Question

For $109\%$ labeled oleum if the number of moles of $H_{2}S{O}_4$ and free $S{O}_3$ be x and y respectively,then what will be the value of $\frac{x+y}{x-y}?$

• A. $1$
• B. $18$
• C. $\frac{1}{3}$
• D. $9.9$

Answer 1

The correct option is D $9.9$

Reaction involved
$S{O}_3+H_{2}O\to H_{2}S{O}_4$
According to the stochiomentry of the reaction
18 g $H_{2}O$ will react with 80 g of $S{O}_3$ to give 98 g of $H_{2}S{O}_4$.
So by unitary method in $109\%$ oleum, 9 g of $H_{2}O$ will react with 40 g of $S{O}_3$ to give 49 g of $H_{2}S{O}_4$.
This implies that in 100 g oleum, 40 g (free) $S{O}_3$ and 60 g $H_{2}S{O}_4$ are present.
Hence moles of $H_{2}S{O}_4$ present $\left(x\right)$ = $\frac{60}{98}=0.6122mol$
Similarly moles of $S{O}_3$$\left(y\right)$ = $\frac{40}{80}=0.5mol$
Hence the value of $\frac{x-y}{x+y}$
$=\frac{0.6122+0.5}{0.6122-0.5}=9.91$