Three vessels (A, B, C) contain H_2O_2 soluti | vedclass

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Question

(A) For vessel $A$: $10 \ vol \ H_2O_2$ means $1 \ mL$ of $H_2O_2$ solution gives $10 \ mL \ O_2$ at $NTP$. $2 \ H_2O_2 \longrightarrow 2 \ H_2O + O_2

Vedclass · Accepted Answer

$C > A > B$

Vedclass · Answer

(A) For vessel $A$: $10 \ vol \ H_2O_2$ means $1 \ mL$ of $H_2O_2$ solution gives $10 \ mL \ O_2$ at $NTP$. $2 \ H_2O_2 \longrightarrow 2 \ H_2O + O_2$. $2 \ mol \ (68 \ g)$ of $H_2O_2$ gives $22400 \ mL \ O_2$ at $NTP$. $O_2$ evolved from $500 \ mL$ of $10 \ vol \ H_2O_2 = 500 	imes 10 = 5000 \ mL$. Weight of $H_2O_2$ in $A = \frac{68 	imes 5000}{22400} \approx 15.18 \ g$. For vessel $B$: $O_2$ evolved from $100 \ mL$ of $30 \ vol \ H_2O_2 = 100 	imes 30 = 3000 \ mL$. Weight of $H_2O_2$ in $B = \frac{68 	imes 3000}{22400} \approx 9.11 \ g$. For vessel $C$: $Molarity = 2 \ M$,$Volume = 250 \ mL = 0.25 \ L$. $Moles \ of \ H_2O_2 = Molarity 	imes Volume = 2 	imes 0.25 = 0.5 \ mol$. Weight of $H_2O_2$ in $C = 0.5 \ mol 	imes 34 \ g/mol = 17 \ g$. Comparing the weights: $C (17 \ g) > A (15.18 \ g) > B (9.11 \ g)$. Thus,the order is $C > A > B$.

Similar Questions

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