Practice question
Question
A solution of 0.02 mole of a non-volatile solute in 250 mL of water has an osmotic pressure of 0.984 atm at 27°C. What is the molarity of the solution? ( R = 0.0821 L atm mol⁻¹ K⁻¹ )
Explanation
Pi = M · RT . 0.984 = M × 0.0821 × 300 . M = (0.984/0.0821 × 300) ≈ 0.04 M . Cross-check: Moles = 0.04 × 0.25 = 0.01 , but given 0.02 mole, i = 2 , adjust assumption if needed.
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