Calculate the mass of silver chloride required to plate 265 mg of pure silver.


Answer 1

Given Mass of pure silver (Ag) = 265 mg

Silver chloride AgCl which is used in plating silver contains 75.27 % Ag

This means that:

A 100 mg of silver chloride contains 75.27 mg of silver

Therefore, the amount of silver chloride required to plate a sample containing 265 mg silver would correspond to:

265 mg Ag * 100 mg AgCl/75.27 mg Ag

= 352.1 mg AgCl

Answer 2


0.35215 grams of silver chloride required to plate 265 mg of pure silver.


Mass of silver = 265 mg = 0.265 g

Moles of silver =

According to reaction, 2 moles of silver are obtained from 2 moles of silver chloride.

Then 0.002454 moles of silver will be obtained from :

of silver chloride

Mass of 0.002454 moles of silver chloride:

= 0.002454 mol × 143.5 g/mol = 0.35215 g

0.35215 grams of silver chloride required to plate 265 mg of pure silver.

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