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.($R=0.0821Latm{K}^{-1}mo{l}^{-1}$, Molar mass of $K_2{SO}_4=174gmo{l}^{-1}$).

• A. $5.276\times {10}^{-3}$
• B. $2.76\times {10}^{-3}$
• C. $1.43\times {10}^{-2}$
• D. $1.22\times {10}^{-4}$

Answer 1

The correct option is A $5.276\times {10}^{-3}$

$\begin{array}{c}\text{we know}\\ \pi =icRT\Rightarrow \pi =\frac{inRT}{V}\end{array}$

$\begin{array}{c}\pi =i\times \frac{W}{M}\times \frac{1}{V}RT\\ \text{Given,}W=2.5\times {10}^{-2}g\\ T={25}^{\circ }C=298K\\ \text{M=molar mass of}{K}_{2}S{O}_4=2\times 39+32+4\times 16\\ =174gmo{l}^{-1}\\ K_{2}S{O}_4\text{Dessociate completly as}\\ K_{2}S{O}_4⟶2K^{+}+S{O}_4^{2-}\\ \text{Ions produced = 3 So, i=3}\\ \text{Hence,}\pi =\frac{3\times 0.025}{174}\times \frac{1}{2}\times 0.0821\times 298\\ \pi =5.27\times {10}^{-3}atm\end{array}$