# Question 983bc

Jan 14, 2018

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 $\text{1 L}$ of the solution.

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

Use the density of the solution to find the mass of the sample. In your case, vinegar is said to have a density of ${\text{1.005 g mL}}^{- 1}$, which tells you that $\text{1 mL}$ of vinegar has a mass of $\text{1.005 g}$.

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 $c$ ${\text{mol L}}^{- 1}$, which tells you that every $\text{1 L" = 10^3 quad "mL}$ of this solution contains $c$ moles of solute.

Since the sample has exactly $\text{1 L}$, 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, $\text{CH"_3"COOH}$, which has a molar mass of ${\text{60.05 g mol}}^{- 1}$.

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 $\left(c \cdot 60.05\right) \quad \text{g}$ of acetic acid in $\text{1005 g}$ of the solution. In order to find the solution's percent concentration by mass, $\text{% m/m}$, you need to figure out how many grams of acetic acid are present in exactly $\text{100 g}$ 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 $c$ ${\text{mol L}}^{- 1}$ and a density of ${\text{1.005 g mL}}^{- 1}$ 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

$c = {\text{1.104 mol L}}^{- 1}$

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.