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

1 Answer
Dec 27, 2015

Answer:

#"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")#