A solution is prepared by dissolving W grams of a non-electrolyte polymer (molar mass = 6000 g mol ⁻¹ ) in enough water to make 500 mL of solution. This solution is found to be isotonic with a solution containing 0.18 g of glucose ( C ₆ H ₁₂ O ₆ ) per 100 mL at the same temperature. The value of W is: (Given atomic masses in g~mol ⁻¹ : C : 12, H : 1, O : 16 )

Step-by-step solution

For isotonic solutions at the same temperature, their osmotic pressures are equal. Since both the polymer and glucose are non-electrolytes, their van 't Hoff factors are i=1 , which implies their molar concentrations must be equal. Molar mass of glucose ( C ₆ H ₁₂ O ₆ ) = 6 12 + 12 1 + 6 16 = 180 g mol ⁻¹ Concentration of glucose solution, C₂ = 0.18 180 1000 100 = 0.01 mol L ⁻¹ Concentration of polymer solution, C₁ = W 6000 1000 500 = W 3000 mol L ⁻¹ Equating the concentrations C₁ = C₂ : W 3000 = 0.01 W = 30 g Answ