Question

Question #13e7f

Answer 1

`"14.7 M"`

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 `"1.0-L"` sample of this phosphoric acid solution. Since you know that every `"mL"` of this solution has a mass of `"1.689 g"`, you can say that the sample will have a mass of

`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 `"85 g"` of phosphoric acid for every `"100 g"` of solution. This means that your sample will contain

`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 `"1.0 L"`, the molarity of the solution will be

`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.