0.702 mol / dm³.
Consider a 1,000 mL sample of this solution. Mass of the sample:
[tex]m = \rho \cdot V = 1.167 \;\text{g}\cdot\text{mL}^{-1}\times1,000\;\text{mL} = 1,167\;\text{g}[/tex].
The mass ratio of copper(II) sulfate pentahydrate in this sample is 15 %. In other words, [tex]1,167 \times 15\;\% = 175.1\;\text{g}[/tex] among that 1,167 grams of the solution is [tex]\text{Cu}\text{SO}_4\cdot(\text{H}_2\text{O})_5[/tex].
Number of moles of [tex]\text{Cu}\text{SO}_4\cdot(\text{H}_2\text{O})_5[/tex] in that [tex]175.1\;\text{g}[/tex] of the hydrate:
[tex]n(\text{hydrate}) = \dfrac{m(\text{hydrate})}{M(\text{hydrate})} = \dfrac{175.1}{249.5} = 0.702\;\text{mol}[/tex].
Concentration of the hydrate:
[tex]c(\text{hydrate}) = \dfrac{n(\text{hydrate})}{V} = \dfrac{0.702\;\text{mol}}{1,000\;\text{mL}}=\dfrac{0.702\;\text{mol}}{1\;\text{L}} = 0.702\;\text{mol}\cdot\text{dm}^{-3}[/tex].
