40 gram nonelectrolyte solute having molar mass 180 g mol ⁻¹ dissolved in water has osmotic pressure 2 atm at 300 K . Calculate the volume of solution. ( R = 0.0821 atm mol ⁻¹ K ⁻¹ )

Step-by-step solution

Using the formula for osmotic pressure: = n V RT Rearranging for volume V : V = nRT Number of moles of solute n = 40 180 = 2 9 mol Substituting the given values: V = 2 9 0.0821 300 2 V = 24.63 9 V = 2.736 L Since 1 L = 1 dm ^3 , the volume of the solution is 2.74 dm ^3 . Answer: 2.74 dm ^3