Osmosis calculation at 0°C: One gram-mole of a non-electrolyte is dissolved to make 22.4 litres of solution at 0°C. What is the osmotic pressure (in atm) of this solution?

Difficulty: Easy

Correct Answer: 1

Explanation:


Introduction / Context:
Osmotic pressure is central to membrane separations, food science, and bioprocessing. For dilute solutions of non-electrolytes, it mirrors the ideal-gas law, enabling quick head calculations for reverse osmosis or dialysis design.


Given Data / Assumptions:

  • Moles of solute n = 1 mol.
  • Solution volume V = 22.4 L.
  • Temperature T = 0°C = 273 K.
  • R = 0.082057 L·atm·mol^-1·K^-1; ideal dilute solution.


Concept / Approach:
For dilute, non-electrolyte solutions, osmotic pressure π follows πV = nRT, analogous to PV = nRT for gases.


Step-by-Step Solution:

π = n * R * T / V.Substitute: π = 1 * 0.082057 * 273 / 22.4.Compute numerator: 0.082057 * 273 ≈ 22.4.Thus π ≈ 22.4 / 22.4 = 1.0 atm.


Verification / Alternative check:
This setup intentionally mirrors one mole of ideal gas at STP occupying about 22.4 L and exerting ~1 atm, hence the neat numerical result for π.


Why Other Options Are Wrong:

  • 0.5, 1.5, 2: Do not match the direct calculation from π = nRT/V.


Common Pitfalls:
For electrolytes, include van’t Hoff factor i. Also ensure volume is solution volume and temperature is in Kelvin for consistency.


Final Answer:
1

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