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The percentage labelling (mixture of H(2...

The percentage labelling (mixture of `H_(2)SO_(4)` and `SO_(3))` refers to the total mass of pure `H_(2)SO_(4)`. The total amount of `H_(2)SO_(4)` found after adding calculated amount of water to `100 g` oleum is the percentage labelling of oleum. The higher the percentage lebeling of oleum higher is the amount of free `SO_(3)` in the oleum sample.
The percent free `SO_(3)` is an oleum is 20%. Label the sample of oleum in terms of percent `H_(2) SO_(4)`.

A

1.135

B

1.045

C

1.0675

D

1.2

Text Solution

Verified by Experts

The correct Answer is:
B
% of free `SO_(3) = 20%`
i.e., `20 g SO_(3)` is present in `100 g` oleum.
`:' 80 g SO_(3)`requires `18 g H_(2) O`
`20 g SO_(3)` requires `= (18)/(80) xx 20 = 4.5 g H_(2) O`
`:.` % labelling of oleum `= 100 + 4.5 = 104.5%
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