Question #37542

1 Answer
Stefan V.
Jun 17, 2016

"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.

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