If molarity moles/liter, then how many grams of NaCl are needed to make 3.550 liter of a 1.30 M solution?

Molarity is indeed defined as moles of solute per liter of solution, but it can also be expressed as grams of solute per liter of solution.

As you know, a `"1 M"` solution contains exactly `1` mole of solute in `"1 L"` of solution.

In your case, a `"1.30 M"` sodium chloride solution will contain `1.30` moles of sodium chloride, the solute, for every `"1 L"` of solution.

You can use the molar mass of sodium chloride to convert this to grams of sodium chloride for every `"1 L"` of solution.

`1.30 color(red)(cancel(color(black)("moles NaCl"))) * "58.44 g"/(1color(red)(cancel(color(black)("mole NaCl")))) = "75.97 g"`

You can now say that a `"1.30 M"` sodium chloride solution must contain `"75.97 g"` of sodium chloride for every `"1 L"` of solution. Now all you have to do is use this concentration to figure out how many grams would be needed to make your sample

`3.550 color(red)(cancel(color(black)("L solution"))) * "75.97 g NaCl"/(1color(red)(cancel(color(black)("L solution")))) = color(darkgreen)(ul(color(black)("270. g NaCl")))`

The answer is rounded to three sig figs.