Question

0.85% aqueous solution of NaNO${}_3$ is apparently 90% dissociated at 27${}^o$C. Calculate its osmotic pressure.

[R $=$ 0.082/atm K${}^{-1}$ mol${}^{-1}$]

• A. $4.74$ atm
• B. $4.67$ atm
• C. $4.44$ atm
• D. $4.24$ atm

Answer 1

Answer: (B)

$4.67$ atm

The expression for the osmotic pressure of the solution is $\pi =iCRT$

The dissociation of one molecule of $NaN{O}_3$ gives two ions. The solution is 90% dissociated.

Thus $i=\frac{10+2\left(90\right)}{100}=1.9$

0.85% solution means 0.85 g of $NaN{O}_3$ in 100 mL water or 8.5 g of $NaN{O}_3$ in 1 L of water.

The molecular weight of $NaN{O}_3$ is 85 g/L.

Thus $C=8.5g/L=\frac{8.5}{85}=0.1M$

Substitute values in the above expression:

$\pi =1.9\times 0.1\times 0.082\times 300=4.64atm$

Hence, option $B$ is correct.