Question

how many ml of 12 molar HCl solution contains 3 moles of HCl?

Answer 1

Answer:

This means that if you know how many moles of solute you have in the target solution, you also know how many moles of solute were present in the stock solution sample.

Use the molarity and volume of the target solution to determine how many moles of hydrochloric acid,

HCl

, you need in that solution

c

=

n

V

⇒

n

=

c

⋅

V

n

HCl

=

0.10 M

⋅

500

⋅

10

−

3

L

=

0.050 moles HCl

Now the question is - what volume of stock solution would contain this many moles of hydrochloric acid?

c

=

n

V

⇒

V

=

n

c

V

stock

=

0.050

moles

12

moles

L

=

0.0041667 L

I'll leave the answer rounded to two sig figs, despite the fact that you only gave one sig fig for the volume of the target solution

V

stock

=

4.2 mL