An isotonic solution of glucose with urea contains 18 g in 1 dm ^3 . Calculate the mass of urea in 500 mL solution. [Molar mass of urea = 60 g mol ⁻¹ . Molar mass of glucose = 180 g mol ⁻¹ ]

Step-by-step solution

For isotonic solutions, the molar concentrations of the solutes are equal. C_ glucose = C_ urea The concentration of the glucose solution is given by: C_ glucose = 18 180 1 = 0.1 mol dm ⁻³ Let W be the mass of urea in 500 mL ( 0.5 dm ^3 ) of the solution. The concentration of the urea solution is: C_ urea = W 60 0.5 Equating the concentrations: W 30 = 0.1 W = 3 g Answer: 3 g .