Question

What is the osmotic pressure of a solution made from 22.3 g of methanol (MM = 32.04 g/mol) that was added to water to make 321 mL of solution at 25.0 °C?

Answer 1

Answer: The osmotic pressure of a solution is 53.05 atm

Explanation:

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

$\pi=iMRT$

Or,

$\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT$

where,

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

i = Van't hoff factor = 1 (for non-electrolytes)

Mass of solute (methanol) = 22.3 g  

Volume of solution = 321 mL

R = Gas constant = $0.0821\text{ L.atm }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[273+25]=298K$

Putting values in above equation, we get:

$\pi=1\times \frac{22.3\times 1000}{32.04\times 321}\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K$

$\pi=53.05atm$

Hence, the osmotic pressure of a solution is 53.05 atm