Home
Class 11
CHEMISTRY
Calculate the % of free SO(3) in oleum (...

Calculate the `%` of free `SO_(3)` in oleum ( a solution of `SO_(3)` in `H_(2)SO_(4)`) that is labelled `109% H_(2)SO_(4)` by weight.

Text Solution

AI Generated Solution

To calculate the percentage of free \( SO_3 \) in oleum that is labeled as \( 109\% H_2SO_4 \) by weight, we will follow these steps: ### Step 1: Understand the Meaning of \( 109\% H_2SO_4 \) The label \( 109\% H_2SO_4 \) indicates that for every 100 grams of solution, there are 109 grams of \( H_2SO_4 \). This means that there is an excess of \( H_2SO_4 \) compared to the standard concentration. ### Step 2: Calculate the Amount of Water Since \( 109\% H_2SO_4 \) means there is 9 grams of water (because \( 109 - 100 = 9 \)), we can conclude that in 100 grams of this oleum solution, there are: - 100 grams of \( H_2SO_4 \) ...
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)|7 Videos

  • QUANTUM NUMBER & PERIODIC TABLE

    ALLEN|Exercise J-ADVANCED EXERCISE|24 Videos

Similar Questions

Explore conceptually related problems

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

The density of a 3.6 M H_(2)SO_(4) solution that is 29% H_(2)SO_(4) by mass will be

For preparing 0.1 M solution of H_(2)SO_(4) in one litre, we need H_(2)SO_(4)

Calculate the normality of solution of 4.9 % (w/v) of H_(2)SO_(4) .

In 109% H_(2)SO_(4) labelled oleum, the percent of free SO_(3) and H_(2)SO_(4) are

Electrolysis of 50% H_(2)SO_(4) gives

Calculate the pH of following solutions. N//100 H_(2)SO_(4)

The molality of 15 % by wt solution of H_2SO_4 is

Oleum is considered as a solution of SO_(3) in H_(2)SO_(4) , which is obtained by passing SO_(3) in solution of H_(2)SO_(4) When 100 g sample of oleum is diluted with desired mass of H_(2)O then the total mass of H_(2)SO_(4) obtained after dilution is known is known as % labelling in oleum. For example, a oleum bottle labelled as 109% H_(2)SO_(4) means the 109 g total mass of pure H_(2)SO_(4) will be formed when 100 g of oleum is diluted by 9 g of H_(2)O which combines with all the free SO_(3) present in oleum to form H_(2)SO_(4) as SO_(3)+H_(2)O to H_(2)SO_(4) What is the % of free SO_(3) in an oleum that is labelled as 104.5% H_(2)SO_(4) ?

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%'.