20 , mL of 0.02 , M hypo solution is used for | vedclass

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(B) From the stoichiometry of the reactions:$2 Cu^{2+} \equiv I_{2} \equiv 2 S_{2}O_{3}^{2-}$.Therefore,$n_{	ext{eq.}}$ of $Cu^{2+} = n_{	ext{eq.}}$

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) From the stoichiometry of the reactions:$2 Cu^{2+} \equiv I_{2} \equiv 2 S_{2}O_{3}^{2-}$.Therefore,$n_{	ext{eq.}}$ of $Cu^{2+} = n_{	ext{eq.}}$ of $S_{2}O_{3}^{2-}$.$n_{	ext{eq.}}$ of $S_{2}O_{3}^{2-} = M 	imes V 	imes n_{	ext{factor}} = 0.02 	imes 20 	imes 1 = 0.4 \, 	ext{mmol}$.Since $n_{	ext{eq.}}$ of $Cu^{2+} = 0.4 \, 	ext{mmol}$ and $n_{	ext{factor}}$ for $Cu^{2+}$ is $1$ (change in oxidation state from $+2$ to $+1$ in $Cu_{2}I_{2}$),we have $n_{	ext{mol}}$ of $Cu^{2+} = 0.4 \, 	ext{mmol}$.$[Cu^{2+}] = \frac{n_{	ext{mol}}}{V_{	ext{solution}}} = \frac{0.4 \, 	ext{mmol}}{10 \, mL} = 0.04 \, M = 4 	imes 10^{-2} \, M$.Thus,the value is $4$.

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