Question

Determine the osmotic pressure of a solution prepared by dissolving $2.5\times {10}^{-2}g$ of $K_2{SO}_4$ in 2L of water at 25, assuming that it is completely dissociated.
(R = 0.0821 L atm $K^{-1}{mol}^{-1}$, Molar mass of $K_2{SO}_4$ = 174 g ${mol}^{-1}$).

Answer 1

$Concentration\left(C\right)=\frac{Moles}{Volume}=\frac{\frac{2.5\times {10}^{-3}}{174}}{2}=7.183\times {10}^{-5}M$

$R=0.0821Latm{K}^{-1}mo{l}^{-1}$ and $T=298K.$

$\therefore$ Osmotic pressure $\left(\pi \right)=CRT=7.183\times {10}^{-5}\times 0.0821\times 298$

$\therefore$ Osmotic pressure $\left(\pi \right)=1.75\times {10}^{-3}atm$