Question

0.85% (w/v) aqueous solution of $NaN{O}_3$ is apparently 90% dissociated at ${27}^{\circ }$. Calculate its osmotic pressure. $(R=8.82latm{K}^{-1}mo{l}^{-1})$

• A. $2.32atm$
• B. $4.64atm$
• C. $9.28atm$
• D. None of these

Answer 1

The correct option is B $4.64atm$

The molar mass of sodium nitrate is $85$ g/mol.
$0.85\%$ w/v means $\frac{0.85}{100}\times \frac{1000}{85}=0.1$ M
$NaN{O}_3\rightleftharpoons N{a}^{+}+N{O}_3^{-}$
The degree of dissociation, $\alpha =0.9$
The vant Hoff's factor, $i=[1+(n^{\prime }-1\left)\alpha \right]=[1+(2-1\left)0.9\right]=1.9$
The osmotic pressure, $\Pi =iCRT=1.9\times 0.1\times 0.08206\times 300=4.64atm$