Calculate the osmotic pressure of a 6.0 × 10-2 M solution of NaCl at 20°C (293 K).
Π = iMRT
where
Π = osmotic pressure in atm
i = van 't Hoff factor of the solute.
((Broken Link Removed)
M = molar concentration in mol/L
R = universal gas constant = 0.08206 L·atm/mol·K
T = absolute temperature in K
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There are 2 particles in the equation. One for Na and one for Cl. so it would be i=2 M=6.0x10^-2 R=.0821 and T=293. The rest is up to you.
To calculate the osmotic pressure of a solution, you can use the equation:
Π = iMRT
Where:
Π is the osmotic pressure (in atm),
i is the van 't Hoff factor (a measure of the number of particles into which a compound dissolves),
M is the molarity of the solution (in mol/L),
R is the ideal gas constant (0.0821 L·atm/mol·K), and
T is the temperature in Kelvin.
For NaCl, the van 't Hoff factor is 2, as it dissociates into two particles (Na+ and Cl-) in the solution.
Let's plug the given values into the equation:
Π = (2) × (6.0 × 10-2 M) × (0.0821 L·atm/mol·K) × (293 K)
First, let's calculate the multiplication inside the brackets:
Π = (2) × (0.00492) × (76.253)
Now, multiply the values:
Π ≈ 0.942 atm
Therefore, the osmotic pressure of a 6.0 × 10-2 M solution of NaCl at 20°C (293 K) is approximately 0.942 atm.