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Question

The process of eudiometry is used to calculate the volume of the various reacting gases present together in a closed vessel, it is done by sparking the content and noting the volume change or pressure change depending upon the conditions mentioned. Various reagents like alkaline pyrogallol, turpentine oil, caustic potash, conc. H2SO4 are used to absorb the selective gases like O2,O3,CO2 and H2O(v) respectively. A 2.0 mole mixture of H2(g),O2(g) and He(g) are placed together in a closed container at pressure equal to 50 atm. An electric spark is passed and pressure is noted as 12.5 atm after cooling the content to original temperature (room temp). Oxygen gas is introduced for pressure to change to 25 atm, keeping volume and temperature constant. Again electrical spark is passed and pressure drops to 10 atm under original temperature conditions. The moles of He(g) present in the mixture sample is:

A
0.2
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B
0.1
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C
0.5
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D
1.4
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Solution

The correct option is B 0.1
Let the number of moles of H2,O2 and He are x, y and z respectively
x+y+x=2.0
Pressure exerted by 2 moles of gaseous mixture is 50 atm.
After the first electric spark decrease in pressure = 5012.5=37.5 atm
decrease in no. of moles of gaseous mixture is 1.5
But from the given information limiting reagent in first step is O2
from H2(g)+12O2(g)H2O(l)
H2 and O2 reacted are in the ratio of 2:1
0.5 moles of O2 should be reacted with 1.0 mole of H2
No. of mole of O2 in the mixture (y)=0.5
When again O2 is passed, pressure increases from 12.5 to 25.0 atm
Change in pressure = 12.5 atm
No. of moles of O2 added = 12.5×250=0.5 mole
Now, H2 will be completely reacted
No. of moles H2 actually left after first electric spark = x1
H2+12O2H2O
(x1)(x12)
Now change in pressure in 2510=15 atm
change in no of moles is 15×250=0.6
(x1)+(x12)=0.6
x=1.4
no. of moles of H2=1.4
no. of moles of He=2.0(1.4+0.5)=0.1


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