Question

One gm metal M3+ was discharging by the passage of 1.81 *1023 electrons.What is the atomic weight of metal?

Answer 1

For discharging of metal the reaction is :M3+ + 3e- => M

Thus 1 mol of the metal is deposited by 3F = 3×96500 C = 289500

CCharge on 1.81×1023 electrons = 1.602×10-19 ×1.81​×1023 C = 2.9×104 C289500 C deposit = 1 mol of metal

Thus 2.9×104 C will deposit = 1289500×2.9×104moles metal = 0.100 moles0.100 moles = 1 g metal

Thus atomic mass of metal = 10 g/mol

Answer 2

Answer:10 g/mole

Explanation:For discharging of metal the reaction is :M3+ + 3e- => M

Thus 1 mol of the metal is deposited by 3F = 3×96500 C = 289500

CCharge on 1.81×1023 electrons = 1.602×10-19 ×1.81​×1023 C = 2.9×104 C289500 C deposit = 1 mol of metal

Thus 2.9×104 C will deposit = 1289500×2.9×104moles metal = 0.100 moles0.100 moles = 1 g metal

Thus atomic mass of metal = 10 g