$1 L$ of a mixture of $C O$ and $C O_{2}$ becomes $1.6 L$ when passes through a red-hot charcoal. The $C O_{2}$ in the mixture is:

35 %

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40 %

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50 %

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60 %

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Solution

The correct option is D $60 %$
Let the volume of $C O$ is $= a$ litre
And the volume of ${C O}_{2}$ is $= b$ litre
Now given, $a + b = 1$
Given when it pass through red-hot charcoal only ${C O}_{2}$ is reacted
${C O}_{2} + C = 2 C O, 2 b l i t r e s$
$a + 2 b = 1.6 & a + b = 1$
So, $a = 0.4$ litre and $b = 0.6$ litre.
Or $a = 0.4$ (40%) and $b = 0.6$ (60%)
$C O = 40$ % and ${C O}_{2} = 60$ %
Hence, option $D$ is correct.

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1 L of a mixture of CO and CO2 becomes 1.6 L when passes through a red-hot charcoal. The CO2 in the mixture is:

$1 L$ of a mixture of $C O$ and $C O_{2}$ becomes $1.6 L$ when passes through a red-hot charcoal. The $C O_{2}$ in the mixture is: