The mole fraction of urea in an aqueous urea solution containing 900 g of water is 0.05. If the density of the solution is 1.2 g cm-3, the molarity of urea solution is ____.
(Given data: Molar masses of urea and water are 60 g mol–1 and 18 g mol–1, respectively)
Mole fraction of urea in aqueous solution = 0.05
0.05 = Moles of urea / Moles of urea + 900 / 18
0.05 (Moles of urea) + 2.5 = Moles of urea
Moles of Urea = 2.63 moles
Mass of urea = 2.63 × 60 = 158 gm
Mass of solution = 158 + 900 = 1058 gm
Density = 1.2 g cm3
Volume of solution = 1058 / 1.2
= 881.67 ml
Molarity = nsolute / Vsolution(innL) × 1000
= 2.63 / 881.67 × 1000
= 2.98
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