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

##### 1 Answer

#### Explanation:

As you know, molarity is defined as the number of *moles of solute*, which in your case is sodium chloride, *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

In your case, the volume of the solution is said to be equal to *per liter*, you can say that you will get **in one liter** of this solution.

Since you have *a little less than half* of a liter, i.e. *fewer than half* of

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"^(-)#.*