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Standard XII

Chemistry

Concentration Terms

Take 1 L of...

Question

Take $1 L$ of a mixture of $C O$ and $C O_{2}$ . Pass this mixture through a tube containing red hot charcoal. The volume now becomes $1.6 L$ . The volumes are measured under the same conditions. Find the composition of the mixture by volume.

0.3 L

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0.6 L

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0.9 L

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None of the above

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Solution

The correct option is A 0.6 L
Let the volume of $C O$ in the mixture be $V$ .
The volume of $C O_{2}$ in the mixture is $(1 - V)$ .
$C O_{2} + C \to 2 C O$
$(1 - V) L$ $C O_{2}$ forms $2 (1 - V) L$ $C O$ .
Total $C O :$
$V + 2 (1 - V) = 1.6$
or, $V = 0.4$
$C O = 0.4 L$
$C O_{2} = 1 - V = 0.6 L = 600 m l$

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Take 1L of a mixture of CO and CO2. Pass this mixture through a tube containing red hot charcoal. The volume now becomes 1.6L. The volumes are measured under the same conditions. Find the composition of the mixture by volume.

The correct option is A 0.6 LLet the volume of CO in the mixture be V.The volume of CO2 in the mixture is (1−V).CO2+C→2CO(1−V)L CO2 forms 2(1−V)L CO.Total CO:V+2(1−V)=1.6or, V=0.4CO=0.4LCO2=1−V=0.6L=600ml

Take $1 L$ of a mixture of $C O$ and $C O_{2}$ . Pass this mixture through a tube containing red hot charcoal. The volume now becomes $1.6 L$ . The volumes are measured under the same conditions. Find the composition of the mixture by volume.