Question #bf497

You know that the molarity of a solution tells you the number of moles of solute present in exactly #"1 L" = 10^3color(white)(.)"mL"# of the solution, so your starting point here will be to find the number of moles of copper(II) sulfate present in the sample.

To do that, sue the molar mass of the compound.

#12.7 color(red)(cancel(color(black)("g"))) * "1 mole CuSO"_4/(159.62color(red)(cancel(color(black)("g")))) = "0.07956 moles CuSO"_4#

So, you know that #"750.0 mL"# of this solution contain #0.07956# moles of copper(II) sulfate, the solute. You can use the known composition of the solution as a conversion factor to figure out how many moles of solute would be present in #10^3color(white)(.)"mL = 1 L"# of this solution.

#10^3 color(red)(cancel(color(black)("mL solution"))) * "0.07956 moles CuSO"_4/(750.0color(red)(cancel(color(black)("mL solution")))) = "0.10608 moles CuSO"_4#

Since this represents the number of moles of solute present in #"1 L"# of this solution, you can say that the molarity of the solution is equal to

#color(darkgreen)(ul(color(black)("molarity = 0.106 mol L"^(-1))))#

The answer is rounded to three sig figs, the number of sig figs you have for the mass of copper(II) sulfate.