A 0.6 % solution of urea (molecular mass = 60 | vedclass

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(A) The molarity of a $0.6 \%$ solution of urea is calculated as: $	ext{Molarity} = \frac{0.6 \ g}{60 \ g/mol} 	imes \frac{1000 \ mL}{100 \ mL} = 0.

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$0.1 \ M$ glucose solution

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(A) The molarity of a $0.6 \%$ solution of urea is calculated as: $	ext{Molarity} = \frac{0.6 \ g}{60 \ g/mol} 	imes \frac{1000 \ mL}{100 \ mL} = 0.1 \ M$. Two solutions are isotonic if they have the same molar concentration of solute particles. For $0.1 \ M$ glucose solution,the concentration is $0.1 \ M$. For $0.6 \%$ glucose solution,the molarity is $\frac{0.6}{180} 	imes 10 = 0.033 \ M$. For $KCl$ solutions,the van't Hoff factor $i = 2$ due to dissociation,so the effective concentration is higher than the molarity. Therefore,a $0.6 \%$ urea solution $(0.1 \ M)$ is isotonic with a $0.1 \ M$ glucose solution.

$A$ $0.6 \%$ solution of urea (molecular mass $= 60$) would be isotonic with

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