Question

A mixture is prepared by mixing $10g{H}_{2}S{O}_4$ and $40gS{O}_3$. FInd the % labeling of oleum.

Answer 1

Formulae for formation of oleum

$H_{2}S{O}_4\left(aq\right)+S{O}_3\left(g\right)⟶H_2{S}_2{O}_7\left(l\right)$

The molar mass of oleum=$2+2\times 32+7\times 16$

$=2+64+112$

$=178g$

then, Mass of oleum formed=$10+40$

$=50g$

$\therefore$ The percentage labelling of oleum$=\frac{50}{178}\times 100=28.09$%