The amount of Aluminium deposited when 0.1 Faraday current is passed through aluminium chloride will be (R=27)
The amount of Aluminium deposited when 0.1 Faraday current is passed through aluminium chloride will be (R=27)
A. 0.9g
B. 0.3g
C. 0.27g
D. 2.7g
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Correct Answer - A
1 Faraday will deposit `Al=(27)/(3)g`
0.1 Faraday will deposit `Al=(27)/(3)xx0.1=0.9g`
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