Question

A $0.85\%$ aqueous solution of $NaN{O}_3$ is apparently $90\%$ dissociated at ${27}^{o}C$. The approximate osmotic pressure is:

• A. 5 atm
• B. 4 atm
• C. 6 atm
• D. 3 atm

Answer 1

The correct option is A 5 atm

Number of moles, $NaN{O}_3=\frac{0.85}{85}=0.01mol$

Here 0.85 gm of NaNO3 is dissolved in 100 ml of solution= 0.1L of solution,

R=gas constant=R = 0.0821 L.atm/mol.K

Temperature = 27 °C = 273 +27 = 300 K

$\pi =\frac{n}{V}RT=\frac{0.01\times 0.082\times 300}{0.1}$

$=4.674\simeq 5atm$.

Hence, the correct option is $A$