a) Molarity of acid
Assume that you have 1 L of the solution.
#"Mass of acid" = 1000 color(red)(cancel(color(black)("mL solution"))) × "1.73 g solution"/(1 color(red)(cancel(color(black)("mL solution")))) = "1730 g solution"#
#"Mass of H"_2"SO"_4 = 1730 color(red)(cancel(color(black)("g solution"))) × ("80 g H"_2"SO"_4)/(100 color(red)(cancel(color(black)("g solution")))) = "1380 g H"_2"SO"_4#
#"Moles of H"_2"SO"_4 = 1380 color(red)(cancel(color(black)("g H"_2"SO"_4))) × (1 "mol H"_2"SO"_4)/(98.08 color(red)(cancel(color(black)("g H"_2"SO"_4)))) = "14.1 mol H"_2"SO"_4#
#"Molarity" = "moles"/"litres" = "14.1 mol"/"1 L" = "14 mol/L"#
Note: The answer can have only 2 significant figures, because that is all you gave for the percentage composition of the acid.
b) Volume of acid
To solve this problem, we can use the dilution formula
#color(blue)(bar(ul(|color(white)(a/a)c_1V_1 = c_2V_2color(white)(a/a)|)))" "#
We can rearrange the formula to get
#V_1 = V_2 × c_2/c_1#
In this problem,
#c_1 = "14.1 mol/L"; V_1 = ?#
#c_2 = "0.50 mol/L"; V_2 = "0.500 L"#
#V_1 = "0.500 L" × (0.50 color(red)(cancel(color(black)("mol/L"))))/(14.1 color(red)(cancel(color(black)("mol/L")))) = "0.018 L" = "18 mL"#
You would need to use 18 mL of the 80 % acid.
