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200 cm³ of 0.2 M BaCl2(aq) is mixed with 600 cm³ of 0.1 M Na2SO4(aq). Find the osmotic pressure (in atm) of the resulting solution, rounded to the nearest integer. (R = 0.0821 L·atm·K⁻¹·mol⁻¹, T = 37°C = 310 K)
- 2
- 4
- 6
- 8
Correct answer: 4
Solution
BaCl2 (0.04 mol) reacts with Na2SO4 (0.06 mol). Ba²+ is limiting. 0.04 mol BaSO4 precipitates. Remaining: 0.02 mol Na2SO4 (excess) -> 0.04 mol Na⁺, 0.02 mol SO4²-; and 0.08 mol Cl⁻ from the 0.04 mol BaCl2. Total moles of ions = 0.04 + 0.02 + 0.08 = 0.14 mol in 800 cm³ = 0.8 L. pi = (n/V)*RT = (0.14/0.8)*0.0821*310 = 0.175 * 25.451 = 4.45 ≈ 4 atm.
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