On passing a current through a molten aluminium chloride for some time, produced `11.2` lit of `Cl_(2)` at NTP at anode, the quantity of aluminium deposited at cathode is

Options

A. 9g
B. 18g
C. 27g
D. 36g

Answer

A

Explanation

`2Cl^(-)(aq) rarr Cl_(2)(g)+ 2e^(-)`
Mol. wt. of `Cl_(2)= 71g`
Eq. wt. of `Cl_(2)= 35.5g`
`because` 71g of `Cl_(2)` would give 22.4L of `Cl_(2)` at NTP
`therefore` 35g of `Cl_(2)` would give `(22.4 xx 35.5)/(71)= 11.2L`
Thus charge required to 11.2L of `Cl_(2)` at NTP will be 1 Faraday. Therefore, 1F will also deposit 1 equivalent aluminimum `(27)/(3)= 9g`