\( 2.5 \mathrm{~g} \) sample of copper is dissolved in excess of \( \mathrm{H}_{2} \mathrm{SO}_{4} \) to prepare \( 100 \mathrm{~mL} \) of \( 0.02 \mathrm{M} \mathrm{CuSO}_{4}(\mathrm{aq}) .10 \mathrm{~mL} \) of \( 0.02 \mathrm{M} \) solution of \( \mathrm{CuSO}_{4} \) (aq) is mixed with excess of KI to show the following changes.
\[
\begin{aligned}
\mathrm{CuSO}_{4}+2 \mathrm{KI} & \longrightarrow \mathrm{K}_{2} \mathrm{SO}_{4}+\mathrm{CuI}_{2} \\
2 \mathrm{CuI}_{2} & \longrightarrow \mathrm{Cu}_{2} \mathrm{I}_{2}+\mathrm{I}_{2}
\end{aligned}
\]
The liberated iodine is titrated with hypo \( \left(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\right) \) and requires \( V \mathrm{~mL} \) of \( 0.1 M \) hypo solution for its complete reduction. The amount of \( \mathrm{I}_{2} \) liberated in the reaction of \( 10 \mathrm{~mL} \) of \( 0.02 \mathrm{M} \) sol. I i with \( \mathrm{KI}_{\text {excess }} \) is :
(A) \( 0.051 \mathrm{~g} \)
(B) \( 0.0254 \mathrm{~g} \)
(C) \( 0.102 \mathrm{~g} \)
(D) \( 0.204 \mathrm{~g} \)
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