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Answer :
Final answer:
When an external pressure of 3 atm is applied to a 0.1 M aqueous solution of NaCl, osmosis will continue because the applied pressure is less than the osmotic pressure of 5 atm.
Explanation:
To determine what happens when a 0.1 M aqueous solution of NaCl is separated by a semi-permeable membrane from pure water and an external pressure of 3 atm is applied to the solution side, we need to first calculate the osmotic pressure of the NaCl solution. Since NaCl dissociates into Na+ and Cl- ions, producing two particles per formula unit in solution, the effective molarity becomes 0.2 M. Using the osmotic pressure formula Π = MRT, where M is the molarity, R is the gas constant (1/12 atm mol⁻¹ K⁻¹ as given), and T is the temperature in Kelvin (27°C = 300K), we find the osmotic pressure.
Π = (0.2 mol/L) × (1/12 atm mol⁻¹ K⁻¹) × (300 K) = 5 atm. Since the applied external pressure (3 atm) is less than the calculated osmotic pressure (5 atm), osmosis will not stop, and reverse osmosis will not occur effectively as long as the external pressure is less than the osmotic pressure. Thus, option A (Osmosis will stop) is incorrect because the applied pressure is not enough to halt osmosis. Options B (Osmosis will continue) is correct, and C (Reverse osmosis will occur) and D (Solute will move from solution to solvent) are incorrect as the applied pressure does not exceed the osmotic pressure to cause reverse osmosis or movement of solute against its concentration gradient.
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