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)`

The same quantity of electrical charge deposited `0.583g` of Ag when p

1.	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» 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

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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)`

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