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If the percent free SO(3) in an oleum is...

If the percent free `SO_(3)` in an oleum is 20% then label the sample of oleum in terms of percent `H_(2) SO_(4)`,

Text Solution

Verified by Experts

Let `x g H_(2) O` combine with all the free `SO_(3)` in `100 g` of the oleum to give a total `(100 + x)g H_(2) SO_(4)`.
`18 g H_(2) O` combines with `80 g SO_(3)`
`x g H_(2) O` combines with `= (x)/(18) xx 80 = 20 g SO_(3)`
`:.x = (20 xx 18)/(80) = 4.5 g`
Labelling of oleum in terms of percent `H_(2) SO_(4)`
`= (100 + x) = (100 + 4.5) = 1.4.5% H_(2) SO_(4)`
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Percent Free So3 In An Oleum And Percent Yield

Calculate the % of free SO_(3) in an oleum, that is labelled 118%.

Knowledge Check

  • 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

  • A 110% sample of oleum contains

    A
    44.4% of `SO_(3)`
    B
    55.6% of sulphuric acid
    C
    55.6% of `SO_(3)`
    D
    44.4% of sulphuric acid

  • Comprehension # 6 The percentage labelling of oleum is a unique process by means of which, the percentage composition of H_(2)SO_(4), SO_(3) (free) and SO_(3) (combined) is calculated. Oleum is nothing but it is a mixture of H_(2)SO_(4) and SO_(3) i.e., H_(2)S_(2)O_(7) , which is obtained by passing. SO_(3) in solution of H_(2)SO_(4) . In order of dissolve free SO_(3) in oleum, dilution of oleum is done, in which oleum converts into pure H_(2)SO_(4) . It is shown by the reaction as under : H_(2)SO_(4)+SO_(3)+H_(2)Orarr2H_(2)SO_(4)("pure") or " " SO_(3)+H_(2)OrarrH_(2)SO_(4)("pure") When 100g sample of oleum is diluted with desired weight of H_(2)O("in" g) , then the total mass of pure H_(2)SO_(4) obtained after dilution is known as percentage labelling in oleum. For example, if the oleum sample is labelled as ""109%H_(2)SO_(4)" it means that 100 g of oleum on dilution with 9m of H_(2)O provides 109g pure H_(2)SO_(4) , in which all free SO_(2) in 100g of oleum is dissolved. For 109% labelled oleum if the number of moles of H_(2)SO_(4) and free SO_(3) be x and y respectively, then what will be the value of (x+y)/(x-y) ?

    A
    `18`
    B
    `18`
    C
    `(1)/(3)`
    D
    `9.9`

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    Calculate the percent free SO_(3) in an oleum which is labelled '118% H_(2) SO_(4)' .

    Calculate the percentage of free SO_(3) in an oleum ( considered as a solution of SO_(3) in H_(2)SO_(4) ) that is labelled 109 % H_(2)SO_(4) .

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    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. 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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