Question

Osmotic pressure of a 0.0103 molar solution of an electrolyte was found to be 0.75 atm at 27c. Calculate vant hoff factor.

Answer 1

we know
$\pi = icrt$
therefore,

$0.75 = i \times 0.0103 \times 0.082 \times 300 \\ i = 2.95$

Answer 2

Answer:

The vant hoff factor is 2.95.

Explanation:

Osmotic pressure of the solution  $\pi$= 0.75 atm

Concentration of the solution = 0.0103 Molar

Temperature of the solution =27 °C = 300 K

van'tHoff factor of the electrolyte = i

$\pi=icRT$

$0.75 atm=i\times 0.0103 mol/l\times 0.0820 L atm/ mol k\times 300 K$

$i=\frac{0.75 atm}{0.0103 mol/l\times 0.0820 L atm/ mol k\times 300 K}$

i = 2.95

The vant hoff factor is 2.95.