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

Dec 12, 2015

$\text{0.144 moles}$

#### Explanation:

As you know, molarity is defined as the number of moles of solute, which in your case is sodium chloride, $\text{NaCl}$, divided by liters of solution.

$\textcolor{b l u e}{\text{molarity" = "moles of solute"/"liters of solution}}$

Simply put, dissolving one mole of a solute in one liter of solution will produce a $\text{1.00-M}$ solution.

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

Since you have a little less than half of a liter, i.e. $\text{467 mL}$, you can expect to have fewer than half of $\text{0.244 moles}$.

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

$\textcolor{b l u e}{c = \frac{n}{V} \implies n = c \cdot V}$

$n = 0.244 \text{moles"/color(red)(cancel(color(black)("L"))) * 467 * 10^(-3) color(red)(cancel(color(black)("L"))) = "0.1139 moles NaCl}$

Rounded to three sig figs, the answer will be

${n}_{N a C l} = \textcolor{g r e e n}{\text{0.114 moles}}$

SIDE NOTE Because sodium chloride dissociates completely in aqueous solution, you wouldn't actually say that you have a $\text{0.244 M}$ sodium chloride solution.

Instead, you would say that you have a solution that is $\text{0.244 M}$ in sodium cations, ${\text{Na}}^{+}$, and $\text{0.244 M}$ in chloride anions, ${\text{Cl}}^{-}$.