# Question 8099a

Nov 11, 2015

$\text{1.3 g/mL}$

#### 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 $\text{1.00-L}$ sample of this concentrated phosphoric acid solution.

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.

$\textcolor{b l u e}{\text{molarity" = "moles of solute"/"liters of solution}}$

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

$c = \frac{n}{V} \implies n = C \cdot V$

$n = {\text{12.2 M" * "1.00 L" = "12.2 moles H"_3"PO}}_{4}$

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 90% phosphoric acid by mass, which is equivalent to saying that you get $\text{90 g}$ of phosphoric acid for every $\text{100 g}$ of solution.

The mass of the solution that contains $\text{1195.5 g}$ of phosphoric acid will thus be

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 $\text{1.00 L}$, and the mass of the solution, which is $\text{1328.3 g}$, so the density of the solution will be - expressed in grams per mililiter

$\textcolor{b l u e}{\text{density" = "mass"/"volume}}$

$\rho = \text{1328.3 g"/(1.00color(red)(cancel(color(black)("L")))) * (1color(red)(cancel(color(black)("L"))))/"1000 mL" = "1.3283 g/mL}$

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

$\rho = \textcolor{g r e e n}{\text{1.3 g/mL}}$

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.