Question

$100$ gm oleum of $109\%$ labelling present in a container. Which is/are correct statements?

• A. $40\%$ free $S{O}_3$ present in oleum
• B. $60\%$ combined $S{O}_4$ present in oleum
• C. New $\%$ labelling of oleum is $104.3\%$ when $4.5$gm of water added in $100$gm oleum
• D. Minimum mass of $H_{2}S{O}_4$ is $30$gm in $50$ gm oleum

Answer 1

The correct option is A $40\%$ free $S{O}_3$ present in oleum

Oleum as mixture of $H_{2}S{O}_4$ and $S{P}_3$ gas

when we and water to $S{O}_3$ if will form $H_{2}S{O}_4$

$S{O}_3+H_{2}O\to H_{2}S{O}_4$

will say $x$gm of $S{O}_3$ and $100-X$gm A $H_{2}S{O}_4$

number of mole $S{O}_3=\frac{x}{molemassofS{O}_3}=\frac{x}{80}$

Amount of $H_{2}S{O}_4$ formed $=\frac{x}{80}\times molarmassof{H}_2^{\prime }S{O}_4$

$\frac{x}{80}\times 98=\frac{98x}{80}gm.$

mass of $H_{2}S{O}_4$ frmed + mass of $H_{2}S{O}_4$ already present in oleum = total mass of oleum

$\frac{98x}{80}+100-x=109$

$\frac{18x}{80}=109-100$

$x=\frac{9\times 80}{18}=40$

do $40t$ free $S{O}_3$ present i oleum.