19 g molten `SnCl_(2)` was electrolysed by two inert electrodes. On electrolysis, 0.119 g of Sn was deposited at cathode. Find the ratio of weights of `SnCl_(2)` and `SnCl_(4)` after electrolysis, assuming, there is no loss of materíal during electrolysis.

Explanation

`Sn^(2+)+2e rarrSn,` electricity required to produce
`"0.11 g Sn "=(2xx96500)/(119)xx0.199=193C`
`2SnCl_(2)rarrSn+SnCl_(4)or, 2Sn^(2+)rarr Sn+Sn^(4+)`
`Sn^(2+)rarrSn^(4+)+2e" Hence, "2xx96500C "-="1 mol "Sn^(4+)`
`therefore" "193C-=10^(-3)" mol "Sn^(4+)`
`therefore` Amount of `SnCl_(4)` produced
`=10^(-3)xx(119+142)=0.261g`
`"0.119 g Sn "=10^(-3)" mol Sn "-=10^(-3)" mol "SnCl_(2) and 10^(-3)" mol "SnCl_(4)-=10^(-3)" mol "SnCl_(2)`
`therefore` To produce `10^(-3)` mol Sn and `10^(-3)` mol `SnCl_(2)2xx10^(-3)` mol `SnCl_(2)` is required.
`therefore" "SnCl_(2)` left after electrolysis `=(19-2xx10^(-3)xx190)=18.62g, SnCl_(4)` produce
`=10^(-3)xx(119+4xx35.5)=0.261g.`
`therefore" Ratio of the amounts of "SnCl_(2)" to "SnCl_(4)=(18.62)/(0.261)=71.34`