# Question e6e4b

Feb 4, 2015

The solution's osmolarity will be $\text{0.0078 osmol/L}$.

Osmolarity is defined as the number of osmoles of solute per liter of solution. One osmol represents 1 mole of particles that contributes to the osmotic pressure of the solution. In your case,

$C u S {O}_{4 \left(a q\right)} r i g h t \le f t h a r p \infty n s C {u}_{\left(a q\right)}^{2 +} + S {O}_{4 \left(a q\right)}^{2 -}$

One mole of copper (II) sulfate will dissociate and produce 2 osmoles of ions in solution.

Now you need to determine the number of moles of copper (II) sulfate that your solution will contain. You can do this by using the given solubility of $\text{5 g/L}$. Solubility is defined as

$\text{solubility" = "mass"/"volume" => "mass" = "solubility" * "volume}$

So,

${m}_{C u S {O}_{4}} = \text{5 g/L" * 25 * 10^(-3)"L" = "0.125 g}$

Now use the given molar mass to go from moles to grams

$\text{0.125 g" * ("1 mole")/("159.62 g") = "0.00078 moles}$

Use this to determine the molarity of the final solution

C = n/V = ("0.00078 moles")/(200*10^(-3)"L") = "0.0039 mol/L"#

This means that the solution's osmolarity will be

$0.0039 \text{moles"/"L" * ("2 osmol")/("1 mol") = "0.0078 osmol/L}$