How many moles of NaCl are contained in 467 mL of a .244M solution of NaCl?

As you know, molarity is defined as the number of moles of solute, which in your case is sodium chloride, `"NaCl"`, divided by liters of solution.

`color(blue)("molarity" = "moles of solute"/"liters of solution")`

Simply put, dissolving one mole of a solute in one liter of solution will produce a `"1.00-M"` solution.

In your case, the volume of the solution is said to be equal to `"467 mL"` and the molarity of the solution to `"0.244 M"`. Since molarity tells you number of moles per liter, you can say that you will get `"0.244 moles"` of sodium chloride in one liter of this solution.

Since you have a little less than half of a liter, i.e. `"467 mL"`, you can expect to have fewer than half of `"0.244 moles"`.

More specifically, the solution will contain - do not forget to convert the volume of the solution from milliliters to liters!

`color(blue)(c = n/V implies n = c * V)`

`n = 0.244 "moles"/color(red)(cancel(color(black)("L"))) * 467 * 10^(-3) color(red)(cancel(color(black)("L"))) = "0.1139 moles NaCl"`

Rounded to three sig figs, the answer will be

`n_(NaCl) = color(green)("0.114 moles")`

SIDE NOTE Because sodium chloride dissociates completely in aqueous solution, you wouldn't actually say that you have a `"0.244 M"` sodium chloride solution.

Instead, you would say that you have a solution that is `"0.244 M"` in sodium cations, `"Na"^(+)`, and `"0.244 M"` in chloride anions, `"Cl"^(-)`.