An oleum sample contains 35.0% free SO3 by mass, the rest being H2SO4. Oleum is labelled by the mass of pure H2SO4 obtainable from 100 g of it on adding enough water. Find the labelling percentage to the nearest integer (H = 1, O = 16, S = 32).

JEE Main levelNumerical answerDifficulty 2 of 3pn157

An oleum sample contains $35.0%$ free $SO X_{3}$ by mass, the rest being $H X_{2} SO X_{4}$ . Oleum is labelled by the mass of pure $H X_{2} SO X_{4}$ obtainable from $100 g$ of it on adding enough water. Find the labelling percentage to the nearest integer (H = 1, O = 16, S = 32).

Numerical answer type, answer in %, tolerance 0.5.

Free SO3 reacts as

SO X_{3} + H X_{2} O H X_{2} SO X_{4}

(80 g gives 98 g). From 100 g oleum: H2SO4 present = 65.0 g; from SO3 = 35.0 x 98/80 = 42.875 g. Total = 107.875 g, so the label is about 108%.