Determine the osmotic pressure of a solution prepared by dissolving $25 \, mg$ of $K_{2}SO_{4}$ in $2 \, L$ of water at $25^{\circ} C$,assuming that it is completely dissociated.

(N/A) When $K_{2}SO_{4}$ is dissolved in water,it dissociates as follows:
$K_{2}SO_{4} \longrightarrow 2K^{+} + SO_{4}^{2-}$
Since it is completely dissociated,the van't Hoff factor $i = 3$.
Given:
$w = 25 \, mg = 0.025 \, g$
$V = 2 \, L$
$T = 25^{\circ} C = 298 \, K$
$R = 0.0821 \, L \, atm \, K^{-1} \, mol^{-1}$
Molar mass of $K_{2}SO_{4} (M) = (2 \times 39) + 32 + (4 \times 16) = 174 \, g \, mol^{-1}$
Using the formula for osmotic pressure:
$\pi = i \times \frac{w}{M \times V} \times R \times T$
$\pi = 3 \times \frac{0.025}{174 \times 2} \times 0.0821 \times 298$
$\pi = 3 \times 1.4368 \times 10^{-4} \times 24.4658$
$\pi \approx 1.05 \times 10^{-2} \, atm$