Question

Question #20ec1

Answer 1

`"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.

`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 `"1.0-L"` sample of this solution and use the information given to determine how many moles of nitric acid, `"HNO"_3`, you have in this sample.

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

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, `"% w/w"`, which essentially tells you how many grams of solute you get per `"100 g"` of solution.

In your case, the solution is said to be `40%` nitric acid by mass. This means that every `"100 g"` of solution will contain `"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 `"63 g mol"^(-1)` means that one mole of nitric acid has a mass of `"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 `"1.0-L"` sample, you an say that the solution's molarity will be equal to

`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.