Question

What volume of 10.0 M HCl is needed to contain 4.00 moles of HCl?

Answer 1

Volume of 10M HCl containing 4 moles HCl = 400 ml

Step-by-step explanation:

Molarity x Volume (L) = moles

10M·V = 4 moles HCl

V = 4moles/10Molar = 4moles/10moles/Liter = 0.40L = 400ml