Question

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

Answer 1

Given that: Mass of $K_{2}S{O}_4,w=25mg=0.025g$

Volume $V=2L$

$T=25+273=298K$

The reaction of dissociation of $K_{2}S{O}_4$:

$K_{2}S{O}_4\to 2K^{+}+S{O}_4^{2-}$

Number if ions produced = 2+ 1 = 3

So van’t Hoff factor $i=3$

Use the formula of Osmotic pressure as:

$\pi =i.\frac{n}{V}RT$

$\pi =i.\frac{w}{MV}RT$

where $R$= Gas constant = $0.0821Latm{K}^{-1}mol-1$

Molar mass of $K_{2}S{O}_4,M=2\times 39+1\times 32+4\times 16=174g/mol$

Upon substitution we get:

$\pi =3\times \frac{0.025}{174\times 2}\times 0.0821\times 298$

$\pi =5.27\times {10}^{-3}atm$