Question

Calculate the osmotic pressure (in torr) of 6.00 L of an aqueous 0.958 M solution at 30.°C, if the solute concerned is totally ionized into three ions (e.g., it could be Na2SO4 or MgCl2).

Answer 1

Answer: The osmotic pressure is 54307.94 Torr.

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iCRT$

where,

$\pi$ = osmotic pressure of the solution = ?

i = Van't hoff factor = 3

C = concentration of solute = 0.958 M

R = Gas constant = $62.364\text{ L Torr }mol^{-1}K^{-1}$

T = temperature of the solution = $30^oC=[30+273]K=303K$

Putting values in above equation, we get:

$\pi=3\times 0.958mol/L\times 62.364\text{ L. Torr }mol^{-1}K^{-1}\times 303K\\\\\pi=54307.94Torr$

Hence, the osmotic pressure is 54307.94 Torr.