# Question #8099a

##### 1 Answer

#### Explanation:

The first thing to do here is pick a sample of this solution. To make calculations easier, let's say that we pick a

Now, a solution's molarity is defined as the number of moles of solute, which in your case is phosphoric acid, divided by the volume of the solution - expressed in liters.

#color(blue)("molarity" = "moles of solute"/"liters of solution")#

This means that you can use this volume sample and the solution's molarity to determine how many moles of phosphoric acid you have

#c = n/V implies n = C * V#

#n = "12.2 M" * "1.00 L" = "12.2 moles H"_3"PO"_4#

Use phosphoric acid's *molar mass*, which tells you what the exact mass of **one mole** of a compound is, in order to determine how many grams of phosphoric acid you have in this sample

#12.2color(red)(cancel(color(black)("moles"))) * "97.995 g"/(1color(red)(cancel(color(black)("mole")))) = "1195.5 g H"_3"PO"_4#

You know that this solution is **for every**

The mass of the solution that contains

#1195.5color(red)(cancel(color(black)("g H"_3"PO"_4))) * "100 g solution"/(90color(red)(cancel(color(black)("g H"_3"PO"_4)))) = "1328.3 g solution"#

Now you know the volume of the solution, which we've chosen to be equal to *grams per mililiter*

#color(blue)("density" = "mass"/"volume")#

#rho = "1328.3 g"/(1.00color(red)(cancel(color(black)("L")))) * (1color(red)(cancel(color(black)("L"))))/"1000 mL" = "1.3283 g/mL"#

I'll leave the answer rounded to two sig figs, despite the fact that you only have one sig figs for the concentration of the solution

#rho = color(green)("1.3 g/mL")#

**SIDE NOTE** *I recommend doing the calculations with different samples of the solution, the result must come out the same regardless of what volume you pick.*