Commercially availiable concentrated hydrochloric acid contains `38% HCl` by mass. (a) What is the molarity of this solution? The density is `1.19g mL^(-1)` ?
(b) What volume of concentrated `HCl` is required to make `1.00 "litre"` of `0.10M HCl`?

Explanation

Calculation of molarity. `38%` HCl by mass means that 38 g of HCl are present in 100 g of the solution.
`"Volume of 100 g of the solution"=("Mass ")/("Density")=(100g)/(1.19"g cm"^(-3))=84.03cm^(3)=0.0840L`
Molar mass of HCl = `36.5"g mol"^(-1)`
`therefore" 38 g HCl"=(38)/(36.5)"moles"=1.04" moles"`
`"Molarity of solution "=(1.04"moles")/(0.0840L)=12.38"mol L"^(-2)`, i.e., Molarity of the solution = 12.38 M
(b) Calculation of volume of conc. HCl for 1.00 L of 0.10 M HCl.
Applying molarity equaton, we have
`{:(M_(1)xxV_(1),=," "M_(2)xxV_(2)),("(conc. HCl)",,"(1.0 L of 0.10 M HCl)"),(12.38xxV_(1),=," "0.10xx1.0):}`
`"or "V_(1)=(0.1)/(12.38)L=(0.1)/(12.38)xx1000cm^(3)=8.1cm^(3)`