Question

If an aqueous solution is 4.48% (w/v) in copper(II) sulfate, CuSO₄ , what is the osmolarity of the solution?

Osmolarity = _______ osmol/L

Answer 1

Ans. Osmolarity is the number of osmoles of solute per liter of solution.

Osmole is the number of moles of a solute that contributes to osmolarity of the solution.

For an electrolyte (produces two or more ions in aqueous solution), number of osmoles is equal to number of ions produced per mol electrolyte.

For non-electrolytes like sucrose, van’t Hoff factor, i=1.

So,

Osmolarity = (i x Molarity) osmole/L ; [where, i = Van’t Hoff factor]

# Given,          [CuSO₄] = 4.48 % (w/v)

Or, [CuSO₄] = 4.48 g CuSO₄ / 100.0 mL                            ; [MW CuSO₄ = 159.6096 g/mol]

                        = (4.48 / 159.6096) mol / 100.0 mL

                        = 0.02807 mol / 0.100 L                              ; [100.0 mL = 0.100 L]

                        = 0.2807 mol/ L

                        = 0.2807 M

Hence, [CuSO₄] = 4.48 % (w/v) = 0.2807 M

# Dissociation of CuSO₄:    

                        CuSO₄(s) --------> Cu²⁺(aq) + SO₄^2-(aq)

Stoichiometry: 1 mol CuSO₄ produces 2 moles of ions.

So, van’t Hoff factor, i = 2

Now,

Osmolarity = (2 x 0.2807) osmole/L = 0.5614 osmole/ L