Question

What mass of CH3COOH is required to prepare 100.00ml of 2.0 M acetic acid solution

Answer 1

12 g.

Step-by-step explanation:

We have to make 100 ml of 2.0M acetic acid.

no. of moles of acetic acid required for this will be calculated by the formula,

Molarity =
`(no. of moles)/(volume of solution)`

Here,

volume of solution = 0.1 L

Molarity = 2.0M

So, no. of moles = 0.2

Molecular weight of acetic acid = 60 g.

So,

Weight of acetic acid required = 0.2×60 g = 12 g.