Determine the osmotic pressure of the solution prepared by dissolving 25 mg of K₂SO₄ in 2L of water at 25^oC. (Assuming it is to be completely dissociated)

K₂SO₄ dissolved = 25 mg = 0.025 g
Volume = 2 L; Temperature, T = 298 K
Molar mass of K₂SO₄ = 174 g mol^-1
K₂SO₄ dissociates into 2 K⁺ and ${\mathrm{SO}}_4^{2-}$ions, so i = 3
$\begin{array}{l}\mathrm{\pi }=\mathrm{iCRT}\\ \mathrm{\pi }=\mathrm{i}\times \frac{\mathrm{w}}{\mathrm{M}}\times \frac{\mathrm{RT}}{\mathrm{V}}\\ =3\times \frac{0.025}{174}\times \frac{1}{2}\times 0.0821\times 298\\ =5.27\times {10}^{-3}\mathrm{atm}\end{array}$