How do you make 20 mL of 25 M HCl given 1.00 M HCI?

Add_mL of acid

To prepare 20 mL of 0.25 M HCl from 1.00 M HCl, 5 mL of the 1.00 M HCl is measured and transferred in a volumetric flask or beaker containing about 10 mL of water. Then, water is added to make it up to the 20mL mark.

Explanation:

The question is not correct, because the molarity of HCl cannot be greater than 12.2 M.

Also, a higher concentration of an acid can not be prepared by any dilution method,  but can only be prepared by distillation procedures requiring great care and expertise.

Therefore,  the following assumptions are made about the question :

The initial acid concentration, M1 = 1.00 M

Required acid concentration, M2 = 0.25 M

Volume of required acid solution, V2 = 20 mL

Using the dilution formula: M1V1 = M2V2

Volume of initial acid solution, V1 must be found then.

Making V1 subject of the formula; V1 = M2V2/M1

V1 = 0.25 × 20 / 1.00

V1 = 5 mL

To prepare 20 mL of 0.25 M HCl from 1.00 M HCl, 5 mL of the 1.00 M HCl is measured and transferred in a volumetric flask or beaker containing about 10 mL of water. Then, water is added to make it up to the 20mL mark.

i think so it’s will be b yes i believe its