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`1 L` of a gaseous mixture is effused in `5 min 11s`, while `1 L` of oxygen takes `10 min`. The gaseous mixture contains methane and hydrogen. Calculate (`a`) The density of gaseous mixture. (`b`) The percentage by volume of each gas in mixture. |
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Answer» `{:(O_(2),"Mixture"),(d_(1)=16, d_(2)=?),(t_(1)=10xx60=600 s, t_(2)=5xx60+11=311 s):}` Volume is same, so `(t_(2))/(t_(1))=sqrt((d_(2))/(d_(1)))` `(311)/(600)=sqrt((d_(2))/(16))` `implies ((311)^(2))/((600)^(2))=(d_(2))/(16)` `implies d_(2)=4.3` `d_(H_(2))=1` and `d_(CH_(4))=8`. So `d_(mix)=(d_(CH_(4))xxx+(100-x)d_(H_(2)))/(100)` `4.3xx100=8x+(100-x)xx1` `430=8x+100-x` `330=7x` `x=47.14%` `%CH_(4)=47.14`, `%H_(2)=52.86` |
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