Question

What mass of CH3COOH is required to prepare 100 ml of 0.2 M of acetic acid solution?

Answer 1

Answer:

1.2 g

Explanation:

Molarity = No. of Moles of Solute/ Volume of solution in litres

Let the no. of moles be x.

then 0.2 M = x/0.1 L

so x =0.02 M

Mass = Molar Mass * No. of Moles

= 60 mol/g * 0.02 mol

= 1.2 g