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.
