What is the osmotic pressure, in atm, of a 0.161 M solution of MgCl₂ at 37.0 °C? (assume complete dissociation).

Answer 1

This is a colligative property question. Based on the osmotic pressure formula:

p = iMRT

Where: p = osmotic pressure

i = Van’t Hoff’s factor (the ratio of particles after dissociation to the particles before dissociation)

M = molarity of the solution (mol L^-1)

R = universal gas constant (0.0821 L atm mol^-1K^-1)

T = Kelvin temperature (K) = Celsius temperature (^oC) + 273

As 1 MgCl₂ can dissociate to make 1 Mg²⁺ and 2 Cl^-, i.e. 1 MgCl₂ particle produces 3 particles (1 from Mg²⁺ and 2 from Cl^-) after dissociation;  i = (2+1)/1 = 3

Molarity (M) = 0.161 M = 0.161 mol L^-1

T (K) = T (^oC) + 273 = 37 + 273 = 310 K

Then, plug in the numbers to obtain

p = 3 × 0.161 mol L^-1 × 0.0821 L atm mol^-1K^-1 × 310 K = 12.3 atm