Correct Answer - `a.2.155g`,`b.1336.15s`
`CrO_(3)+6H^(o+)+6e^(-) rarr Cr+3H_(2)O(` Atomic weight `=52)`
`52g` of `Cr` is deposited by passing electric current of `6xx96500C`.
`a.` In the given data `24000C` current is passed , so amount of deposited `Cr=(52xx24000)/(6xx96500)g=2.155g`
`b.` `m=Zxx2xxt(Z of Cr=(52)/(6xx96500))`
`1.5=(52)/(6xx96500)xx12.5xxt`
`:.t=(1.5xx6xx96500)/(52xx12.5)s`
`=1336.15s`