Question

a solution 0:1 m of na2so4 ,is dissolved to the extent of 95%. what would be its osmotic pressure at 27 °c​

Answer 1

dissociation of $Na_2SO_4$ is...

$Na_2SO_4\Leftrightarrow 2Na^++SO_4^{--}$

here, one molecule sodium sulphate dissociates to form three molecules.

so, n = 3

now, Van't Hoff's factor, i = $1-\alpha+n\alpha$, where $\alpha$ is degree of dissociation.

here, $\alpha=0.95$

so, $i=1-0.95+3\times0.95=2.9$

now, $\pi=iCRT$

where, C denotes concentration, R denotes universal gas constant and T is temperature in Kelvin.

here, i = 2.9 , C = 0.1M, R = 0.082 and T = 27°C or 300K

so, π = 2.9 × 0.1 × 0.082 × 300K = 7.134 atm

hence, osmotic pressure is 7.134 atm.