Question

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

Answer 1

`"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, `"MgCl"_2`, has a molar mass of `"95.211 g/mol"`. This means that every mole of magnesium chloride will have a mass of `"95.211 g"`.

You are interested in finding out what volume of your `"2.5-M"` magnesium chloride solution would contain `"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

`color(blue)("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

`color(blue)(c = n/V implies V = 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

`"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 = color(green)("74 mL")`