A solution containing 8.6 g of urea per lite | vedclass

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(B) Since the solutions are isotonic,their osmotic pressures are equal: $\pi_1 = \pi_2$.Using the formula $\frac{n_1}{V_1} = \frac{n_2}{V_2}$,where $n

Vedclass · Accepted Answer

$34.89$

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(B) Since the solutions are isotonic,their osmotic pressures are equal: $\pi_1 = \pi_2$.Using the formula $\frac{n_1}{V_1} = \frac{n_2}{V_2}$,where $n$ is the number of moles and $V$ is the volume in liters.For urea: $w_1 = 8.6 \ g$,$M_1 = 60 \ g/mol$,$V_1 = 1 \ L$.For the organic solute: $w_2 = 0.5 \ g$ (in $100 \ mL$ or $0.1 \ L$),$V_2 = 0.1 \ L$.Substituting the values: $\frac{8.6}{60 	imes 1} = \frac{0.5}{M_2 	imes 0.1}$.$\frac{8.6}{60} = \frac{5}{M_2}$.$M_2 = \frac{5 	imes 60}{8.6} = \frac{300}{8.6} \approx 34.8837 \ g/mol$.Rounding to the nearest option,$M_2 = 34.89$.

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

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