1.

A current strength of 9.65 amperes is passed through excess fused AlCl_(3) for 5 hours . How many litres of chlorine will be liberated at STP ? (F = 96500 C)

Answer»

`2.016`
`1.008`
`11.2`
`20.16`

Solution :Quantity of electricity = `I xx t`
`= 96.5 xx 5 xx 60 xx 60`
`= 173,700 C `
Faraday of electricity = `(173,700)/(96500)= 1.8 F `
`2 Cl^(-) - 2e^(-) to Cl_(2)`
1 F of electricity liberates 35.5 G or 11.2 L of `Cl_2` at S.T.P.
`therefore` Volume of `Cl_2` LIBERATED by 1.8 F electricity .
`= 1.8 xx 11.2 = 20.16 L`


Discussion

No Comment Found

Related InterviewSolutions