Question #c79f3

For starters, you know that an #"85% m/v"# phosphoric acid solution contains #"85 g"# of phosphoric acid for every #"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# #"mL"#

of this solution. Use the mass by volume percent concentration to calculate the mass of phosphoric acid present in #10^3# #"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# #"mL"# of solution, you can say that its molarity is equal to

#color(darkgreen)(ul(color(black)("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.