0.5 g of fuming H2SO4 (oleum) is diluted with water. This solution is completely neutralised by 26.7 ml of 0.4 N NaOH. The percentage of free SO3 in the sample is

Question

0.5 g of fuming H2SO4 (oleum) is diluted with water. This solution is completely neutralised by 26.7 ml of 0.4 N NaOH. The percentage of free SO3 in the sample is (A) 30.6 % (B) 40.6 % (C) 20.6 % (D) 50 %

Tardigrade Admin · Accepted Answer

Meq. of H2SO4 + Meq. of SO3 = Meq. of NaOH ∴ ((0.5 - x)/98/2)×1000 + (x/80/2)×1000 = 26.7 ×0.4 ∴ x = 0.103 ∴ % of SO3 = (0.103/0.5)×100 = 20.6 %

Q. $0.5 g$ of fuming $H_{2} S O_{4}$ (oleum) is diluted with water. This solution is completely neutralised by $26.7 m l$ of $0.4 N$ $N a O H$ . The percentage of free $S O_{3}$ in the sample is

$30.6%$

$40.6%$

$20.6%$

$50%$

Meq. of $H_{2} S O_{4} +$ Meq. of $S O_{3} =$ Meq. of $N a O H$
$∴ \frac{( 0.5 - x )}{98/2} \times 1000 + \frac{x}{80/2} \times 1000$
$= 26.7 \times 0.4$
$∴ x = 0.103$
$∴ %$ of $S O_{3} = \frac{0.103}{0.5} \times 100 = 20.6%$

Q. 0.5g of fuming H2​SO4​ (oleum) is diluted with water. This solution is completely neutralised by 26.7ml of 0.4N NaOH. The percentage of free SO3​ in the sample is

30.6%

40.6%

20.6%

50%