A solution containing 10 g per dm^3 of urea (molecular mass = 60 g mol^-1 ) is isotonic with a 5\% solution of a non-volatile solute. The molecular ma - Chemistry MCQ Chemistry | ExamTest.live

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A solution containing 10 g per dm^{3} of urea (molecular mass = 60 g mol^{- 1}) is isotonic with a 5% solution of a non-volatile solute. The molecular mass of this non-volatile solute is :

A $25 g mol^{- 1}$
B $300 g mol^{- 1}$
C $350 g mol^{- 1}$
D $200 g mol^{- 1}$

Solution & Step-by-step Explanation

Isotonic solutions have the same molar concentration (C_{1} = C_{2}). C_{1} (urea) = \frac{10}{60} mol/L .A 5% solution means 5 g in 100 mL or 50 g in 1000 mL (1 dm^{3}). C_{2} (non-volatile solute) = \frac{50}{M} mol/L .Equating the two: \frac{10}{60} = \frac{50}{M} \Rightarrow M = \frac{50 \times 60}{10} = 300 g mol^{- 1} .

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A solution containing 10 g per dm^{3} of urea (molecular mass = 60 g mol^{- 1}) is isotonic with a 5% solution of a non-volatile solute. The molecular mass of this non-volatile solute is :

25 g mol^{- 1}

300 g mol^{- 1}

350 g mol^{- 1}

200 g mol^{- 1}

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A solution containing $10 g per dm^{3}$ of urea (molecular mass = $60 g mol^{- 1}$ ) is isotonic with a $5%$ solution of a non-volatile solute. The molecular mass of this non-volatile solute is :

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A solution containing 10g per dm3 of urea (molecular mass = 60g mol−1) is isotonic with a 5% solution of a non-volatile solute. The molecular mass of this non-volatile solute is :

Solution & Step-by-step Explanation

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A solution containing $10 g per dm^{3}$ of urea (molecular mass = $60 g mol^{- 1}$ ) is isotonic with a $5%$ solution of a non-volatile solute. The molecular mass of this non-volatile solute is :