# How would you prepare 250 mL of 0.158 M solution of MgCl_2 from a 3.15 M stock solution?

Nov 20, 2015

Here's what I got.

#### Explanation:

The idea here is that you need to figure out how many moles of magnesium chloride, ${\text{MgCl}}_{2}$, you need to have in the target solution, then use this value to determine what volume of the stock solution would contain this many moles.

As you know, molarity is defined as the number of moles of solute, which in your case is magnesium chloride, divided by liters of solution.

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

So, how many moles of magnesium chloride must be present in the target solution?

$c = \frac{n}{V} \implies n = c \cdot V$

$n = {\text{0.158 M" * 250.0 * 10^(-3)"L" = "0.0395 moles MgCl}}_{2}$

Now determine what volume of the target solution would contain this many moles of magnesium chloride

$c = \frac{n}{V} \implies V = \frac{n}{c}$

V = (0.0395color(red)(cancel(color(black)("moles"))))/(3.15color(red)(cancel(color(black)("moles")))/"L") = "0.01254 L"

Rounded to three sig figs and expressed in mililiters, the volume will be

$V = \text{12.5 mL}$

So, to prepare your target solution, use a $\text{12.5-mL}$ sample of the stock solution and add enough water to make the volume of the total solution equal to $\text{250.0 mL}$.

This is equivalent to diluting the $\text{12.5-mL}$ sample of the stock solution by a dilution factor of $20$.