Question

$0.85\%$ aqueous solution of $NaN{O}_3$ is apparently $90\%$ dissociated at ${27}^0$C. Calculate its osmotic pressure.

Answer 1

$NaN{O}_3\rightleftharpoons N{a}^{+}+N{O}_3^{-}$

As the number of products is 2 so $n=2$.

$\beta =\frac{i-1}{\frac{1}{n}-1}$

$\beta =$degree of dissociation

$i=$Van't Hoff factor

$0.9=\frac{i-1}{\frac{1}{2}-1}$

$i=0.55$

As concentration of 0.85% aqueous solution, in 1 L 0.85 gram are present.

Thus, concentration is $\frac{0.85}{85}$ moles per litre.

$\pi =iCRT=0.55\times 0.01moles{L}^{-1}\times 0.0821atm-Lmo{l}^{-1}{K}^{-1}\times 300=187.065J/L=0.135atm$

Osmotic pressure is 0.135 atm.