800 mL of a 0.5 M HNO ₃ at 300 K is heated to a temperature of 350 K , its volume is reduced to one half and 10.5 g of HNO ₃ is evaporated. Assuming complete ionization, the osmotic pressure of the remaining solution is: [Given R =0.082 ~L

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800 mL of a 0.5 M HNO ₃ at 300 K is heated to a temperature of 350 K , its volume is reduced to one half and 10.5 g of HNO ₃ is evaporated. Assuming complete ionization, the osmotic pressure of the remaining solution is: [Given R =0.082 ~L ~atm ~K ⁻¹ ~mol ⁻¹ ]

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Correct answer: D. 33.5 atm

Initial moles of HNO₃ : n₁ = 0.5 0.8 = 0.4 mol Moles evaporated: n_ evap = 10.5 63 = 1 6 mol Remaining moles: n₂ = 0.4 - 1 6 = 12 30 - 5 30 = 7 30 mol Final volume: V₂ = 800 2 = 400 mL = 0.4 L Final molarity: C₂ = 7/30 0.4 = 7 12 M HNO₃ H^+ + NO₃^- , so van't Hoff factor i = 2 = i C₂ R T = 2 7 12 0.082 350 = 7 6 28.7 = 200.9 6 33.5 atm

800 mL of a 0.5 M HNO ₃ at 300 K is heated to a temperature of 350 K , its volume is reduced to one half and 10.5 g of HNO ₃ is evaporated. Assuming complete ionization, the osmotic pressure of the remaining solution is: [Given R =0.082 ~L ~atm ~K ⁻¹ ~mol ⁻¹ ]

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