# Question #13e7f

##### 1 Answer

#### Answer:

#### Explanation:

Before doing anything else, make sure that you understand what it is you're looking for here.

A solution's **molarity** will tell you how many *moles of solute* you get **per liter of solution**.

This means that if you know the number of moles of solute and the volume of the solution, you can find the molarity of the solution by *dividing* these two values.

#color(blue)(c = n_"solute"/V_"solution")#

Your strategy here will be to use pick a sample of this solution and use its **density** to determine its **mass**. Once you know the sample's mass, you can use the given **percent concentration by mass** to find the mass of solute present in the sample.

To keep the calculations as simple as possible, pick a **every**

#1 color(red)(cancel(color(black)("L"))) * (10^3color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * overbrace("1.689 g"/(1color(red)(cancel(color(black)("mL")))))^(color(purple)("density")) = "1689 g"#

The solution's percent concentration by mass tells you that you get **for every**

#1689 color(red)(cancel(color(black)("g solution"))) * overbrace(("85 g H"_3"PO"_4)/(100color(red)(cancel(color(black)("g solution")))))^(color(purple)(="85% w/w")) = "1435.65 g H"_3"PO"_4#

Now all you need to do is use the **molar mass** of phosphoric acid to determine how many *moles* you'd get in this many grams

#1435.65 color(red)(cancel(color(black)("g"))) * ("1 mole H"_3"PO"_4)/(97.995color(red)(cancel(color(black)("g")))) = "14.65 moles H"_3"PO"_4#

Since the sample had a volume of

#c = "14.65 moles"/"1.0 L" = color(green)("14.7 M")#

I'll leave the answer rounded to three **sig figs**, despite the fact that you only have two sig figs for the percent concentration by mass.

**SIDE NOTE** *It is very important to realize that the result must be the same regardless of the volume of the sample.*

*I highly recommend redoing the calculations using a different starting volume.*