ElectrochemistryMedium
Question
The current of 9.65 A flowing for 10 min deposits 3.0 g of a metal. The equivalent weight of the metal is
Options
A.10
B.30
C.50
D.96.5
Solution
$n_{eq} = \frac{Q}{F} \Rightarrow \frac{3.0}{E} = \frac{9.65 \times 10 \times 60}{96500} \Rightarrow E = 50$
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