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.

Related questions
See all questions in Factors Affecting Solubility
Impact of this question
8202 views around the world
You can reuse this answer
Creative Commons License