The mass of silver deposited when a current of 10a is passed through a solution of silver salt for 4830s is [Ag = 108, F = 96500 c mol-1]

Answer Details

The question is asking about the amount of silver deposited when an electric current is passed through a solution of a silver salt for a certain amount of time. The formula for calculating the amount of substance produced by electrolysis is given as: Amount of substance (in mol) = Current (in A) × Time (in s) / Faraday's constant (in C mol^-1) In this case, we are given the current (10 A), time (4830 s), and Faraday's constant (96500 C mol^-1), and we know that the substance being produced is silver (Ag), with a molar mass of 108 g mol^-1. Using these values and the formula, we can calculate the amount of silver produced in mol and then convert it to grams using its molar mass. Amount of substance (in mol) = 10 A × 4830 s / 96500 C mol^-1 = 0.5 mol Mass of silver (in g) = 0.5 mol × 108 g mol^-1 = 54.0 g Therefore, the mass of silver deposited when a current of 10 A is passed through a solution of silver salt for 4830 s is 54.0 g. The answer is.