Question

2 gm of fuming sulphuric acid is diluted with water to make the volume 1 litre, 100 ml of this solution is then titrated with 0.1 M $NaOH$ solution, the volume of $NaOH$ consumed at equivalence point is 43.25 ml . Which of the following is/are correct for the above titration?

• A. % of oleum is 105.96
• B. % of free ${SO}_3$ is 26.5
• C. Equivalent of ${SO}_3$ are 0.013
• D. wt of $H_2{SO}_4$ is 1.85
• E. None of these

Answer 1

Answer: (A)

% of oleum is 105.96

Let wt of $H_2{SO}_4$ and ${SO}_3$ are x and y respectively.

Then meq of $H_2{SO}_4$ + meq of ${SO}_3$ = meq of $NaOH$

$\left(x/49\right)\times 1000+\left(y/40\right)\times 1000=0.1\times \left(1000/100\right)\times 43.25$

$40x+49y=84.77$

And $x+y=2$

So, $x=1.47gm$ $y=0.53gm$

Equivalent of ${SO}_3$ = 0.53/40 = 0.013

80 gm ${SO}_3$ react with 18 gm water

So, 0.53 gm ${SO}_3$ will react with $18/80\times 0.53=>0.11925gm$

Hence 2 gm oleum gain 0.11925 gm weight

$100gm$ oleum gain 5.9625 gm wt

So, oleum is 105.96%