Chemistry · Electrolytic and metallic conduction, conductance in electrolytic solutions, molar conductivities and their variation with concentration, Kohlrausch's law and its applications
In the refining of silver by electrolytic method what will be the weight of 100
In the refining of silver by electrolytic method what will be the weight of 100 g Ag anode if 5 ampere current is passed for 2 hours? Purity of silver is \( 95 \% \) by weight. in gram is:
- A. 57
- B. 32
- C. 60
- D. 13
Step-by-step solution
Total charge passed: Q = 5 A × 2 h × 3600 s/h = 36000 C. Mass of silver dissolved from anode: m = (M × Q)/(n × F) = (108 g/mol × 36000 C)/(1 × 96485 C/mol) ≈ 40.3 g. Initial silver in anode: 100 g × 0.95 = 95 g. Remaining silver: 95 - 40.3 = 54.7 g. Impurities (5 g) remain, so total weight = 54.7 + 5 = 59.7 g ≈ 60 g.
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