What is the osmotic pressure of a solution containing 0.2 moles of solute in 1 l

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Question: What is the osmotic pressure of a solution containing 0.2 moles of solute in 1 liter of solution at 25 °C? (R = 0.0821 L·atm/(K·mol))

Options:

Correct Answer: 4.92 atm

Solution:

Osmotic pressure = (n/V)RT = (0.2/1) * 0.0821 * 298 = 4.92 atm.

What is the osmotic pressure of a solution containing 0.2 moles of solute in 1 liter of solution at 25 °C? (R = 0.0821 L·atm/(K·mol))

What is the osmotic pressure of a solution containing 0.2 moles of solute in 1 liter of solution at 25 °C? (R = 0.0821 L·atm/(K·mol))

Step 1: Identify the formula for osmotic pressure, which is π = (n/V)RT.
Step 2: Determine the values needed for the formula: n (number of moles of solute) = 0.2 moles, V (volume of solution in liters) = 1 liter, R (ideal gas constant) = 0.0821 L·atm/(K·mol), and T (temperature in Kelvin) = 25 °C + 273 = 298 K.
Step 3: Calculate the concentration of the solution (n/V), which is 0.2 moles / 1 liter = 0.2 moles/L.
Step 4: Substitute the values into the osmotic pressure formula: π = (0.2) * (0.0821) * (298).
Step 5: Perform the multiplication: π = 0.2 * 0.0821 * 298 = 4.92 atm.

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