Question

What volume of 1M NaOH will be required to react completely with 100g of oleum which is 109% labelled

Answer 1

109% oleum means 9g of H2O is required to convert all SO3 present in 100gm of it to H2SO4
SO3+h20=H2SO4
that is 100X88.98/40=2224.5ml

Answer 2

Hey dear,

● Answer -
V(NaOH) = 2.225 L

● Explanation -
109 % oleum means 9 g H2O is added to 100 g of oleum.
SO3 + H2O --> H2SO4

1 mole (80 g) of SO3 reacts with 1 mole (18 g) of water.
∴ 40 g SO3 reacts with 9 g H2O.

H2SO4 present in 100 g oleum = 100 - 40 = 60 g
NaOH required to neutralize 40 g SO3 = 40×80/80 = 40 g
NaOH required to neutralize 60 g H2SO4 = 60×80/98 = 49 g
Total NaOH required = 40 + 49 = 89 g

Volume of 1 M NaOH -
V = 1 × 89 / 40
V = 2.225 l

Therefore, 2.225 L of 1 M NaOH will be required to react with 100 g oleum.

Thanks for asking...