Question #983bc

As you know, the molarity of a solution tells you the number of moles of solute present for every #"1 L"# of the solution.

In order to make the calculations easier, pick a sample of this solution that has exactly #"1 L" = 10^3 quad "mL"#.

Use the density of the solution to find the mass of the sample. In your case, vinegar is said to have a density of #"1.005 g mL"^(-1)#, which tells you that #"1 mL"# of vinegar has a mass of #"1.005 g"#.

This means that the sample has a mass of

#10^3 color(red)(cancel(color(black)("mL solution"))) * "1.005 g"/(1color(red)(cancel(color(black)("mL solution")))) = "1005 g"#

Now, let's say that the solution has molarity of #c# #"mol L"^(-1)#, which tells you that every #"1 L" = 10^3 quad "mL"# of this solution contains #c# moles of solute.

Since the sample has exactly #"1 L"#, you can say that it will contain #c# moles of solute.

Next, use the molar mass of the solute to convert the number of moles to grams. In this case, the solute is acetic acid, #"CH"_3"COOH"#, which has a molar mass of #"60.05 g mol"^(-1)#.

The sample contains #c# moles of acetic acid, the equivalent to

#c color(red)(cancel(color(black)("moles CH"_3"COOH"))) * "60.05 g"/(1color(red)(cancel(color(black)("mole CH"_3"COOH")))) = (c * 60.05)quad "g"#

So, you know that you have #(c * 60.05) quad "g"# of acetic acid in #"1005 g"# of the solution. In order to find the solution's percent concentration by mass, #"% m/m"#, you need to figure out how many grams of acetic acid are present in exactly #"100 g"# of this solution.

#100 color(red)(cancel(color(black)("g solution"))) * ((c * 60.05) quad "g CH"_3"COOH")/(1005 color(red)(cancel(color(black)("g solution")))) = (c * 5.9751) quad "g CH"_3"COOH"#

So, you can say that a vinegar solution that has a molarity of #c# #"mol L"^(-1)# and a density of #"1.005 g mL"^(-1)# has a percent concentration by mass of

#color(darkgreen)(ul(color(black)("% m/m CH"_3"COOH" = (c * 5.9751)%)))#

So, if the solution has a molarity of

#c = "1.104 mol L"^(-1)#

you can say that its percent concentration by mass is equal to

#"% m/m CH"_3"COOH" = (1.104 * 5.9751)% #

#"% m/m CH"_3"COOH" = 6.597%#

The answer is rounded to four sig figs.