On electrical decomposition of 150 mL dry an | vedclass

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(A) The chemical reaction for the formation of ozone is: $3O_2(g) 	o 2O_3(g)$.Given initial volume of $O_2 = 150 \ mL$.$10\%$ of $O_2$ is converted t

Vedclass · Accepted Answer

$145 \ mL, 135 \ mL$

Vedclass · Answer

(A) The chemical reaction for the formation of ozone is: $3O_2(g) 	o 2O_3(g)$.Given initial volume of $O_2 = 150 \ mL$.$10\%$ of $O_2$ is converted to $O_3$,so volume of $O_2$ reacted = $150 	imes 0.10 = 15 \ mL$.According to the stoichiometry,$3 \ mL$ of $O_2$ produces $2 \ mL$ of $O_3$.Therefore,$15 \ mL$ of $O_2$ will produce $\frac{2}{3} 	imes 15 = 10 \ mL$ of $O_3$.Volume of remaining $O_2 = 150 - 15 = 135 \ mL$.Total volume of the gaseous mixture = $135 \ mL \ (O_2) + 10 \ mL \ (O_3) = 145 \ mL$.Turpentine oil absorbs $O_3$ gas,so the volume of the remaining gas $(O_2)$ = $135 \ mL$.

On electrical decomposition of $150 \ mL$ dry and pure $O_2$,$10\%$ of $O_2$ gets changed to $O_3$. The volume of the gaseous mixture after the reaction and the volume of the remaining gas left after passing through turpentine oil will be,respectively:

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