One mole of a mixture of $C O$ and $C O_{2}$ requires exactly $20$ g of $N a O H$ in solution for complete conversion of all the $C O_{2}$ into $N a_{2} C O_{3}$ . If the mixture (one mole) is completely oxidised to $C O_{2}$ , then weight of $N a O H$ to be used is:

60

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80

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40

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20

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Solution

The correct option is D $80$ g
$2 N a O H + C O_{2} \to N a_{2} C O_{3} + H_{2} O$
$80$ g of $N a O H$ ( $2$ moles of $N a O H$ ) required to convert $44$ g of $C O_{2}$ ( $1$ mole of $C O_{2}$ )

If $20$ g of NaOH ( $\frac{1}{2}$ mole) was used, then there was $11$ g of $C O_{2}$ ( $\frac{1}{4}$ mole) in mixture.

$∴$ Mole of $C O$ in mixture $=$ $\frac{3}{4}$
If this $C O$ is completely oxidised to $C O_{2}$ then, mole of $C O_{2}$ formed $=$ $\frac{3}{4}$
$∴$ Total moles of $C O_{2}$ $=$ $\frac{1}{4}$ + $\frac{3}{4}$ $=$ $1$
$∴$ Moles of $N a O H$ required $= 2$ $\times$ moles of $C O_{2}$ $=$ $2$ $\times$ $1$ $=$ $2$
$∴$ Weight of $N a O H$ required $=$ $2$ $\times$ $40$ $=$ $80$ g

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