Question

Find the freezing point of a solution containing \(27\ \text{g}\) urea in \(1\ \text{dm}^{3}\) of the aqueous solution. [\(K_{f}=1.8\ \text{K kg mol}^{-1}\) and molar mass of urea \(=60\ \text{g mol}^{-1}\)]

• A. \(-0.81\ ^{\circ }\text{C}\)
• B. \(-0.78\ ^{\circ }\text{C}\)
• C. \(-0.84\ ^{\circ }\text{C}\)
• D. \(-0.87\ ^{\circ }\text{C}\)

Hint:

Urea does not dissociate, so i = 1 and delta Tf = Kf times molality.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Depression in freezing point: \(\Delta T_{f}=K_{f}\,m\). Urea is a non-electrolyte, so \(i=1\).

Step 2: Molality
Moles of urea \(=\dfrac{27}{60}=0.45\).
The solution is dilute, so 1 dm\(^{3}\) holds about 1 kg of water. Molality \(\approx 0.45\ \text{m}\).

Step 3: Calculation
\(\Delta T_{f}=1.8\times 0.45=0.81\ \text{K}\)

Step 4: Freezing point
\(T_{f}=0-0.81=-0.81\ ^{\circ}\text{C}\)

Final Answer:
The solution freezes at minus 0.81 degrees Celsius. \[ \boxed{\text{(A) }-0.81\ ^{\circ}\text{C}} \]