COMEDK · Chemistry · 18. Electrochemistry
The quantity of Ca that can be produced from molten CaCl\(_2\), with the same quantity of electricity (in coulombs) required to produce 4.8 g of Mg from molten MgCl\(_2\) is:
[Atomic mass of Mg=24 u; Atomic mass of Ca=40 u]
- A \(4.8\) g
- B \(8.0\) g
- C \(5.2\) g
- D \(6.0\) g
Answer & Solution
Correct Answer
(B) \(8.0\) g
Step-by-step Solution
Detailed explanation
According to Faraday's second law of electrolysis, when the same quantity of electricity is passed through different electrolytes, the masses of the substances deposited are proportional to their equivalent weights. \(\dfrac{W_{Ca}}{W_{Mg}} = \dfrac{E_{Ca}}{E_{Mg}}\) Equivalent…
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