Question

The volume of 6.00M HCL needed to make 0.32L of 3.0M HCL is

Answer 1

Explanation: the answer is in the picture

Answer 2

`\huge\boxed{\sf V_1=0.16 \ L}`

Step-by-step explanation:

Given Data:

Initial Molarity =
`M_1` = 6.00 M

Final Volume =
`V_2` = 0.32 L

Final Molarity =
`M_2` = 3.0 M

Required:

Initial Volume =
`V_1` = ?

Formula:

`M_1V_1=M_2V_2`

Solution:

Put the given data in the above formula,

Finding initial volume.

`6 * V_1=0.32 * 3\\\\6 * V_1 = 0.96\\\\Divide \ both \ sides \ by \ 6\\\\V_1=0.96/6\\\\V_1=0.16 \ L\\\\\rule[225]{225}{2}`