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What is the osmotic pressure of a 0.2 M NaCl solution at 25 °C? (R = 0.0821 L at
What is the osmotic pressure of a 0.2 M NaCl solution at 25 °C? (R = 0.0821 L atm/(K mol))
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Q1
What is the osmotic pressure of a 0.2 M NaCl solution at 25 °C? (R = 0.0821 L atm/(K mol))
4.92 atm
2.46 atm
1.23 atm
0.61 atm
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Osmotic pressure = iCRT = 2 * 0.2 * 0.0821 * 298 = 4.92 atm (i = 2 for NaCl)
Questions & Step-by-step Solutions
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Q
Q: What is the osmotic pressure of a 0.2 M NaCl solution at 25 °C? (R = 0.0821 L atm/(K mol))
Solution:
Osmotic pressure = iCRT = 2 * 0.2 * 0.0821 * 298 = 4.92 atm (i = 2 for NaCl)
Steps: 8
Show Steps
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), R (ideal gas constant), and T (temperature in Kelvin).
Step 3: For NaCl, the van 't Hoff factor (i) is 2 because it dissociates into two ions: Na+ and Cl-.
Step 4: The concentration (C) is given as 0.2 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 = 2 * 0.2 * 0.0821 * 298.
Step 8: Calculate the osmotic pressure: π = 2 * 0.2 * 0.0821 * 298 = 4.92 atm.
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