissociation. Here, \(n = 2\).
\[i = 1 + 0.2(2 - 1) = 1 + 0.2 = 1.2.\]
Step 4: Depression in Freezing Point (\(\Delta T_f\))
The formula for freezing point depression is:
\[\Delta T_f = iK_fm,\]
where \(K_f\) is the molal depression constant.
\[\Delta T_f = 1.2 \times 1.86 \, \text{K mol}^{-1} \times \frac{1}{6} \, \text{mol/kg}.\]
\[\Delta T_f = 0.372 \, \text{K}.\]
Since the change in temperature in Kelvin and Celsius is the same:
\(\Delta T_f = 0.372^\circ\text{C} = 372 \times 10^{-3} \, {C}.\)
Final Answer:
The depression in freezing point is \(0.372^\circ\text{C}\).

Answer 2

The accurate solution is 372.
By using the formula \(i=1+(n−1)α\), where \(n=2\) and \(α=0.2\), we get \(i=1.2\).
Then, we can use the formula \(ΔTf=iKf\times m\) to calculate the freezing point depression, where \(Kf\text{ is }1.86\) and m is calculated using the given information.
Substituting the values, we get \(ΔTf=1.2\times1.86\times\frac{(5\times1000)}{(60\times500)}=3.72\).
Converting the result to Celsius, we get \(ΔTf=372\times10^-2\).