Given that the molar mass of NaCl 58.44 g/mol, what is the molarity of a solution that contains 87.75 g of NaCl in 500. mL of solution?

1 Answer
Feb 24, 2017

#"3.00 mol L"^(-1)#


Molarity is a measure of how many moles of solute you get in exactly #"1 L"# of a given solution.

Your goal here is to use the molar mass of sodium chloride to find the number of moles present in the sample.

#87.75 color(red)(cancel(color(black)("g"))) * "1 mole NaCl"/(58.44color(red)(cancel(color(black)("g")))) = "1.5015 moles NaCl"#

Now, you know that this solution contains #1.5015# moles of sodium chloride, the solute, in #"500. mL"# of solution. Since

#500.0 color(red)(cancel(color(black)("mL"))) * "1 L"/(10^3color(red)(cancel(color(black)("mL")))) = "0.500 L"#

you can use this known composition of the solution as a conversion factor to help you find the number of moles of solute present in #"1 L"# of solution

#1 color(red)(cancel(color(black)("L solution"))) * "1.5015 moles NaCl"/(0.500color(red)(cancel(color(black)("L solution")))) = "3.003 moles NaCl"#

You can now say that the molarity of the solution is

#color(darkgreen)(ul(color(black)("molarity = 3.00 mol L"^(-1))))#

The answer is rounded to three sig figs, the number of sig figs you have for the volume of the solution.