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Question: The equivalent weight of KCl is: A. \({\dfrac{74.5}{2}}\) B. \(74.5\) C. \({\dfrac{74.5}{3}}\)...

The equivalent weight of KCl is:
A. 74.52{\dfrac{74.5}{2}}
B. 74.574.5
C. 74.53{\dfrac{74.5}{3}}
D. 74.54{\dfrac{74.5}{4}}

Explanation

Solution

The first step in calculation of equivalent weight is estimation of n-factor of a compound. It is also known as the valence factor, which indicates the total electron exchange occurring in the compound. Then, calculate the molar mass of the compound. Using these two quantities, calculate the equivalent weight.

Complete step by step answer:
Let us start this question by understanding the concept of equivalent weight first.
Equivalent weight is the ratio of molar mass and the valence factor.
Molar mass of a compound is the sum of atomic weight of its individual atoms.
Valence factor is the number of electrons gained or lost by a compound to achieve stability. Valence factor is also known as n-factor.

KCl is called Potassium Chloride. It is a salt made by an exchange of 1 electron. Potassium loses one electron and chlorine gains one electron, to complete their octet, respectively.
Now, let us calculate the molar mass of KCl.
Molar mass of KCl = Molar mass of K + Molar mass of Cl
Molar mass of KCl = 39.09 u + 35.45 u = 74.54 u = 74.5 u (by approximation)
Valence factor of KCl = 1 (exchange of 1 electron between K and Cl)
Equivalent weight = Molar mass / Valence factor
Equivalent weight = 74.51{\dfrac{74.5}{1}}

Therefore, the answer is – option (B) – The equivalent weight of KCl is 74.5.

Additional Information:
Equivalent weight is unit less, because it is a relative quantity.

Note:
Equivalent weight of any compound can be related to another compound by using its atomic mass. For example – any compound ‘x’ will combine with or displace directly or indirectly 1.008 parts by weight of hydrogen, 8 parts by weight of oxygen, 35.5 parts by weight chlorine, etc.