At a constant temperature,the osmotic pressur | vedclass

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(D) The osmotic pressure $\pi$ is given by the formula $\pi = i \cdot C \cdot R \cdot T$,where $i$ is the van't Hoff factor,$C$ is the molar concentra

Vedclass · Accepted Answer

$0.6$

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(D) The osmotic pressure $\pi$ is given by the formula $\pi = i \cdot C \cdot R \cdot T$,where $i$ is the van't Hoff factor,$C$ is the molar concentration,$R$ is the gas constant,and $T$ is the temperature.Since the osmotic pressures are equal at the same temperature,we have $i_1 \cdot C_1 = i_2 \cdot C_2$.For $NH_4NO_3$,the dissociation is $NH_4NO_3 ightarrow NH_4^+ + NO_3^-$,so $i_1 = 2$.Given $C_1 = 1.5 \ M$,then $i_1 \cdot C_1 = 2 	imes 1.5 = 3.0$.For $Al_2(SO_4)_3$,the dissociation is $Al_2(SO_4)_3 ightarrow 2Al^{3+} + 3SO_4^{2-}$,so $i_2 = 5$.Given $C_2 = x \ M$,then $i_2 \cdot C_2 = 5 	imes x = 5x$.Equating the two: $5x = 3.0$.Therefore,$x = \frac{3.0}{5} = 0.6 \ M$.

At a constant temperature,the osmotic pressures of a $1.5 \ M \ NH_4NO_3$ aqueous solution and an $x \ M \ Al_2(SO_4)_3$ aqueous solution are approximately equal. Find the value of $x$. (Assume $100 \%$ dissociation of the solute in the solution.)

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