Oleum is H2SO4 containing dissolved SO3. A sample labelled 118% oleum means that 100 g of it yields 118 g of pure H2SO4 when diluted with sufficient water (H = 1, S = 32, O = 16). The mass percentage of free SO3 in the sample is:

JEE Advanced levelSingle correct optionDifficulty 3 of 3mc031

Oleum is $H X_{2} SO X_{4}$ containing dissolved $SO X_{3}$ . A sample labelled $118%$ oleum means that $100 g$ of it yields $118 g$ of pure $H X_{2} SO X_{4}$ when diluted with sufficient water (H = 1, S = 32, O = 16). The mass percentage of free $SO X_{3}$ in the sample is:

Let

100

g contain

x

SO X_{3}

and

(100 - x)

H X_{2} SO X_{4}

. Since

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

converts

80

g to

98

g, the total is

100 - x + 98 x /80 = 118

. Then

0.225 x = 18

and

x = 80

. The sample is

80%

SO X_{3}