Home
Class 11
CHEMISTRY
An oleum sample contains 10 g SO(3) and ...

An oleum sample contains `10 g SO_(3)` and `15 g H_(2) SO_(4)`
Answer the following questions on the basis of above information :
Find new `%` labeling of `0.45g` of `H_(2)O` is added to the above oleum sample

A

`100%`

B

`102.83%`

C

`107.07%`

D

`109%`

Text Solution

Verified by Experts

The correct Answer is:
(C )
Promotional Banner

Topper's Solved these Questions

  • MOLE CONCEPT

    ALLEN|Exercise EXERCISE - 01|49 Videos

  • MOLE CONCEPT

    ALLEN|Exercise EXERCISE - 02|46 Videos

  • IUPAC NOMENCLATURE

    ALLEN|Exercise Exercise - 05(B)|8 Videos

  • QUANTUM NUMBER & PERIODIC TABLE

    ALLEN|Exercise J-ADVANCED EXERCISE|26 Videos

Similar Questions

Explore conceptually related problems

An oleum sample contains 10 g SO_(3) and 15 g H_(2) SO_(4) Answer the following questions on the basis of above information : % labelling of oleum sample is .

A 50 gm oleum sample contains (400/49) gm of combined SO_(3). Find percent label of the oleum sample.

Which one of gas dissolves in H_(2)SO_(4) to give oleum?

5.0 g of a sample containing (NH_(4))_(2) SO_(4) & inert impurity (X) was boiled with excess of NaOH and the NH_(3)(g) so produced was absorbed in 200ml of 0.5 M H_(2)SO_(4) solution in another vessel. The unreacted H_(2)SO_(4) required 400ml of 0.4 M solution of NaOH for exact neutralization. Calculate weight % of nitrogen in given sample

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. In the above question number 1 , what will be the percentage of combined SO_(3) in the given oleum sample?

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