
On electrical decomposition of 150ml dry pure ${{O}_{2}}$, 10% of ${{O}_{2}}$ 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
Answer
556.2k+ views
Hint: 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 ${{O}_{2}}$ gas, 10% o it is getting converted into O atom and we have to find the remaining ${{O}_{2}}$ 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 ${{O}_{2}}$=150ml
Now it is said that 10% of 150 ml is getting decomposed to O.So the amount of ${{O}_{2}}$ remaining should be calculated.
Percentage of ${{O}_{2}}$ converted =10%
Volume of ${{O}_{2}}$ converted = 10% of 150ml = $\dfrac{10}{100}\times 150$
Volume of ${{O}_{2}}$ decomposed to O = 15ml
Hence the volume of ${{O}_{2}}$ remaining = 150-15 ml = 135ml
Now we know that ozone is also producing in the reaction so let’s take that reaction as,
$3{{O}_{2}}\to 2{{O}_{3}}$
So by this equation we can say that 3 moles of ${{O}_{2}}$ are undergoing decomposition to produce 2 moles of ${{O}_{3}}$ molecule.
Now let’s do the calculation,
$3V\to 2V$
$1V=\dfrac{2}{3}V$
V is the volume of remaining oxygen gas present.
$1V=\dfrac{2}{3}\times 135$
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 $C$
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.
Complete answer:
Here in the question it is given that, on electrical decomposition of 150ml of dry pure ${{O}_{2}}$ gas, 10% o it is getting converted into O atom and we have to find the remaining ${{O}_{2}}$ 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 ${{O}_{2}}$=150ml
Now it is said that 10% of 150 ml is getting decomposed to O.So the amount of ${{O}_{2}}$ remaining should be calculated.
Percentage of ${{O}_{2}}$ converted =10%
Volume of ${{O}_{2}}$ converted = 10% of 150ml = $\dfrac{10}{100}\times 150$
Volume of ${{O}_{2}}$ decomposed to O = 15ml
Hence the volume of ${{O}_{2}}$ remaining = 150-15 ml = 135ml
Now we know that ozone is also producing in the reaction so let’s take that reaction as,
$3{{O}_{2}}\to 2{{O}_{3}}$
So by this equation we can say that 3 moles of ${{O}_{2}}$ are undergoing decomposition to produce 2 moles of ${{O}_{3}}$ molecule.
Now let’s do the calculation,
$3V\to 2V$
$1V=\dfrac{2}{3}V$
V is the volume of remaining oxygen gas present.
$1V=\dfrac{2}{3}\times 135$
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 $C$
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.
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