Commercially available concentrated hydrochloric acid contains 38 % HCl by mass. (i) What is the molarity of the solution if its density is `1.19 "g cm"^(-2)` ?
(ii) What volume of concentrated HCl is needed to make 1.0 L of 0.2 M HCl solution ?

Explanation

Step I. Calculation of volume of 100 g of `HCl` solution
Mass of HCl solution = 100 g
Density of HCl solution `= 1.10 "g cm"^(-3)`
Volume of HCl solution `= ("Mass")/("Density") = ((100 g))/(("1.19 g cm"^(-3)))=84.03 cm^(3)=0.08403 "dm"^(3)`
Step II. Calculation of molarity of the solution
Molarity of solution `= ("Mass of HCl/Molar mass")/("Volume of solution in dm"^(3))=(("38 g")//(36.5"g mol"^(-1)))/((0.08304 "dm"^(3)))`
`=12.38 "mol dm"^(-3)=12.38 M`
Step III. Calculation of volume of concentrated `HCl` needed
Molarity of conc. `HCl (M_(1))=12.38 M`
Let the volume of the conc. HCl needed `= V_(1)`
Molarity of dilute `HCl (M_(2))=0.2 M`
Volume of dilute `HCl (V_(2))=1.0 L`
Applying molarity equation : `M_(1)V_(1)=M_(2)V_(2)`
`(12.38 M)xxV_(1)=(0.2 M)xx(1.0 L)`
`V_(1)=((0.2 M)xx(1.0L))/((12.38M))=0.01615=16.15mL`.