Silver chloride, often used in silver plating, contains 75.27 % Ag. Calculate the mass of silver...
Question:
Silver chloride, often used in silver plating, contains {eq}75.27 \ \% \ Ag. {/eq} Calculate the mass of silver chloride required to plate {eq}195 \ mg {/eq} of pure silver.
Mass:
In physics, the 'mass' term is used to denote the resistance applied by an object to its acceleration when the object is moving due to an applied force. It is also used to show the total amount of matters available in an object. By using {eq}\;\rm{kg} {/eq} or {eq}\;\rm{g} {/eq} units, the mass of an object can be denoted in the international system of units (SI).
Answer and Explanation: 1
Given information
- The mass percentage of the silver is {eq}M\% = 75.27\% {/eq}
- The mass of the pure silver to be plated is {eq}{M_{{\rm{silver}}}} = 195\;{\rm{mg}} {/eq}
Write the expression of the mass percentage of the silver.
{eq}M\% = \left( {\dfrac{{{M_{{\rm{silver}}}}}}{{{M_{{\rm{silver}}\;{\rm{chloride}}}}}}} \right) \times 100\% {/eq}
Here, {eq}{M_{{\rm{silver}}\;{\rm{chloride}}}} {/eq} is the mass of required silver chloride.
Substitute all given values in the above equation and calculate the mass of the required silver chloride.
{eq}\begin{align*} \left({75.27\% } \right) &= \left( {\dfrac{{195\;{\rm{mg}}}}{{{M_{{\rm{silver}}\;{\rm{chloride}}}}}}} \right) \times 100\% \\ {M_{{\rm{silver}}\;{\rm{chloride}}}} &= \left( {\dfrac{{195\;{\rm{mg}}}}{{75.27}}} \right) \times 100\\ {M_{{\rm{silver}}\;{\rm{chloride}}}} &= 259.06\;{\rm{mg}} \end{align*} {/eq}
Thus, the mass of the required silver chloride is {eq}{259.06\;{\rm{mg}}} {/eq}.
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