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Q.

Total number of electrons present is 4.4 gm oxalate ion $(C_{2} {O_{4}}^{2 -})$ is:

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a

$\begin{aligned} 0 .05 N_{A} \end{aligned}$

b

$2 .1 N_{A}$

c

$\begin{aligned} 2 .3 N_{A} \end{aligned}$

d

$2 .2 N_{A}$

answer is B.

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Detailed Solution

Total number of electron,

$= \frac{4.4}{88} \times [6 \times 2 + 8 \times 4 + 2] \times N_{A}$
$= \begin{aligned} 2 .3 N_{A} \end{aligned}$ .

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