# How many grams of manganese (IV) oxide are needed to make 5.6 liters of a 2.1 M solution?

May 6, 2017

Are you sure that $M n {O}_{2}$ is that soluble............?

#### Explanation:

Ordinarily, we would perform the calculation........:

$\text{Concentration"="Moles of solute"/"Volume of solution}$;

and thus $\text{moles of solute"="volume"xx"concentation}$; and

$\text{mass of solute"="volume"xx"concentration"xx"molar mass}$

$= 5.6 \cdot L \times 2.1 \cdot m o l \cdot {L}^{-} 1 \times 86.94 \cdot g \cdot m o {l}^{-} 1 \cong 1000 \cdot g$.

The problem is that $M n {O}_{2}$ is as soluble as a brick, and you would not be able to achieve these concentrations.......This insolubility lent itself to an important (preindustrial) applications, i.e. as a pigment in oil painting. Umber (from Umbria in Italy where the ore was mined) featured in the great paintings of Rembrandt and Caravaggio as a deep orange-brown.