# Question 90394

May 27, 2015

The volume by volume percent concentration of your solution will be equal to 0.991%.

You know that your acetic acid solution has a density equal to that of water, more specifically to ${\text{1.00 g/cm}}^{3}$. Moreover, you know the molarity of the solution to be $\text{0.165 mol/L}$. This will be your starting point.

Since molariy is defined as moles of solute, in your case acetic acid, per liters of solution, you can use it to determine how many moles of acetic acid you have.

To make the calculations easier, you can assume a 1.00-L sample of solution.

$C = \frac{n}{V} \implies n = C \cdot V$

${n}_{\text{acetic acid" = 0.165"mol"/cancel("L") * 1.00cancel("L") = "0.165 moles}}$

Use acetic acid's molar mass to determine how many grams you'd get in your sample

0.165cancel("moles") * "60.05 g"/(1cancel("mole")) = "9.91 g" $\text{acetic acid}$

Now you can use the solution's density to see what volume this much acetic acid would occupy

9.91cancel("g") * (1cancel("cm"^3))/(1cancel("g")) * "1 L"/(1000cancel("cm"^3)) = 9.91 * 10^(-3)"L"

I've converted the volume to liters because it will make the calculation of the v/v percent concentration easier.

So, you know that for every 1.00 L of solution, you get 9.91 mL of acetic acid. This means that its v/v percent concentration will be

$\text{%v/v" = "volume of acetic acid"/"volume of solution} \cdot 100$

"%v/v" = (9.91 * 10^(-3)cancel("L"))/(1.00cancel("L")) * 100 = color(green)("0.991%")#

On average, regular vinegar is about 5% v/v, which means that your solution is roughly 5 times more dilute than that.