Commercially available concentrated hydrochlorc acid contains 38% HCI by mass. (i) What is the molarity of this solution? The density is 1.19 g cm^(-3). (ii) What volume of concentrated HCI is required to make 1.00 L of 0.10 M HCI?

Commercially available concentrated hydrochlorc acid contains 38% HCI

1.	Commercially available concentrated hydrochlorc acid contains 38% HCI by mass. (i) What is the molarity of this solution? The density is 1.19 g cm^(-3). (ii) What volume of concentrated HCI is required to make 1.00 L of 0.10 M HCI?
Answer» Solution :(i) Suppose, we have 1000 cm (1.0 L) of the given sample of commercial hydrochloric acid. `therefore` Density `=("Mass")/("Volume")` `therefore` Mass of `1000 cm^(3)` of the sample `=1000 xx 1.19 = 1190 g` The given sample CONTAINS 38% HCl by mass. Hence, the mass of HCl present in `1000 cm^3` (1190) g `=(1190 xx 38)/100 = 452.2 g` Hence, `w = 452.2 g, V = 1000 cm^(3), M = 35.45 + 1.008 = 36.46` `therefore 452.2 = (M xx 36.46 xx 1000)/1000` or M = 12.40 Hence, the molarity of the given commercial sample is 12.40 M. ANS. (ii) The mass of pure HCl required the make 1.0 L (`1000 cm^3`) of a solution of 0.10 M HCl is given by `w =(0.10 xx 36.46 xx 1000)/1000 = 3.6 g` `therefore` 3.6 g of pure HCl will be present in the sample having mass `=100/38 xx 3.6 = 9.5 g` Since, Volume `=("Mass")/("Density")` The volume of sample having mass 9.5 g `=9.5/1.19 = 8.0 cm^(3)` Hence, `8.0 cm^(3)` of concentrated HCl are required to make 1.00 L of 0.10 M solution.