The volume strength of 8.9 , M , H_{2}O_{2} s | vedclass

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(A) The decomposition of $H_{2}O_{2}$ is given by: $2H_{2}O_{2}(aq) \rightarrow 2H_{2}O(l) + O_{2}(g)$.At $STP$ ($273 \, K$ and $1 \, atm$),$1 \, mole

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$100$

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(A) The decomposition of $H_{2}O_{2}$ is given by: $2H_{2}O_{2}(aq) ightarrow 2H_{2}O(l) + O_{2}(g)$.At $STP$ ($273 \, K$ and $1 \, atm$),$1 \, mole$ of $O_{2}$ gas occupies $22.4 \, L$.From the stoichiometry,$2 \, moles$ of $H_{2}O_{2}$ produce $1 \, mole$ of $O_{2}$.Therefore,$1 \, mole$ of $H_{2}O_{2}$ produces $11.2 \, L$ of $O_{2}$ at $STP$.Volume strength $= Molarity 	imes 11.2 = 8.9 	imes 11.2 = 99.68$.Rounding off to the nearest integer,we get $100$.

The volume strength of $8.9 \, M \, H_{2}O_{2}$ solution calculated at $273 \, K$ and $1 \, atm$ is.........
$(R = 0.0821 \, L \, atm \, K^{-1} \, mol^{-1})$ (rounded off to the nearest integer)

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The volume strength of $8.9 \, M \, H_{2}O_{2}$ solution calculated at $273 \, K$ and $1 \, atm$ is.........$(R = 0.0821 \, L \, atm \, K^{-1} \, mol^{-1})$ (rounded off to the nearest integer)