Oleum is considered to be a mixture of H2SO4 and free SO3. A sample of oleum contains 30% SO3 by mass. On adding enough water, all SO3 is converted to H2SO4 (SO3 + H2O -> H2SO4). Taking SO3 = 80 g/mol and H2SO4 = 98 g/mol, find the total mass of H2SO4 (in g, two decimals) obtainable from 100 g of this oleum, i.e. its percentage labelling.

JEE Advanced levelNumerical answerDifficulty 3 of 3pn214

Oleum is considered to be a mixture of $H X_{2} SO X_{4}$ and free $SO X_{3}$ . A sample of oleum contains 30% $SO X_{3}$ by mass. On adding enough water, all $SO X_{3}$ is converted to $H X_{2} SO X_{4}$ ( $SO X_{3} + H X_{2} O H X_{2} SO X_{4}$ ). Taking $SO X_{3}$ = 80 g/mol and $H X_{2} SO X_{4}$ = 98 g/mol, find the total mass of $H X_{2} SO X_{4}$ (in g, two decimals) obtainable from 100 g of this oleum, i.e. its percentage labelling.

Numerical answer type, answer in g, tolerance 0.01.

100 g oleum has 70 g

H X_{2} SO X_{4}

and 30 g

SO X_{3}

. The 30 g

SO X_{3}

is 0.375 mol and forms 0.375 mol

H X_{2} SO X_{4}

0.375 \times 98 = 36.75

g. Total

H X_{2} SO X_{4}

70 + 36.75 = 106.75

g, so the oleum is labelled 106.75%.