# Question #20ec1

##### 1 Answer

#### Answer:

#### 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.

#color(blue)(|bar(ul(color(white)(a/a)"molarity" = "moles of solute"/"one liter of solution"color(white)(a/a)|)))#

To make the calculations easier, you can pick a *nitric acid*,

The first thing to do with this **density** to find its mass. A density of **every**

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

#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 L" = 10^3"mL")color(white)(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**, **per**

In your case, the solution is said to be **every**

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 **one mole** of nitric acid has a mass of

#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

#color(blue)(|bar(ul(color(white)(a/a)c = n_"solute"/V_"solution"color(white)(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**.