A 6.84 % ( w/v ) solution of cane sugar (molar mass = 342 g mol ⁻¹ ) is isotonic with a 1.2 % ( w/v ) solution of an unknown non-electrolyte substance. The molar mass of the unknown substance is:

Step-by-step solution

For isotonic solutions at a given temperature, their osmotic pressures are equal ( ₁ = ₂ ). Since both solutes are non-electrolytes, their molar concentrations must be equal ( C₁ = C₂ ). A 6.84 % ( w/v ) solution means 6.84 g of solute is present in 100 mL ( 0.1 L ) of solution. C₁ = Mass Molar mass Volume in L = 6.84 342 0.1 = 0.2 mol L ⁻¹ Similarly, for the 1.2 % ( w/v ) unknown solution: C₂ = 1.2 M₂ 0.1 = 12 M₂ mol L ⁻¹ Equating C₁ and C₂ : 0.2 = 12 M₂ M₂ = 12 0.2 = 60 g mol ⁻¹ Answer: 60 g mol ⁻¹