Calculate the relative lowering of vapour pressure of solution containing 3 g urea in 50 g water. [ molar mass of urea =60 ~g ~mol ⁻¹ ]

Step-by-step solution

Using Raoult's law, the relative lowering of vapor pressure equals the mole fraction of solute: P P^0 = X_ urea . Given 3 g urea (molar mass 60 g mol ⁻¹ ) and 50 g water (molar mass 18 g mol ⁻¹ ), the moles are: n_ urea = 3 60 = 0.05 mol n_ water = 50 18 2.7778 mol The mole fraction of urea is: X_ urea = 0.05 0.05 + 2.7778 = 0.05 2.8278 0.01768 Rounding to three decimal places gives P P^0 0.018 , which corresponds to option C . Final answer: C