500 ml of a solution contains 15 grams of acetic acid and 8 grams of methanol. What will be its active mass?
Answer: molecular weight of acetic acid (CH3COOH) = 2 x 12 + 4 x 1 + 2 x 16 = 60
Active mass of acetic acid (CH3COOH) = weight of (CH3COOH) in grams / atomic mass of (CH3COOH) / full volume in litres
= 15/60/½
= 0.5 mole/litre
Molecule of Methanol (CH3OH) = 12 + 4 x 1 + 16 = 32
Active mass of methanol (CH3OH) = weight of (CH3OH) in grams / atomic mass of (CH3OH) / full volume in litres
= 8/32 / ½
= 0.5 mole / litre