# Question c79f3

May 20, 2017

${\text{8.7 mol L}}^{- 1}$

#### Explanation:

For starters, you know that an $\text{85% m/v}$ phosphoric acid solution contains $\text{85 g}$ of phosphoric acid for every $\text{100 mL}$ of solution.

As you know, molarity is a measure of the number of moles of solute, which in your case would be phosphoric acid, present in

1 color(red)(cancel(color(black)("L"))) * (10^3color(white)(.)"mL")/(1color(red)(cancel(color(black)("L")))) = 10^3 $\text{mL}$

of this solution. Use the mass by volume percent concentration to calculate the mass of phosphoric acid present in ${10}^{3}$ $\text{mL}$ of solution.

10^3 color(red)(cancel(color(black)("mL solution"))) * ("85 g H"_3"PO"_4)/(100color(red)(cancel(color(black)("mL solution")))) = "850 g H"_3"PO"_4

To convert the sample to moles, use the compound's molar mass

850 color(red)(cancel(color(black)("g H"_3"PO"_4))) * ("1 mole H"_3"PO"_4)/(97.99color(red)(cancel(color(black)("g H"_3"PO"_4)))) = "8.67 moles H"_3"PO"_4#

Since this value represents the number of moles of phosphoric acid present in ${10}^{3}$ $\text{mL}$ of solution, you can say that its molarity is equal to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{molarity = 8.7 mol L}}^{- 1}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the percent concentration of the solution.