Q. The number of moles oxygen liberated by electrolysis of $90\, g$ of water is :

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9 moles

4.5 moles

2.5 moles

5 moles

Solution:

$90\, g$ of water is $5$ moles
$2 H _{2} O \longrightarrow 2 H _{2}+ O _{2}$
$2$ moles of $H _{2} O$ liberates $1$ moles of $O _{2}$
Therefore $5$ moles $=1 \times 5 / 2$
No. of moles of $O _{2}=2.5$ moles

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