A mixture ( $15$ ml) of $C O$ and $C O_{2}$ is mixed with $V$ ml (excess) of oxygen and electrically sparked. The volume after explosion was $(V + 12)$ ml. What would be the residual volume if $25$ ml of the original mixture is exposed to alkali?

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Solution

The correct option is C $10$ ml
$C O + C O_{2} + O_{2} \to C O_{2}$
$15 m l$ $v m l$ $V + 12 m l$
$\Rightarrow$ actually only co reach with oxygen to produce
$C O_{2}$
$\to 2 C O + O_{2} \to 2 C O_{2}$
2 moles of CO give 2 moles of $C O_{2}$
And 1 mole of Oxygen with CO give 2 moles of $C o_{2}$
thus,
$22.4 L$ of oxygen produces $2.22.5 L$ of $C O_{2}$
V ml of oxygen gives 24 ml of $C O_{2}$
thus amount of $C O_{2}$ in mixture produced by
$C O = 15 - 2 V$
$\Rightarrow (V + 12) - 2 V$ ml of $C O_{2}$ 1) produced by $C O = 15 - 2 V$
$V + 12 - 2 V = 15 - 2 V$
$V = 3 m l$
Thurs, $25 m l$ of mixture will contain $\frac{2 V}{15} \times 25 m l$ of CO
$\Rightarrow 10 m l$ of CO
Ans 10 ml

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