In this question, we have to find the maximum amount of grams that can be plated, and the informations we have are:
AgNO3
4.8 L of AgNO3 solution
3.4% of Ag by mass
density = 1.01g/mL
Using density and the volume of the solution, we can find the total mass of the solution, using the density formula:
d = m/v
d is going to be 1.01g/mL
m = ?
v = it will be 4800 mL, since we need mL and we have 4.8 L
Adding these values to the formula
1.01 = m/4800
m = 4848 grams
Now, since we have only 3.4% of Ag, we use the value in grams of the total solution to find the final mass of Ag
4848 g = 100%
x grams = 3.4%
x = 164.832 grams of Ag is the maximum amount.