The correct density of concentrated hydrochloric acid is #"1.19 g/cm"^3#.
Step 1. Calculate the molar concentration of the conc. #"HCl"#
Assume that you have #"1 dm"^3# of the concentrated acid.
#"Mass of conc. HCl" = 1000 color(red)(cancel(color(black)("cm"^3))) × "1.19 g"/(1 color(red)(cancel(color(black)("cm"^3)))) = "1190 g"#
#"Mass of HCl" = 1190 color(red)(cancel(color(black)("g conc."))) × (36.00 "g HCl")/(100 color(red)(cancel(color(black)("g conc.")))) = "428.4 g HCl"#
#"Moles of HCl" = 428.4 color(red)(cancel(color(black)("g HCl"))) × ("1 mol HCl")/(35.5 color(red)(cancel(color(black)("g HCl")))) = "12.07 mol HCl"#
#c = "12.07 mol"/"1 dm"^3 = "12.07 mol/dm"^3#
Step 2. Calculate the volume of conc. #"HCl"_2# 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 = "0.215 mol/dm"^3; V_1 = "1 dm"^3#
#c_2 = "12.07 mol/dm"^3; V_2 = ?#
∴ #V_2 = "1 dm"^3 × (0.215 color(red)(cancel(color(black)("mol/dm"^3))))/(12.07 color(red)(cancel(color(black)("mol/dm"^3)))) = "0.001 78 dm"^3 = "17.8 cm"^3#
You would place about #"400 cm"^3# of distilled water in a #"1 dm"^3# volumetric flask, slowly add 17.8 mL of the conc. #"HCl"# from a pipet with swirling, and then add enough distilled water to make the solution up to the #"1 dm"^3# mark.
