# How many moles of NaOH are present in 19.0 mL of 0.150 M NaOH?

Nov 7, 2015

$\text{0.00285 moles}$

#### Explanation:

Molarity is defined as moles of solute, which in your case is sodium hydroxide, $\text{NaOH}$, divided by liters of solution.

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

SImply put, a $\text{1-M}$ solution will have $1$ mole of solute dissolved in $1$ liter of solution.

Now, you know that your solution has a molarity of $\text{0.150 M}$ and a volume of $\text{19.0 mL}$.

Since the volume is much smaller than $\text{1 L}$, you can expect to have fewer moles of sodium hydroxide in this sample than you would have had in a full liter of solution.

Convert the volume of the sample from mililiters to liters by using the conversion factor

$\text{1 L" = 10^3"mL}$

to get

19.0color(red)(cancel(color(black)("mL"))) * "1 L"/(10^3color(red)(cancel(color(black)("mL")))) = 19.0 * 10^(-3)"L"

This means that the number of moles of solute you get in this sample will be equal to

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

n = 0.150"moles"/color(red)(cancel(color(black)("L"))) * 19.0 * 10^(-3)color(red)(cancel(color(black)("L"))) = color(green)("0.00285 moles NaOH")