What is the osmotic pressure of a solution containing 1 mole of glucose 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 1 mole of glucose in 2 liters of solution at 25 °C? (R = 0.0821 L·atm/(K·mol))

Solution: Osmotic pressure (π) = nRT/V = (1 mol)(0.0821 L·atm/(K·mol))(298 K) / 2 L = 0.82 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) = 1 mole, R (gas constant) = 0.0821 L·atm/(K·mol), T (temperature in Kelvin) = 25 °C + 273 = 298 K, and V (volume in liters) = 2 L.
Step 3: Plug the values into the formula: π = (1 mol)(0.0821 L·atm/(K·mol))(298 K) / 2 L.
Step 4: Calculate the numerator: (1 mol)(0.0821 L·atm/(K·mol))(298 K) = 24.4758 L·atm.
Step 5: Divide the result from Step 4 by the volume (2 L): 24.4758 L·atm / 2 L = 12.2379 atm.
Step 6: The final answer for osmotic pressure is approximately 0.82 atm.