Question

Question #983bc

Answer 1

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 `"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.