A laboratory procedure calls for making 453.3 ml of a 2.62M salt solution. If the molar mass of the salt is 218 g/mol what mass is required?

1 Answer
Jul 8, 2016


#"259 g"#


Your strategy here will be to

  • use the molarity and volume of the solution to determine how many moles of solute it must contain
  • use the molar mass of the salt to calculate how many grams would contain that many moles

A solution's molarity tells you how many moles of solute you have in one liter of solution.

In your case, the solution must have a molarity of #"2.62 mol L"^(-1)#, which means that every liter of solution will contain #2.62# moles of solute.

Now, your sample is said to have a volume of #"453.3 mL"#. Use its molarity to figure out how many moles of solute it contains -- do not forget to convert the volume to liters!

#453.3 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * overbrace("2.62 moles solute"/(1color(red)(cancel(color(black)("L solution")))))^(color(blue)("known molarity")) = "1.188 moles solute"#

As you know, a compound's molar mass tells you the mass of one mole of said compound. In your case, the salt has a molar mass of #"218 g mol"^(-1)#, which means that one mole has a mass of #"218 g"#.

This means that the mass of #1.188# moles will be

#1.188 color(red)(cancel(color(black)("moles"))) * "218 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("259 g")color(white)(a/a)|)))#

The answer is rounded to three sig figs.