Question

Determine the osmotic pressure of a solution prepared by dissolving $2.5\times {10}^{-2}g$ of $K_{2}S{O}_4$ in $2L$ of water at ${25}^{\circ }C$, assuming that it is completely dissociated.
$(R=0.0821Latm{K}^{-1}mo{l}^{-1}$, Molar mass of $K_{2}S{O}_4=174gmo{l}^{-1})$.

Answer 1

Given:

Weight of $K_{2}S{O}_4\left(W_2\right)=2.5\times {10}^{-2}g$

Mass of $K_{2}S{O}_4\left(M_2\right)=174gmo{l}^{-1}$

Volume $=2L$

$R=0.0821$ L $at{m}^{-1}Kmo{l}^{-1}$

$T={25}^{o}C=298$K

Osmotic pressure$=\frac{W_2\times R\times T}{M_2\times V}$

$\pi =\frac{2.5\times {10}^{-2}\times 0.0821\times 298}{174\times 2}$

$\pi =\frac{61.1645\times {10}^{-2}}{348}$

$\pi =1.76\times {10}^{-3}$ atm

Therefore, osmotic pressure is $1.76\times {10}^{-3}$ atm.