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Find the % labelling of 100 gm oleum sam...

Find the % labelling of 100 gm oleum sample if it contains 20 gm
`SO_(3)`

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To find the % labeling of a 100 gm oleum sample that contains 20 gm of SO₃, we can follow these steps: ### Step-by-Step Solution: 1. **Understand the Composition of Oleum**: Oleum is a solution of sulfur trioxide (SO₃) in sulfuric acid (H₂SO₄). The formula can be represented as: \[ \text{Oleum} = \text{SO}_3 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_4 ...
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Oleum is mixture of H_(2)SO_(4) and SO_(3) i.e. H_(2)S_(2)O_(7) which is obtained by passing SO_(3) is solution of H_(2)SO_(4) . In order to dissolve SO_(3) in oleum, dilution of oleum is done by water in which oleum is converted into pure H_(2)SO as shown below: H_(2)SO_(4)+SO_(3)+H_(2)Oto2H_(2)SO_(4) (pure) When 100 gm oleum is diluted with water then total mass of diluted oleum is known as percentage labelling in oleum. For example: 109% H_(2)SO_(4) labelling of oleum sample means that 109 gm pure H_(2)SO_(4) is obtained on diluting 100 gm oleum with 9 gm H_(2)O which dissolves al free SO_(3) in oleum. If the number of moles of free SO_(3), H_(2)SO_(4) , and H_(2)O be x, y and z respectively in 118% H_(2)SO_(4) labelled oleum, the value of (x+y+z) is

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. Higher the percentage lebeling of oleum higher is the amount of free SO_(3) in the oleum sample. What is the amount of free SO_(3) in an oleum sample labelled as '118%'.

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.Higher the percentage lebeling of oleum higher is the amount of free SO_(3) in the oleum sample. 100 g sample of '149%' oleum was taken and calculated amount of H_(2)O was added to make H_(2)SO_(4) . 500 mL solution of x MKOH solution is required to neutralize the solution. The value of x is.

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