What mass of urea (in g ) is required to prepare 500 mL of an aqueous solution that is isotonic with a 3.6 % (w/v) aqueous solution of glucose at the same temperature? (Given: Molar mass of urea = 60 g mol ⁻¹ , Molar mass of glucose = 180 g mol ⁻¹ )

Step-by-step solution

For isotonic solutions at the same temperature, the osmotic pressures are equal ( ₁ = ₂ ). Since both urea and glucose are non-electrolytes, their molar concentrations must be equal ( C₁ = C₂ ). A 3.6 % (w/v) glucose solution contains 3.6 g of glucose in 100 mL of solution. Molarity of glucose solution = 3.6 180 1000 100 = 0.2 M Therefore, the molarity of the urea solution must also be 0.2 M . Moles of urea required for 500 mL ( 0.5 L ) of solution = 0.2 mol L ⁻¹ 0.5 L = 0.1 mol Mass of urea required = 0.1 mol 60 g