What is the osmotic pressure of a solution containing 0.5 moles of solute in 2 liters of solution at 25 °C? (R = 0.0821 L·atm/(K·mol))

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Q: What is the osmotic pressure of a solution containing 0.5 moles of solute in 2 liters of solution at 25 °C? (R = 0.0821 L·atm/(K·mol))

Solution: Osmotic pressure (π) = nRT/V = (0.5 moles * 0.0821 * 298) / 2 = 12.21 atm.

Steps: 6

Step 1: Identify the formula for osmotic pressure, which is π = nRT/V.
Step 2: Determine the values needed for the formula: n (number of moles) = 0.5 moles, R (ideal gas constant) = 0.0821 L·atm/(K·mol), T (temperature in Kelvin) = 25 °C + 273 = 298 K, and V (volume in liters) = 2 liters.
Step 3: Plug the values into the formula: π = (0.5 moles * 0.0821 L·atm/(K·mol) * 298 K) / 2 liters.
Step 4: Calculate the numerator: 0.5 * 0.0821 * 298 = 12.21.
Step 5: Divide the result by the volume: 12.21 / 2 = 6.105 atm.
Step 6: The osmotic pressure is approximately 6.105 atm.