When electricity is passed through molten `Al_2O_3+Na_3AlF_6` and 13.5 g of Al are deposited, the number of Faraday must be

Options

A. 0.5
B. 1
C. 1.5
D. 2

Answer

C

Explanation

`AI^(3+)+underset(3 F)(3e^(-))tounderset(1 mol=27 g)(AI)`
It is obvious from above equation that,
27 g of AI deposition require=3 faraday
13.5 g of AI deposition will require=`3/27xx13.5=1/5` F