# How many milliliters of a 2.5 M MgCl_2 solution contain 17.5 g MgCl_2?

Dec 27, 2015

$\text{74 mL}$

#### Explanation:

Your strategy here will be to

• use the molar mass of magnesium chloride to determine how many moles you would have in that sample

• use the solution's molarity to find how many liters would contain that many moles

• convert the volume from liters to milliliters

So, magnesium chloride, ${\text{MgCl}}_{2}$, has a molar mass of $\text{95.211 g/mol}$. This means that every mole of magnesium chloride will have a mass of $\text{95.211 g}$.

You are interested in finding out what volume of your $\text{2.5-M}$ magnesium chloride solution would contain $\text{17.5 g}$ of magnesium chloride.

This mass of magnesium chloride will be equivalent to

17.5 color(red)(cancel(color(black)("g"))) * "1 mole MgCl"_2/(95.211 color(red)(cancel(color(black)("g")))) = "0.1838 moles MgCl"_2

Now, molarity is defined as moles of solute, which in your case is magnesium chloride, divided by liters of solution

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

This means that the volume of your solution that contains $0.1838$ moles of magnesium chloride will be

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

V = (0.1838 color(red)(cancel(color(black)("moles"))))/(2.5color(red)(cancel(color(black)("moles")))/"L") = "0.07352 L"

Finally, convert this value to milliliters by using the conversion factor

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

This will get you

0.07352color(red)(cancel(color(black)("L"))) * (10^3"mL")/(1color(red)(cancel(color(black)("L")))) = "73.52 mL"

Rounded to two sig figs, the number of sig figs you have for the molarity of the solution, the answer will be

$V = \textcolor{g r e e n}{\text{74 mL}}$