Question #91322

1 Answer
May 13, 2016

Answer:

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.