# Question b81e1

Oct 2, 2015

You use the density and percent concentration by mass of the stock solution.

#### Explanation:

You know that you're dealing with a $\text{1-mol/L}$ nitric acid, ${\text{HNO}}_{3}$, solution.

To get the solution's percent concentration by volume, you would need to have some information about the stock solution of nitric acid used to make the $\text{1-mol/L}$ solution.

The idea here is that your $\text{1-mol/L}$ nitric acid solution was prepared by diluting a sample taken from a stock solution of nitric acid. This means that you need to find the volume of the stock solution used to prepare the $\text{1-mol/L}$ solution in order to be able to find its volume by volume percent concentration.

Usually, nitric acid stock solutions have a percent concentration by mass of nitric acid equal to 70.2% and a density of $\text{1.41 g/mL}$.

Use nitric acid's molar mass to determine how many grams of acid you get in your $\text{1-mol/L}$ solution

1color(red)(cancel(color(black)("mole HNO"""_3))) * "63.013 g"/(1color(red)(cancel(color(black)("mole HNO"""_3)))) = "63.013 g"

Now focus on finding what volume of the stock solution will contain 63.013 g of nitric acid.

To make calculations easier, assume that you have a $\text{1-L}$ sample of the stock solution. This sample will have a mass of

1color(red)(cancel(color(black)("L"))) * (1000color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * "1.41 g"/(1color(red)(cancel(color(black)("mL")))) = "1410 g"

Since nitric acid makes up 70.2% of this mass, you know that the $\text{1-L}$ stock sample will contain

1410color(red)(cancel(color(black)("g solution"))) * ("70.2 g HNO"""_3)/(100color(red)(cancel(color(black)("g solution")))) = "989.82 g HNO"""_3

Now, the volume of this solution that will contain 63.013 g of nitric acid is

63.013color(red)(cancel(color(black)("g HNO"""_3))) * "1000 mL stock"/(989.92color(red)(cancel(color(black)("g HNO"""_3)))) = "63.7 mL"

So, you prepared your $\text{1-mol/L}$ nitric acid solution by taking a $\text{63.7-mL}$ sample of the stock solution and adding enough water to make the total volume of th solution equal to $\text{1-L}$.

This means that the volume by volume, $\text{%v/v}$, percent concentration of the $\text{1-mol/L}$ solution is

$\text{%v/v" = V_(HNO_3)/V_"sol} \times 100$

"%v/v" = (63.7color(red)(cancel(color(black)("mL"))))/(1000color(red)(cancel(color(black)("mL")))) xx 100 = 6.37 = color(green)("6.4% v/v")#

SIDE NOTE The answer will depend on the percent concentration by mass, and implicitly on the density, of the stock solution of nitric acid.