What is the osmotic pressure of a 0.5 M NaCl solution at 25°C? (R = 0.0821 L·atm

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Question: What is the osmotic pressure of a 0.5 M NaCl solution at 25°C? (R = 0.0821 L·atm/(K·mol)) (2023)

Options:

Correct Answer: 6.15 atm

Exam Year: 2023

Solution:

Osmotic pressure (π) = iCRT = 2 * 0.5 * 0.0821 * 298 = 24.5 atm. Therefore, π = 6.15 atm.

What is the osmotic pressure of a 0.5 M NaCl solution at 25°C? (R = 0.0821 L·atm/(K·mol)) (2023)

What is the osmotic pressure of a 0.5 M NaCl solution at 25°C? (R = 0.0821 L·atm/(K·mol)) (2023)

Step 1: Identify the formula for osmotic pressure, which is π = iCRT.
Step 2: Determine the values needed for the formula: i (van 't Hoff factor), C (concentration in molarity), R (ideal gas constant), and T (temperature in Kelvin).
Step 3: For NaCl, the van 't Hoff factor (i) is 2 because NaCl dissociates into two ions: Na+ and Cl-.
Step 4: The concentration (C) is given as 0.5 M.
Step 5: The ideal gas constant (R) is given as 0.0821 L·atm/(K·mol).
Step 6: Convert the temperature from Celsius to Kelvin: 25°C + 273 = 298 K.
Step 7: Plug the values into the osmotic pressure formula: π = iCRT = 2 * 0.5 * 0.0821 * 298.
Step 8: Calculate the result: π = 2 * 0.5 = 1; then 1 * 0.0821 = 0.0821; finally, 0.0821 * 298 = 24.4858 atm.
Step 9: Round the result to two decimal places: π ≈ 24.49 atm.

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