Question

An aqueous solution of acetic acid has a molality of \(0.1\) m. If the percent dissociation of acetic acid in water is \(1.35\%\), calculate the depression in freezing point of the solution. (\(K_{f}\) for water \(=1.86\) \({}^{\circ }\)C kg mol\({}^{-1}\))

• A. \(0.189\) \({}^{\circ }\)C
• B. \(0.084\) \({}^{\circ }\)C
• C. \(-0.189\) \({}^{\circ }\)C
• D. \(0.118\) \({}^{\circ }\)C

Hint:

Find the van't Hoff factor i = 1 + alpha, then use the depression formula with i.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Acetic acid dissociates into 2 particles, so \(i = 1 + \alpha\). The depression is \(\Delta T_f = i K_f m\).

Step 2: Van't Hoff factor
\(\alpha = 0.0135\), so \(i = 1.0135\).

Step 3: Calculation
\(\Delta T_f = 1.0135\times 1.86\times 0.1 = 0.1885 \approx 0.189\) \(^{\circ}\)C.

Final Answer:
The depression in freezing point is about 0.189 degrees Celsius. \[ \boxed{\text{(A) }0.189^{\circ}\text{C}} \]