How many grams of #"CuSO"_4# would be dissolved in 5.1 L of 0.5 M #"CuSO"_4# solution?

It is a molarity question.

1 Answer
Nov 21, 2016

Answer:

#"410 g"#

Explanation:

Molarity is simply a measure of how many moles of solute you have in one liter of a given solution. In order for a solution to have a molarity of #"1 M"#, it must contain #1# mole of solute for every #"1 L"# of solution.

In your case, you have a #"0.5 M"# solution of copper(II) sulfate, #"CuSO"_4#. This tells you that #"1 L"# of this solution contains #0.5# moles of copper(II) sulfate.

Since you have more than #"1 L"# of this solution, you are going to get more than #0.5# moles of solute. More specifically, you will get

#5.1 color(red)(cancel(color(black)("L solution"))) * "0.5 moles CuSO"_4/(1color(red)(cancel(color(black)("L solution")))) = "2.55 moles CuSO"_4#

Finally, use the molar mass of copper(II) sulfate to convert the number of moles to grams

#2.55 color(red)(cancel(color(black)("moles CuSO"_4))) * "159.61 g"/(1color(red)(cancel(color(black)("mole CuSO"_4)))) = color(darkgreen)(ul(color(black)("410 g")))#

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for the molarity of the solution.