Question

Question #37542

Answer 1

`"76.9 L"`

Explanation:

You can find the volume of water needed to dissolve `"1.00 g"` of calcium carbonate, `"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 `"0.013 g/L"` of water at `25^@"C"`.

This tells you that at that temperature, you can only hope to dissolve `"0.013 g"` of calcium carbonate per liter of water, i.e. a saturated solution of calcium carbonate will contain `"0.013 g"` of dissolved calcium carbonate at `25^@"C"`.

You can thus use the compound's solubility as a conversion factor to calculate the volume of water needed to dissolve `"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^@"C"`, you need `"76.9 L"` of water in order to dissolve `"1.00 g"` of calcium carbonate.