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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. H_(2)SO_(4) are used to absorb theselective gases like O_(2), O_(3), CO_(2) and H_(2)O (v) respectively. A 2.0 mole mixture of H_(2)(g),O_(2)(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 temp. (room temp). Oxygen gas is introduced for pressure to change to 25 atm, keeping volume and temp constant. Again electrical spark is passed and pressure drops to 10 atm under original temp. conditions. The mole fraction of H_(2)(g) present in the initial mixture is: |
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Answer» 0.05 `therefore x+y+z=20` Pressure exerted by 2 moles of gaseous MIXTURE is 50 atm. After the first electric spark decrease in pressure =50-12.5=37.5 atm `therefore` decrease in no. of moles of gaseous mixture is 1.5 But from the given information limiting reagent in first step is `O_(2)` `therefore" from "H_(2)(g)+(1)/(2)O_(2)(g) RARR H_(2)O(l)` `therefore H_(2) and O_(2)` REACTED are in the ratio of 2:1 `therefore` 0.5 mole of `O_(2)` should be reacted with 1.0 mole of `H_(2)` `therefore` No. of mole of `O_(2)` in the mixture (y) =0.5 When again `O_(2)` is passed, pressure increased from 12.5 to 25.0 atm `therefore` Change in pressure =12.5 atm `therefore` No. of moles of `O_(2)` added `=(12.5xx2)/(50)=0.5` mole Now `H_(2)` will be completely reacted No. of moles `H_(2)` actually left after first electric spark =x-1 `H_(2)+(1)/(2)O_(2) rarr H_(2)O` `(x-1)((x-1)/(2))` Now change in pressure in 25-10 =15 atm `therefore` change in no. of moles is `(15xx2)/(50)=0.6` `therefore (x-1)+((x-1)/(2))=0.6` `rArr x=1.4` `therefore` no of moles `H_(2)=1.4` `therefore` no. of moles of `He=2.0-(1.4+0.5)=0.1` `therefore` mole fraction of `H_(2)=(1.4)/(2.0)=0.7` |
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