A 5.25\% solution of a substance is isotonic with a 1.5\% solution of urea (molar mass = 60 g mol^-1 ) in the same solvent. If the densities of both t - Chemistry MCQ AIEEE 2007 2026 | ExamTest.live

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A 5.25% solution of a substance is isotonic with a 1.5% solution of urea (molar mass = 60 g mol^{- 1}) in the same solvent. If the densities of both the solutions are assumed to be equal to 1.0 g cm^{- 3}, molar mass of the substance will be:

Solution & Step-by-step Explanation

Isotonic solutions have the same osmotic pressure (π_{1} = π_{2}).At the same temperature: C_{1} RT = C_{2} RT ⟹ C_{1} = C_{2} .Molar concentration C = \frac{Mass in 1000 mL}{Molar Mass} .Given 5.25% mass by volume (since density \approx 1 g/mL): C_{1} = \frac{52.5}{M} For 1.5% urea solution: C_{2} = \frac{15}{60} Equating C_{1} and C_{2} : \frac{52.5}{M} = \frac{15}{60} M = \frac{52.5 \times 60}{15} = 52.5 \times 4 = 210 g mol^{- 1}

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A 5.25% solution of a substance is isotonic with a 1.5% solution of urea (molar mass = 60 g mol^{- 1}) in the same solvent. If the densities of both the solutions are assumed to be equal to 1.0 g cm^{- 3}, molar mass of the substance will be:

90.0 g mol^{- 1}

115.0 g mol^{- 1}

105.0 g mol^{- 1}

210.0 g mol^{- 1}

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