What mass of ch3cooh is required to prepare 100.00ml of 2.0 m acetic acid solution

12 g.

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 =

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.

the atomic number will stay the same no matter how many electrons it has, so it doesn’t decrease nor increase. it’ll only change with a different number of protons