If `6.65xx10^(-2)g` of metalllic zinc is added to `100ml` saturated solution of `AgCl`, it react with `Ag^(+)` of solution as following reaction.
`Zn(s)+2Ag^(+)(aq)hArrZn^(2+)(aq)+2Ag(s)`
and approximately `10^(-x)` moles of `Ag` will be precipated. Calculate the value of `x` (Given `E_(Zn^(++)//Zn)^(@)=-0.76VE_(Ag^(+)//Ag)^(@)=0.8V, K_(sp)` of `AgCl=10^(-10)`, atomic mass of `Zn=65.3u, 10^(52.8813)=7.61xx10^(52)`)

Answer

6

Explanation

`Zn(s)+2Ag^(+)(aq)hArrZn^(2+)(aq)+2Ag(s)`
`E_("cell")^(@)=0.059/2log_(10)K_(eq)`
`0.8+0.76=0.059/2 log_(10)K_(eq)`
`K_(eq)=7.61xx10^(52)`
As `K_(eq)` is very large the reactioin will tend towards completion
Moles of `Ag^(+)` in solution `=(10^(-5)xx100)/1000=10^(-6)`
As `Ag^(+)` is limiting reagent, the moles of `Ag` precipitate `~~10^(-6)`
Hence `x=6`