A solution is prepared by dissolving 0.3 g of a non-volatile non-electrolyte solute 'A' of molar mass 60 ~g ~mol ⁻¹ and 0.9 g of a non-volatile non-electrolyte solute ' B ' of molar mass 180 ~g ~mol ⁻¹ in 100 ~mL H ₂ O at 27^ C . Osmotic pressure of the solution will be [Given: R =0.082 ~L ~atm ~K ⁻¹ ~mol ⁻¹ ]

Step-by-step solution

Calculate osmotic pressure using = nRT V where n is total moles of solute particles. Moles of solute A: n_A = 0.3 g 60 g/mol = 0.005 mol Moles of solute B: n_B = 0.9 g 180 g/mol = 0.005 mol Total moles: n_ total = 0.005 + 0.005 = 0.01 mol Volume of solution: 100 mL = 0.1 L Temperature: 27°C = 300 K Osmotic pressure: = 0.01 0.082 300 0.1 = 0.246 0.1 = 2.46 atm