Question

1. How many mL of 10.0 M HCl is required to prepare 5.00 L of 1.5 M HCI?

Answer 1

To prepare the dilute solution of any acid or a base, you can use the molarity equation,

M1V1 = M2V2

here M1 = molarity of standard HCl = 10.0 M

V1 = volume of standard HCl required = ?

M2 = molarity of HCl prepared = 1.5 M

V2 = volume of HCl prepared = 5.00 L

So, V1 = M2V2/M1

V1 = (1.5 M × 5.00 L) / 10.0 M

= 0.75 L

Conversion unit: 1 L = 1000 mL

​​​​​​So, 0.75 L = 0.75 × 1000 = 750 mL

​​​​​​So, volume of standard HCl required is 750 mL