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Question: On electrical decomposition of 150ml dry pure \({{O}_{2}}\), 10% of \({{O}_{2}}\) gets changed to O,...

On electrical decomposition of 150ml dry pure O2{{O}_{2}}, 10% of O2{{O}_{2}} gets changed to O, then the volume of gaseous mixture after reaction and volume of remaining gas left after passing in turpentine oil will be:
A) 145ml
B) 149ml
C) 128ml
D) 105ml

Explanation

Solution

First we should know how to convert a percentage of decomposition to volume or amount of the gas decomposed. Turpentine oil is a good absorbent of the ozone molecules formed in a reaction mixture.

Complete answer:
Here in the question it is given that, on electrical decomposition of 150ml of dry pure O2{{O}_{2}} gas, 10% o it is getting converted into O atom and we have to find the remaining O2{{O}_{2}} present.
And it is given that to find the volume of remaining gas after passing through the turpentine oil, so from that statement we get a clue that ozone molecule is also present in the reaction mixture.
Let’s first write the data given in the question
The initial concentration of O2{{O}_{2}}=150ml
Now it is said that 10% of 150 ml is getting decomposed to O.So the amount of O2{{O}_{2}} remaining should be calculated.
Percentage of O2{{O}_{2}} converted =10%
Volume of O2{{O}_{2}} converted = 10% of 150ml = 10100×150\dfrac{10}{100}\times 150
Volume of O2{{O}_{2}} decomposed to O = 15ml
Hence the volume of O2{{O}_{2}} remaining = 150-15 ml = 135ml
Now we know that ozone is also producing in the reaction so let’s take that reaction as,
3O2→2O33{{O}_{2}}\to 2{{O}_{3}}
So by this equation we can say that 3 moles of O2{{O}_{2}} are undergoing decomposition to produce 2 moles of O3{{O}_{3}} molecule.
Now let’s do the calculation,
3V→2V3V\to 2V
1V=23V1V=\dfrac{2}{3}V
V is the volume of remaining oxygen gas present.
1V=23×1351V=\dfrac{2}{3}\times 135
V=90ml
Volume of remaining oxygen in the reaction mixture is 90ml
Volume of the mixture = Volume of oxygen remaining ++ Volume of oxygen decomposed to O CC
Volume of mixture = 90ml+15ml=105ml

Note:
If it is given that the solution is passed through alkaline pyragon, then it means there is formation of oxygen gas, since alkaline pyragon absorbs the oxygen gas.