In silver chloride used for silver plating, 75.27% of Ag is present. Therefore, in 1 mole of AgCl, there are 0.7527 mol of Ag present.
Or, 1 mol of Ag is obtained from [tex]\frac{1}{0.7527} mol[/tex] of AgCl.
Mass of pure silver given is 255 mg, converting mass into number of moles as follows:
[tex]n=\frac{m}{M}[/tex]
here, n is number of moles, m is mass and M is molar mass.
Putting the values,
[tex]n=\frac{255 mg} {107.87 g/mol} (\frac{1 g}{1000 mg})=0.002364 mol[/tex]
Thus, number of moles of Ag are 0.002364 mol.
Now, 0.002364 mol of Ag will be obtained from [tex](0.002364 mol)(\frac{1}{0.7527} )=0.003140[/tex] mol of AgCl.
Molar mass of AgCl is 143.32 g/mol, converting number of moles into mass as follows:
[tex]m=nM=(0.003140 mol)(143.32 g/mol)=0.450 g[/tex]
Converting mass into mg,
[tex]m=0.450 g(\frac{1000 mg}{1 g} )=450 mg[/tex]
Therefore, mass of silver chloride required to plate 255 mg of pure silver will be 450 mg.
