Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K2SO4 in 2 liter of water at 25° C, assuming that it is completely dissociated.
When $\mathrm{K}_{2} \mathrm{SO}_{4}$ is dissolved in water, $\mathrm{K}^{+}$and $\mathrm{SO}_{4}^{2-}$ ions are produced.
$\mathrm{K}_{2} \mathrm{SO}_{4} \longrightarrow 2 \mathrm{~K}^{+}+\mathrm{SO}_{4}^{2-}$
Total number of ions produced = 3
i =3
Given,
w = 25 mg = 0.025 g
V = 2 L
T = 250C = (25 + 273) K = 298 K
Also, we know that:
R = 0.0821 L atm K-1mol-1
M = (2 × 39) + (1 × 32) + (4 × 16) = 174 g mol-1
Appling the following relation,
$\pi=i \frac{n}{v} \mathrm{R} T$
$=i \frac{w}{M} \frac{1}{v} \mathrm{R} T$
$=3 \times \frac{0.025}{174} \times \frac{1}{2} \times 0.0821 \times 298$
$=5.27 \times 10^{-3} \mathrm{~atm}$