Question

When \(9.2 \times 10^{-3} \, kg\) of formic acid is added to \(600 \, ml\) of water, the freezing point of water is depressed. If \(30\%\) of formic acid undergoes dissociation, what would be the freezing point of the solution? Given \(K_f\) of \(H_2O = 1.86 \, K \, kg \, mol^{-1}\) and molar mass of formic acid \(= 46 \, amu\).

• A. \(273.9 \, K\)
• B. \(272.2 \, K\)
• C. \(270.1 \, K\)
• D. \(270.8 \, K\)

Hint:

For electrolytes undergoing dissociation, always include Van't Hoff factor \(i\) in colligative property formulae. For weak acid dissociation into two ions, \(i = 1 + \alpha\).

Answer 1

The Correct Option is B

Step 1: Convert the given mass of formic acid into grams.
Given mass of formic acid is:
\[ 9.2 \times 10^{-3} \, kg \] Since \(1 \, kg = 1000 \, g\), therefore:
\[ 9.2 \times 10^{-3} \, kg = 9.2 \, g \]

Step 2: Calculate moles of formic acid.
Molar mass of formic acid is:
\[ 46 \, g \, mol^{-1} \] Therefore, moles of formic acid are:
\[ \text{Moles} = \frac{9.2}{46} \] \[ \text{Moles} = 0.2 \, mol \]

Step 3: Calculate molality of the solution.
Volume of water is:
\[ 600 \, ml \] For water, \(600 \, ml = 600 \, g = 0.6 \, kg\).
Molality is given by:
\[ m = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} \] \[ m = \frac{0.2}{0.6} \] \[ m = 0.333 \, mol \, kg^{-1} \]

Step 4: Calculate Van't Hoff factor for formic acid.
Formic acid dissociates as:
\[ HCOOH \rightleftharpoons H^+ + HCOO^- \] For dissociation of one molecule into two ions:
\[ i = 1 + \alpha \] Given degree of dissociation is \(30\%\), so:
\[ \alpha = \frac{30}{100} = 0.3 \] \[ i = 1 + 0.3 = 1.3 \]

Step 5: Calculate depression in freezing point.
The formula for depression in freezing point is:
\[ \Delta T_f = iK_fm \] Substituting the values:
\[ \Delta T_f = 1.3 \times 1.86 \times 0.333 \] \[ \Delta T_f \approx 0.806 \, K \]

Step 6: Calculate freezing point of the solution.
Freezing point of pure water is approximately:
\[ 273 \, K \] Therefore, freezing point of solution is:
\[ T_f = 273 - 0.806 \] \[ T_f = 272.194 \, K \] \[ T_f \approx 272.2 \, K \] Final Answer:
The freezing point of the solution is:
\[ \boxed{272.2 \, K} \]