0.60 g of urea is dissolved in 360 g of water at room temperature. The vapour pressure of pure water is 35 mm of Hg , the lowering of vapour pressure at this temperature is: [Molar mass of urea is 60 ~g ~mol ⁻¹ ]

Step-by-step solution

The molar mass of urea is M_ urea = 60 g mol ⁻¹ . The mass of urea is w_ urea = 0.60 g . The number of moles of urea is n_ urea = 0.60 60 = 0.01 mol . The mass of water is w_ water = 360 g . The molar mass of water is M_ water = 18 g mol ⁻¹ . The number of moles of water is n_ water = 360 18 = 20 mol . According to Raoult's law, the relative lowering of vapour pressure is given by P^0 - P_s P^0 = n_ urea n_ urea + n_ water . Substituting the values, P^0 - P_s 35 = 0.01 0.01 + 20 = 0.01 20.01 0.01 20 = 0.0005 . The