Which solution is isotonic with 6 % w/v aque | vedclass

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(B) For two solutions to be isotonic,their molar concentrations (osmolarity) must be equal. First,calculate the molarity of the $6 \% \ w/v$ urea solu

Vedclass · Accepted Answer

$0.5$

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(B) For two solutions to be isotonic,their molar concentrations (osmolarity) must be equal. First,calculate the molarity of the $6 \% \ w/v$ urea solution: $M = \frac{	ext{mass of solute (g)}}{	ext{molar mass (g/mol)} 	imes 	ext{volume of solution (L)}} = \frac{6 \ g}{60 \ g \ mol^{-1} 	imes 0.1 \ L} = 1 \ M$. Since urea is a non-electrolyte,its van't Hoff factor $i = 1$. For $NaCl$,which is a strong electrolyte,$i = 2$. The condition for isotonicity is $i_1 C_1 = i_2 C_2$. $1 	imes 1 \ M = 2 	imes C_2$. $C_2 = 0.5 \ M$. Therefore,a $0.5 \ M \ NaCl$ solution is isotonic with the $1 \ M$ urea solution.

Which solution is isotonic with $6 \% \ w/v$ aqueous solution of urea (in $M \ NaCl$)? [Molar mass of Urea $= 60 \ g \ mol^{-1}$]

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