According to the label on a bottle of concentrated hydrochloric acid, the contents are 36.0% HCl by mass and have a density of 1.18 g/mL.

Question

What volume of it would you need to prepare 927 mL of 2.40 M HCl?

Answer 1

Let's find the molarity of the original HCl solution, and then use V1M1 = V2M2 to make the dilution and end up with 927 ml of 2.40 M HCl.

If we have 1000 ml (1 L) of 36.0% by mass HCl and the density is 1.18 g/ml, that means we have....

1000 ml x 1.18 g/ml = 1180 g of solution

36% of this is HCl, so...

0.36 x 1180 g = 424.8 g HCl are in the solution

How many moles is this? molar mass HCl = 36.46 g/mol

moles HCl = 424.8 g x 1 mol/36.46 g = 11.65 moles

Since this is in 1 L the concentration of the HCl = 11.65 M

V1M1 = V2M2

(x ml)(11.65 M) = (927 ml)(2.40 M)

x = 191 mls needed