The same quantity of electricity that liberated 2.158 g silver was passed through a solution of a gold salt and 1.314 g of gold was deposited. The equivalent mass of silver is 107.9. Calculate the equivalent mass of gold. What is the oxidation state of gold salt ? (At. mass of gold =197)

The same quantity of electrical charge that deposited 0.583g of silver was passed through solution of gold salt and 0.355g of gold was formed. What is the oxidation state of gold in this salt?

The same quantity of electrical charge that deposited 0.583 g of silver was passed through a solution of gold salt and 0.355 g of gold was formed . What is the oxidation state of gold in this salt ?

Same quantity of electrical charge that deposited 0.583 g of silver was passed through a solution of gold salt. If 0.335 g of gold is deposited, calculate the oxidation state of gold in the given salt. At wt of Au = 197.

The same quantity of electricity that liberates 4.316 g of silver from AgNO_(3) solution was passed through a solution of gold salt. If the atomic weight of gold be 197 and its valency in the above-mentioned salt be 3, calculate the weight of gold deposited at the cathode and the quantity of electricity passed.

How many moles and atoms of gold are present in 49-25 g of gold? (Atomic mass of gold=97u)

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)