Colligative Properties – Osmotic Pressure
A 0.025 M aqueous solution of a non-electrolyte solute exerts an osmotic pressure of 0.615 atm at 300 K. The value of the gas constant R used in the calculation is closest to:
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About This Question
- Subject
- chemistry
- Chapter
- solutions
- Topic
- colligative properties – osmotic pressure
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
0.0821 L atm K⁻¹ mol⁻¹
Osmotic pressure π = CRT. Substituting: 0.615 = 0.025 × R × 300, so R = 0.615 / (0.025 × 300) = 0.615 / 7.5 = 0.082 L atm K⁻¹ mol⁻¹. This matches R = 0.0821 L atm K⁻¹ mol⁻¹, which is the appropriate value when pressure is in atm and volume in litres. The other values of R correspond to different unit systems (SI, calorie, torr) and would not give a consistent answer with the given units.
This easy difficulty chemistry question is from the chapter solutions, covering the topic of colligative properties – osmotic pressure. It appeared in the 2025 exam.
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