# Question #983bc

##### 1 Answer

#### Answer:

Here's how you can do that.

#### Explanation:

As you know, the **molarity** of a solution tells you the number of moles of solute present for every

In order to make the calculations easier, pick a sample of this solution that has exactly

Use the **density** of the solution to find the **mass** of the sample. In your case, vinegar is said to have a density of

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 **moles** of solute.

Since the sample has exactly **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, **molar mass** of

The sample contains **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 **percent concentration by mass**,

#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 **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**.