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

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.