The molecular weight of NaCl is 58.44 grams/mole. If you had a 1.0 molar solution (1.0 M), you would have to put 58.44 g of salt in 1.0 liter of solution. How many moles of NaCl would you have in 100 mL of this solution?

The thing to remember about any solution is that the particles of solute and the particles of solvent are evenly mixed.

My own work

This essentially means that if you start with a solution of known molarity and take out a sample of this solution, the molarity of the sample will be the same as the molarity of the initial solution.

In your case, you dissolve `"58.44 g"` of sodium chloride in `"1.0 L"` of water and make a `"1.0 M"` sodium chloride solution.

If you take

`100 color(red)(cancel(color(black)("mL"))) * "1 L"/(10^3color(red)(cancel(color(black)("mL")))) = "0.1 L"`

of this solution, this sample must have the same concentration as the initial solution, i.e. `"1.0 M"`.

Notice that the volume of the sample is

`(1color(red)(cancel(color(black)("L"))))/(0.1color(red)(cancel(color(black)("L")))) = color(blue)(10)`

times smaller than the volume of the initial solution, so it follows that it must contain `color(blue)(10)` times fewer moles of solute in order to have the same molarity.

Sodium chloride was said to have a molar mass of `"58.44 g mol"^(-1)`, which tells you that `1` mole of sodium chloride has a mass of `"58.44 g"`.

Since the initial solution was made by dissolving the equivalent of

`"58.44 g " = " 1 mole NaCl"`

in `"10 L"` of water, it follows that the `"0.1 L"` sample must contain

`"1 mole NaCl"/color(blue)(10) = color(green)(|bar(ul(color(white)(a/a)color(black)("0.1 moles NaCl")color(white)(a/a)|)))`

This is equivalent to saying that the `"0.1 L"` sample contains

`"58.44 g NaCl"/color(blue)(10) = "5.844 g NaCl"`

ALTERNATIVELY

You can get the same result by using the formula for molarity, which is

`color(blue)(|bar(ul(color(white)(a/a)c = n_"solute"/V_"solution"color(white)(a/a)|)))`

Rearrange this equation to solve for `n_'solute"` and plug in your values to find

`c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution"`

`n_"NaCl" = "1.0 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(100 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(blue)("volume in liters")) = color(green)(|bar(ul(color(white)(a/a)color(black)("0.1 moles NaCl")color(white)(a/a)|)))`

Keep in mind that the volume must always be expressed in liters when working with molarity.