The correct density of concentrated sulfuric acid is #"1.84 g/cm"^3#.

**Step 1. Calculate the molar concentration of the conc. #"H"_2"SO"_4#**

Assume that you have #"1 dm"^3# of the concentrated acid.

#"Mass of conc. acid" = 1000 color(red)(cancel(color(black)("cm"^3))) × "1.84 g"/(1 color(red)(cancel(color(black)("cm"^3)))) = "1840 g"#

#"Mass of H"_2"SO"_4 = 1840 color(red)(cancel(color(black)("g conc."))) × (98 "g H"_2"SO"_4)/(100 color(red)(cancel(color(black)("g conc.")))) = "1803 g H"_2"SO"_4#

#"Moles of H"_2"SO"_4 = 1803 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)))) = "18.38 mol H"_2"SO"_4#

#c = "18.38 mol"/"1 dm"^3 = "18.38 mol/dm"^3#

**Step 2. Calculate the volume of conc. #"H"_2"SO"_4# required**

We can use the dilution formula for this calculation.

#color(blue)(bar(ul(|color(white)(a/a)c_1V_1 = c_2V_2color(white)(a/a)|)))"#

where

#c_1# and #c_2# are the molar concentrations and

#V_1# and #V_2# are the corresponding volumes.

We can rearrange this formula to get

#V_2 = V_1 × c_1/c_2#

In this problem,

#c_1 = "0.054 mol/dm"^3; V_1 = "0.500 dm"^3#

#c_2 = "18.38 mol/dm"^3; V_2 = ?#

∴ #V_2 = "0.500 dm"^3 × (0.054 color(red)(cancel(color(black)("mol/dm"^3))))/(18.38 color(red)(cancel(color(black)("mol/dm"^3)))) = "0.0015 dm"^3 = "1.5 cm"^3#

You would place about #"200 cm"^3# of distilled water in a #"500 cm"^3# volumetric flask, slowly add 1.5 mL of the conc. #"H"_2"SO"_4# from a pipet with swirling, and then add enough distilled water to make the solution up to the #"500 cm"^3# mark.