Question

Determine the osmotic pressure of a solution prepared by dissolving 25 mg of $K_2SO_4$ in 2 litre of water at $25\degree C$, assuming that it is completely dissociated.

Answer 1

The correct answer is: $5.27 \times 10^{-3}atm$
When $K_2SO_4$ is dissolved in water $K^+$ and $SO_4^{2-}$ ions are produced.
$K_2SO_4→2K^++SO_4^{2-}$
Total number of ions produced = 3
$\therefore i =3$
Given,
$w = 25 mg = 0.025 g$
$V = 2 L$
$T =$ $25 \degree C$ $= (25 + 273) K = 298 K$
Also, we know that:
$R = 0.0821\, L\, atm\, K^{-1}mol^{-1 }$
$M = (2 × 39) + (1 × 32) + (4 × 16) = 174 g mol^{-1 }$
Appling the following relation,
$π=i\frac{n}{v}RT$
$=i\frac{w}{M}\frac{1}{v}RT$
$=3 \times \frac{0.025}{174} \times \frac{1}{2} \times 0.0821 \times 298$
$=5.27 \times 10^{-3}atm$