One litre of mixture of $C O$ and ${C O}_{2}$ is passed through red hot charcoal in tube. The new volume becomes $1.4 l i t r e$ . Find out composition of original mixture by volume. All measurements are made at same $P$ and $T$ .

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Solution

$C + C O_{2} \to 2 C O$
$\Rightarrow$ Initial volume $= 1 L$
Final volume $= 1.6 L$
Let $V_{C O_{2}} = x$ $V_{C O} = (1 - x)$
$\Rightarrow (1 - x) + 2 x = 1.6$
$1 + x = 1.6$
$x = 0.6$
$∴ % C O_{2} = 60 %$ , $% C O = 40 %$
Hence, solved.

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One litre of mixture of CO and CO2 is passed through red hot charcoal in tube. The new volume becomes 1.4 litre. Find out composition of original mixture by volume. All measurements are made at same P and T.

C+CO2→2CO⇒ Initial volume =1L Final volume =1.6LLet VCO2=x VCO=(1−x)⇒(1−x)+2x=1.6 1+x=1.6 x=0.6∴%CO2=60%, %CO=40%Hence, solved.

One litre of mixture of $C O$ and ${C O}_{2}$ is passed through red hot charcoal in tube. The new volume becomes $1.4 l i t r e$ . Find out composition of original mixture by volume. All measurements are made at same $P$ and $T$ .

$C + C O_{2} \to 2 C O$
$\Rightarrow$ Initial volume $= 1 L$
Final volume $= 1.6 L$
Let $V_{C O_{2}} = x$ $V_{C O} = (1 - x)$
$\Rightarrow (1 - x) + 2 x = 1.6$
$1 + x = 1.6$
$x = 0.6$
$∴ % C O_{2} = 60 %$ , $% C O = 40 %$
Hence, solved.