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(D) The term '$15$ volume' of $H_2O_2$ means that $1 \ mL$ of this $H_2O_2$ solution produces $15 \ mL$ of $O_2$ gas at $STP$ upon complete decomposit

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(D) The term '$15$ volume' of $H_2O_2$ means that $1 \ mL$ of this $H_2O_2$ solution produces $15 \ mL$ of $O_2$ gas at $STP$ upon complete decomposition.Given volume of $H_2O_2$ solution = $2 \ L = 2000 \ mL$.Since $1 \ mL$ of $H_2O_2$ gives $15 \ mL$ of $O_2$,then $2000 \ mL$ of $H_2O_2$ will give $2000 	imes 15 = 30000 \ mL$ of $O_2$.Converting $30000 \ mL$ to litres: $30000 \ mL / 1000 = 30 \ L$.Therefore,$x = 30$.

Two litres of $15$ volume solution of $H_2O_2$ on complete decomposition liberated $x$ litres of $O_2$ at $STP$. What is the value of $x$ (in litres)?

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