Question

0.5 g of fuming H2SO4 is diluted with water. This solution is completely neutralised by 26.7 mL of 0.4N NaOH .The percentage of free SO3 in the sample is
a)30.6% b)40.6% c)20.6% d)50%

Answer 1

H2SO4 + SO3 + H2O -> 2(H2SO4) . 
==> SO3 + H2SO4 -----> H2SO4
∴Eq of SO3 = 1/2 of H​2SO4 
==> 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  ==> mass of H2SO4 = (0.5-x) 
 = (0.5 - x)/49 + x/40 = 10.68/1000 
Solving.... x = 0.1836g 

Percentage = 0.1836/0.5 x 100 = 20.73%

Answer 2

Given:

Quantity of H2SO4 = 0.5g

Quantity of NaOH = 0.4N

Neutrlization quantity = 26.7mL

To Find:

Percentage of free SO3 in the sample

Solution:

Let the mass of SO3  in the sample be = x  

Let the mass of H2SO4 be  = (0.5−x)

Mass of SO3 = 80/2 = 40

Equivalents of SO3 = x/40

Mass of H2SO4 = 98/2 = 49

Therefore,  

Total no of equivalents = x/40 + ( 0.5 - x)/49

26.7 mL of 0.4 N NaOH contains  no. of equivalents of NaOH

= 0.4 / 1000 × 26.7

Hence,

0.4 / 1000 × 26.7 = 49x + ( 40 × 05 - 40x) / 40 × 49

x = 0.9328/9

= 0.1036

Therefore

% of SO3 = 0.1036/ 0.5 × 100

= 20.6

Answer: The percentage of free SO3 in the sample is 20.6%