A solution containing 8.6 g of urea per lite | vedclass

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Question

(B) For isotonic solutions,the molar concentrations are equal: $C_1 = C_2$.Concentration of urea $(C_1)$: $C_1 = \frac{	ext{mass}}{	ext{molar mass}

Vedclass · Accepted Answer

$34.89$

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(B) For isotonic solutions,the molar concentrations are equal: $C_1 = C_2$.Concentration of urea $(C_1)$: $C_1 = \frac{	ext{mass}}{	ext{molar mass} 	imes 	ext{volume in L}} = \frac{8.6 \ g}{60 \ g/mol 	imes 1 \ L} = \frac{8.6}{60} \ mol/L$.Concentration of organic solute $(C_2)$: $0.5\% \ (w/v)$ means $0.5 \ g$ in $100 \ mL$,which is $0.5 \ g$ in $0.1 \ L$.$C_2 = \frac{0.5 \ g}{M 	imes 0.1 \ L} = \frac{5}{M} \ mol/L$,where $M$ is the molar mass.Equating $C_1 = C_2$:$\frac{8.6}{60} = \frac{5}{M}$$M = \frac{5 	imes 60}{8.6} = \frac{300}{8.6} \approx 34.88 \ g/mol$.

$A$ solution containing $8.6 \ g$ of urea per liter is isotonic with a $0.5\% \ (w/v)$ solution of a non-volatile organic solute. What is the molar mass of the organic solute?

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