The molality of an aqueous solution of urea is 3 m . If the density of the solution is 1.18 g/mL , the molarity of the solution is [Given: Molar mass of urea = 60 g mol ⁻¹ ]

Step-by-step solution

Given, molality = 3 m , which means 3 moles of urea are present in 1 kg ( 1000 g ) of solvent (water). Mass of solvent = 1000 g Molar mass of urea = 60 g mol ⁻¹ Mass of urea (solute) = moles molar mass = 3 60 = 180 g Total mass of the solution = Mass of solvent + Mass of solute Total mass of solution = 1000 g + 180 g = 1180 g Given density of solution = 1.18 g/mL Volume of solution = Mass of solution Density = 1180 g 1.18 g/mL = 1000 mL = 1 L Molarity = Number of moles of solute Volume of solution (in L) Molarity =