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Standard XII

Physics

Temperature

1 L of a mixt...

Question

1 L of a mixture of CO and $C O_{2}$ is passed through red hot charcoal in a tube. The new volume becomes 1.4 L. Find out percentage composition of mixture by volume. All measurements are made at constant temperature and pressure.

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Solution

Let $x L$ be volume of $C O,$ then volume of $C O_{2} = (1 - x) L$ .
$C + C O_{2} ⟶ 2 C O$
$(1 - x) L$ $C O_{2}$ reacts with $C$ .
Final volume will be $x + (1 - x) + (1 - x) = 1.6 x = 0.4 L C O$
Volume of $C O_{2} = 0.6 L$

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1 L of a mixture of CO and CO2 is passed through red hot charcoal in a tube. The new volume becomes 1.4 L. Find out percentage composition of mixture by volume. All measurements are made at constant temperature and pressure.

Let x L be volume of CO, then volume of CO2=(1−x)L. C+CO2⟶2CO(1−x)L CO2 reacts with C.Final volume will be x+(1−x)+(1−x)=1.6x=0.4LCOVolume of CO2=0.6L

1 L of a mixture of CO and $C O_{2}$ is passed through red hot charcoal in a tube. The new volume becomes 1.4 L. Find out percentage composition of mixture by volume. All measurements are made at constant temperature and pressure.

Let $x L$ be volume of $C O,$ then volume of $C O_{2} = (1 - x) L$ .
$C + C O_{2} ⟶ 2 C O$
$(1 - x) L$ $C O_{2}$ reacts with $C$ .
Final volume will be $x + (1 - x) + (1 - x) = 1.6 x = 0.4 L C O$
Volume of $C O_{2} = 0.6 L$