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 )

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Correct answer: C. 2.74\ dm ^3

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

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