Two voltameters, one containing a solution of silver salt and the other a solution of a trivalent-metal salt, are connected in series. A current of 2 A is maintained for 1.50 hours. It is observed that 1.00 g of the trivalent metal is deposited. Determine the atomic weight of the trivalent metal.

Step-by-step solution

According to Faraday's first law of electrolysis, the mass of the substance deposited is given by: m = M nF I t where M is the atomic weight, n is the valency, F is Faraday's constant ( 96500 C mol ⁻¹ ), I is the current, and t is the time in seconds. Given: m = 1.00 g n = 3 (for a trivalent metal) I = 2 A t = 1.50 hours = 1.50 3600 s = 5400 s Substituting the values into the formula: 1.00 = M 2 5400 3 96500 M = 1.00 3 96500 2 5400 M = 289500 10800 M = 26.8 g mol ⁻¹