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  4. 0.85%, aqueous solution of NaNO3 is apparently 90% dissociated at 27°C. The osmotic pressure will be (R = 0.082 atm K-1 mol-1)

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Q. 0.85%, aqueous solution of $NaNO_3$ is apparently 90% dissociated at $27^{\circ}C$. The osmotic pressure will be $(R = 0.082 \,atm \,K^{-1}\,\, mol^{-1})$

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Solution:

Molecular weight of $NaNO_3 = 85\, g\, mol^{-1}$
Molarity $= \frac{W\times1000}{M\times V} = \frac{0.85\times1000}{85\times100} = 0.1\,mol\, L^{-1}$
$NaNO_{3}$ solution is $90\%$ dissociated
$\begin{matrix}NaNO_{3}&{\rightleftharpoons}&Na^{+}&+&NO^{-}_{3}\\ 1-0.9&&0.9&&0.9\end{matrix}$
van’t Hoff factor, $i = 1 - 0.9 + 0.9 + 0.9 = 1.9$
$\therefore \quad\pi = 1.9 × 0.1 × 0.082 × 300 = 4.674\, atm$