Question

QUESTION 2.41

Determine the osmotic pressure of a solution prepared by dissolving 25 mg of $K_{2}S{O}_4$ in 2L of water at ${25}^{\circ }C$, assuming that it is completely dissociated.

Answer 1

Amount of $K_{2}S{O}_4$ dissolved = 25 mg = 0.025g
Volume of solution = 2L
$T={25}^{\circ }C=25+273K=298K$
Molar mass of
$K_{2}S{O}_4=(2\times 39)+\left(32\right)+(4\times 16)=174gmo{l}^{-1}$
Since, $K_{2}S{O}_4$ dissociates completely as
$K_{2}S{O}_4\stackrel{aq}{\to }2K^{+}+S{O}_4^{2-}$
Total ions produced after dissociation (per mole) = 3
So, i = 3, $\pi =iCRT=i\times \frac{n}{V}RT=\frac{i\times W\times R\times T}{M\times V}$
$=\frac{3\times \left(0.025g\right)\times \left(0.0821Latm{K}^{-1}mo{l}^{-1}\right)\times \left(298K\right)}{\left(174gmo{l}^{-1}\right)\times \left(2L\right)}$
$=5.27\times {10}^{-3}atm$