4 mL of 'X volume' H_2O_2 on heating gives 8 | vedclass

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(B) The decomposition of $H_2O_2$ is given by the reaction: $2H_2O_2(aq) \rightarrow 2H_2O(l) + O_2(g)$.According to the definition of '$X$ volume' $H

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(B) The decomposition of $H_2O_2$ is given by the reaction: $2H_2O_2(aq) \rightarrow 2H_2O(l) + O_2(g)$.According to the definition of '$X$ volume' $H_2O_2$,$1 \ mL$ of $H_2O_2$ solution produces $X \ mL$ of $O_2$ gas at $STP$.Given that $4 \ mL$ of $H_2O_2$ produces $80 \ mL$ of $O_2$ at $STP$.Therefore,$1 \ mL$ of $H_2O_2$ produces $\frac{80 \ mL}{4 \ mL} = 20 \ mL$ of $O_2$ at $STP$.Thus,the value of $X$ is $20$.

$4 \ mL$ of '$X$ volume' $H_2O_2$ on heating gives $80 \ mL$ of oxygen at $STP$. The value of $X$ is:

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