# The molecular weight of NaCl is 58.44 grams/mole. If you had a 1.0 molar solution (1.0 M), you would have to put 58.44 g of salt in 1.0 liter of solution. How many moles of NaCl would you have in 100 mL of this solution?

Jul 12, 2016

$\text{0.1 moles NaCl}$

#### Explanation:

The thing to remember about any solution is that the particles of solute and the particles of solvent are evenly mixed. This essentially means that if you start with a solution of known molarity and take out a sample of this solution, the molarity of the sample will be the same as the molarity of the initial solution.

In your case, you dissolve $\text{58.44 g}$ of sodium chloride in $\text{1.0 L}$ of water and make a $\text{1.0 M}$ sodium chloride solution.

If you take

100 color(red)(cancel(color(black)("mL"))) * "1 L"/(10^3color(red)(cancel(color(black)("mL")))) = "0.1 L"

of this solution, this sample must have the same concentration as the initial solution, i.e. $\text{1.0 M}$.

Notice that the volume of the sample is

$\left(1 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{L"))))/(0.1color(red)(cancel(color(black)("L}}}}\right) = \textcolor{b l u e}{10}$

times smaller than the volume of the initial solution, so it follows that it must contain $\textcolor{b l u e}{10}$ times fewer moles of solute in order to have the same molarity.

Sodium chloride was said to have a molar mass of ${\text{58.44 g mol}}^{- 1}$, which tells you that $1$ mole of sodium chloride has a mass of $\text{58.44 g}$.

Since the initial solution was made by dissolving the equivalent of

$\text{58.44 g " = " 1 mole NaCl}$

in $\text{10 L}$ of water, it follows that the $\text{0.1 L}$ sample must contain

"1 mole NaCl"/color(blue)(10) = color(green)(|bar(ul(color(white)(a/a)color(black)("0.1 moles NaCl")color(white)(a/a)|)))

This is equivalent to saying that the $\text{0.1 L}$ sample contains

$\text{58.44 g NaCl"/color(blue)(10) = "5.844 g NaCl}$

ALTERNATIVELY

You can get the same result by using the formula for molarity, which is

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} c = {n}_{\text{solute"/V_"solution}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Rearrange this equation to solve for n_'solute" and plug in your values to find

$c = {n}_{\text{solute"/V_"solution" implies n_"solute" = c * V_"solution}}$

n_"NaCl" = "1.0 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(100 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(blue)("volume in liters")) = color(green)(|bar(ul(color(white)(a/a)color(black)("0.1 moles NaCl")color(white)(a/a)|)))

Keep in mind that the volume must always be expressed in liters when working with molarity.