Question

The same quantity of electrical charge deposited `0.583g` of Ag when passed through `AgNO_(3),AuCl_(3)` solution calculate the weight of gold formed. (At weighht of `Au=197gmol)`

Answer 1

Given: `W_(ag)=0.583`gm
Asked: `W_(Au)=?`
Formula used: `(W_(Ag))/(Eq. Wt. of Ag)=(W_(Au))/(Eq. Wt. of au)`
Explanation: `W_(ag)="weight of Ag deposited", W_(Au)="weight of Au deposited"`
`E_(Ag)="equivalent weight of Ag", E_(Au)="Equivalent weight of Au"`
Substitution and calculation
`(W_(Ag))/(Eq. Wt. of Ag)=(W_(Au))/(Eq. Wt. of Au)`
`Eq. wt. of Ag=108`, `Eq. wt. of Au=(197)/(3)`
`(0.583)/(109)=(W_(Au))/(197//3)),W_(Au)=(0.583xx197)/(108xx3)=0.354` g