On passing `0.1 F` of electricity through aluminimum chloride, the amount of aluminimu metal deposited on cathode is `(A1 =27)` .

Options

A. 0.9gm
B. 0.3 gm
C. 0.27 gm
D. 2.7 gm

Answer

A

Explanation

At cathode: `Al^(3+)+3e^(-)toAl`
`E_(Al)=(27)/(3)=9`
`W_(Al)=E_(Al)xx"No. of faradays"=9xx0.1=0.9gm`