Question

Write a note on a) Labelling of oleum b)Volume strength of $H_2_O_2$ .

Write all the reactions and formula involved.​

Answer 1

Answer:

a) Oleum is mixture of H2SO4 and SO3. Labelling of oleum is done for 100 gm sample. For eg. 109% oleum means 100g is the sample and 9g is the water required by sample to completely combine SO3 in the sample.

b) Volume strength of H2O2 is the volume of O2 released on decomposition of one volume of H2O2 at STP. So, if H2O2 is labeled 10Vol decomposing 1L of that solution will give 10L of O2 and if decompose 1mL then give 10mL of O2. ... It is defined as volume of O2 gas liberated at NTP from unit volume of H2O2.

Ur answer is here.....!!!!!

Plz Mark as brainliest....!!!!

Answer 2

Labelling of oleum

$\boxed{\green{Oleum}}$

$\rightarrow$ Oleum is a mixture of sulphuric acid($H_2SO_4$ ) and sulphur trioxide ( $SO_3$). It is also known as fuming sulphuric acid and is volatile in nature.

$\rightarrowThe\:formula\:of\:oleum\:is\:H_2S_2O_7$

$\boxed{\green{ Labelling\:of\:oleum}}$

$\rightarrow$ x% of labelled oleum represents that x grams of $H_2SO_4$ formed when 100 gram of oleum is diluted with water. Therefore, weight of dilute water = (x - 100) gram.

Two types of questions are asked in JEE MAINS and ADVANCED. They are being solved here :-

Que:- Calculate the ℅ of free $SO_3$ in an oleum that is labelled 118% $H_2SO_4$.

Trick :- Weight of water = 118 - 100 = 18 gram.

Moles of free $SO_3$ = Moles of water for dilution = 1 mole.

weight of $SO_3$ in its 1 mole = 80 gram Ans.

Your answer is 80%

Que:- If ℅ of free $SO_3$ is 20% , Calculate labelling.

Trick :- Consider it as 20 gram $SO_3$ . Number of moles = 0.25

Weight of water in its 0.25 moles = 4.5 gram.

Hence, labelling = 100 + 4.5 = 104.5 %

Volume Strength of

$H_2O_2$ .

$\rightarrow$ x volume of $H_2O_2$ means x litre of $O_2$ is liberated at STP by 1 litre of $H_2O_2$ on decomposition.

The reaction that takes place is

2 $H_2O_2 \rightarrow 2H_2O +O_2$ .

$\boxed{\green{ Formulas }}$

• Volume strength = 5.6 × Normality

• Volume strength = 11.2 × Molarity