# Question ba790

Mar 4, 2017

Here's my interpretation of the problem.

#### Explanation:

I'm going to assume that you must determine the number of formula units presents in $\text{1.25 mL}$ of a sodium hydroxide solution of known molarity $c$.

The idea here would be that you can use a solution's molarity as a conversion factor that allows you to figure out the number of moles of solute present in a given volume of solution.

Molarity is defined as the number of moles of solute, which in your case would be sodium hydroxide, present in $\text{1 L}$ of solution.

So, let's say that your sodium hydroxide solution has a molarity of $c$ ${\text{mol L}}^{- 1}$. This means that every $\text{1 L}$ of solution contains $c$ moles of sodium hydroxide.

You can thus use this ratio as a conversion factor to determine the number of moles of sodium hydroxide present in

$1.25 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{mL solution"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * (c color(white)(.)"moles NaOH")/(1color(red)(cancel(color(black)("L solution}}}}$

$= \left(0.00125 \cdot c\right) \textcolor{w h i t e}{.} \text{moles NaOH}$

To convert this to formula units of sodium hydroxide, use Avogadro's constant

(0.00125 * c) color(red)(cancel(color(black)("moles NaOH"))) * overbrace((6.022 * 10^(23)color(white)(.)"formula units NaOH")/(1color(red)(cancel(color(black)("mole NaOH")))))^(color(blue)("Avogadro's constant"))#

$= \left(7.53 \cdot c\right) \cdot {10}^{20} \textcolor{w h i t e}{.} \text{formula units NaOH}$

Keep in mind that this is just my interpretation of what the question would actually ask.