Question

Give me a trick to easily solve these type of questions like,

*0.5g of oleum is diluted with water.tgis solution is comolcompl neutralised by 26.7ml of 0.4 N NaOH .find the percentage of free SO3 in sample.

I know to solve it but it is very lengthy.

Pls give me step explanation also.

Answer 1

Eq. H2SO4 = Eq of NaOH
= (26.7 x 0.4)/1000
= 10.68 mEq.

H2SO4 + SO3 + H2O -----> 2(H2SO4)
>---------< is oleum.
==> SO3 + H2SO4 -----> H2SO4
∴Eq of SO3 = 1/2 of mEq. total H2SO4
==>mEq. of Oleum = Eq of H2SO4 + Eq of SO3
= Eq of total H2SO4
= 10.68/1000 (bec. 1 mEq. = 1/1000 Eq.)
Let x be mass of SO2 in Oleum. ==> mass of H2SO4 = (0.5-x)
Eq. of Oleum = (0.5 - x)/49 + x/40 = 10.68/1000
Solving.... x = 0.1836g

Percentage = 0.1836/0.5 x 100 = 20.73%