Question #91322

1 Answer
Stefan V.
May 13, 2016

Approximately #"330 g"#.

Explanation:

The first thing to do here is figure out how many grams of calcium chloride, #"CaCl"_2#, are needed to make your target solution.

As you know, molarity is defined as number of moles of solute per liter of solution. This basically means that a #"1 molar"# solution will contain #1# mole of solute in one liter of solution.

In your case, you have #3# liters of #"1 molar"# solution, which means that you have

#3 color(red)(cancel(color(black)("L solution"))) * overbrace("1 mole CaCl"_2/(1color(red)(cancel(color(black)("L solution")))))^(color(purple)("= 1 molar")) = "3 moles CaCl"_2#

To convert the number of moles of calcium chloride to grams, use the compound's molar mass

#3 color(red)(cancel(color(black)("moles CaCl"_2))) * "111 g"/(1color(red)(cancel(color(black)("mole CaCl"_2)))) = color(green)(|bar(ul(color(white)(a/a)"330 g"color(white)(a/a)|)))#

I'll leave the answer rounded to two sig figs, despite the fact that you only have one sig fig for the volume and molarity of the target solution.

Related questions
See all questions in Molarity
Impact of this question
1580 views around the world
You can reuse this answer
Creative Commons License