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
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
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.