Question

Silver chloride, often used in silver plating, contains 75.27% ag. calculate the mass of silver chloride required to plate 255 mg of pure silver.

Answer 1

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 $\frac{1}{0.7527} mol$ of AgCl.

Mass of pure silver given is 255 mg, converting mass into number of moles as follows:

$n=\frac{m}{M}$

here, n is number of moles, m is mass and M is molar mass.

Putting the values,

$n=\frac{255 mg} {107.87 g/mol} (\frac{1 g}{1000 mg})=0.002364 mol$

Thus, number of moles of Ag are 0.002364 mol.

Now, 0.002364 mol of Ag will be obtained from $(0.002364 mol)(\frac{1}{0.7527} )=0.003140$ mol of AgCl.

Molar mass of AgCl is 143.32 g/mol, converting number of moles into mass as follows:

$m=nM=(0.003140 mol)(143.32 g/mol)=0.450 g$

Converting mass into mg,

$m=0.450 g(\frac{1000 mg}{1 g} )=450 mg$

Therefore, mass of silver chloride required to plate 255 mg of pure silver will be 450 mg.