# Question 20ec1

Mar 29, 2016

${\text{8.3 mol L}}^{- 1}$

#### Explanation:

Your goal here is to find this solution's molarity, so in essence you're looking for the number of moles of solute present in one liter of solution.

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{molarity" = "moles of solute"/"one liter of solution} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

To make the calculations easier, you can pick a $\text{1.0-L}$ sample of this solution and use the information given to determine how many moles of nitric acid, ${\text{HNO}}_{3}$, you have in this sample.

The first thing to do with this $\text{1.0-L}$ sample is use the solution's density to find its mass. A density of ${\text{1.3 g mol}}^{- 1}$ means that every $\text{mL}$ of this solution will have a mass of $\text{1.3 g}$.

Convert the sample from milliliters to liters by using the conversion factor

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{1 L" = 10^3"mL}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

You will thus have

1.0color(red)(cancel(color(black)("L"))) * (10^3color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * overbrace("1.3 g"/(1color(red)(cancel(color(black)("mL")))))^(color(purple)("the given density")) = "1300 g"

Now, notice that the problem provides you with the solution's percent concentration by mass, $\text{% w/w}$, which essentially tells you how many grams of solute you get per $\text{100 g}$ of solution.

In your case, the solution is said to be 40% nitric acid by mass. This means that every $\text{100 g}$ of solution will contain $\text{40 g}$ of nitric acid.

Use this information to find the mass of nitric acid present in the sample

1300color(red)(cancel(color(black)("g solution"))) * overbrace("40 g HNO"_3/(100color(red)(cancel(color(black)("g solution")))))^(color(brown)("40% w/w HNO"_3)) = "520 g HNO"_3

Now that you know how many grams of nitric acid you have in your solution, you can use nitric acid's molar mass as a conversion factor to find the number of moles of solute.

A molar mass of ${\text{63 g mol}}^{- 1}$ means that one mole of nitric acid has a mass of $\text{63 g}$. In your case, the sample will contain

520color(red)(cancel(color(black)("g"))) * overbrace("1 mole HNO"_3/(63color(red)(cancel(color(black)("g")))))^(color(blue)("molar mass of HNO"_3)) = "8.254 moles HNO"_3

Now that you know how many moles of nitric acid you have in the $\text{1.0-L}$ sample, you an say that the solution's molarity will be equal to

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} c = {n}_{\text{solute"/V_"solution}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

c = "8.254 moles"/"1.0 L" = color(green)(|bar(ul(color(white)(a/a)"8.3 mol L"^(-1)color(white)(a/a)|)))#

The answer is rounded to two sig figs.