# Question 37542

Jun 17, 2016

$\text{76.9 L}$

#### Explanation:

You can find the volume of water needed to dissolve $\text{1.00 g}$ of calcium carbonate, ${\text{CaCO}}_{3}$, by using the compound's solubility in water, which you can find here

https://en.wikipedia.org/wiki/Calcium_carbonate

So, calcium carbonate has a solubility of $\text{0.013 g/L}$ of water at ${25}^{\circ} \text{C}$.

This tells you that at that temperature, you can only hope to dissolve $\text{0.013 g}$ of calcium carbonate per liter of water, i.e. a saturated solution of calcium carbonate will contain $\text{0.013 g}$ of dissolved calcium carbonate at ${25}^{\circ} \text{C}$.

You can thus use the compound's solubility as a conversion factor to calculate the volume of water needed to dissolve $\text{1.00 g}$ of calcium carbonate

1.00color(red)(cancel(color(black)("g CaCO"_3))) * "1 L water"/(0.013 color(red)(cancel(color(black)("g CaCO"_3)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("76.9 L water")color(white)(a/a)|)))#

The answer is rounded to three sig figs.

Therefore, you can say that at ${25}^{\circ} \text{C}$, you need $\text{76.9 L}$ of water in order to dissolve $\text{1.00 g}$ of calcium carbonate.