What is the expected osmotic pressure of a 0.5 M NaCl solution at 25 °C?

What is the expected osmotic pressure of a 0.5 M NaCl solution at 25 °C?

Correct Answer: 24.6 atm

Step 1: Identify the formula for osmotic pressure, which is π = iCRT.
Step 2: Determine the value of 'i' for NaCl. Since NaCl dissociates into two ions (Na+ and Cl-), i = 2.
Step 3: Identify the concentration (C) of the NaCl solution, which is given as 0.5 M.
Step 4: Use the ideal gas constant (R), which is 0.0821 L·atm/(K·mol).
Step 5: Convert the temperature from Celsius to Kelvin. 25 °C is equal to 298 K (25 + 273).
Step 6: Plug the values into the formula: π = (2)(0.5 M)(0.0821 L·atm/(K·mol))(298 K).
Step 7: Calculate the osmotic pressure: π = 2 * 0.5 * 0.0821 * 298.
Step 8: Perform the multiplication: π ≈ 24.6 atm.

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