Question

An aqueous solution is 15.0% by mass of copper(II) sulfate pentahydrate, CuSO4∙5H2O.What

Answer 1

0.702 mol / dm³.

Explanation

Consider a 1,000 mL sample of this solution. Mass of the sample:

$m = \rho \cdot V = 1.167 \;\text{g}\cdot\text{mL}^{-1}\times1,000\;\text{mL} = 1,167\;\text{g}$.

The mass ratio of copper(II) sulfate pentahydrate in this sample is 15 %. In other words, $1,167 \times 15\;\% = 175.1\;\text{g}$ among that 1,167 grams of the solution is $\text{Cu}\text{SO}_4\cdot(\text{H}_2\text{O})_5$.

Number of moles of $\text{Cu}\text{SO}_4\cdot(\text{H}_2\text{O})_5$ in that $175.1\;\text{g}$ of the hydrate:

$n(\text{hydrate}) = \dfrac{m(\text{hydrate})}{M(\text{hydrate})} = \dfrac{175.1}{249.5} = 0.702\;\text{mol}$.

Concentration of the hydrate:

$c(\text{hydrate}) = \dfrac{n(\text{hydrate})}{V} = \dfrac{0.702\;\text{mol}}{1,000\;\text{mL}}=\dfrac{0.702\;\text{mol}}{1\;\text{L}} = 0.702\;\text{mol}\cdot\text{dm}^{-3}$.