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The weight of 600 ml of a mixture of ozone and oxygen is 1 gm at STP. The volume of oxygen in the mixture is |
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Answer» 200 ML `"volume"_("STP")=(1)/(M)xx22400=600impliesM=(112)/(3)` If moles of `O_(2)=X % O_(3)=y` then `M=(32x+48y)/(x+y)=(112)/(3)` `IMPLIES 96x+144y=112x+112y` `(x)/(y)=(1)/(2)` `therefore V_(O_(2))=(1)/(3)xx600=200mL` |
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