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Question: What is the equivalent weight of \[SnC{l_2}\] in the following reaction? \[SnC{l_2} + C{l_2} \to S...

What is the equivalent weight of SnCl2SnC{l_2} in the following reaction?
SnCl2+Cl2SnCl4SnC{l_2} + C{l_2} \to SnC{l_4}
(A) 95
(B) 45
(C) 60
(D) 30

Explanation

Solution

Hint : This is a redox type of reaction. Equivalent weight of a substance is the amount of mass of the substance that can react with or is equivalent to 1gm of Hydrogen or 8gm of Oxygen or 35.5gm of Chlorine.

Complete step by step solution :
As the reaction given in the question is a redox type of reaction, we can find out equivalent weight in this case by following the formula.
Equivalent weight of compound = Molecular weight of the compoundno. of electrons lost or gained{\text{Equivalent weight of compound = }}\dfrac{{{\text{Molecular weight of the compound}}}}{{{\text{no}}{\text{. of electrons lost or gained}}}}……..(1)
SnCl2+2+Cl2SnCl4+4\mathop {SnC{l_2}}\limits_{ + 2} + C{l_2} \to \mathop {SnC{l_4}}\limits_{ + 4}
We can see in the reaction that in SnCl2SnC{l_2}, Sn is in +2 oxidation state and in SnCl4SnC{l_4}, Sn is in +4 oxidation state. So, the number of electrons lost is 2 that means we will divide its molecular weight by 2 in order to obtain its equivalent weight.
Now Molecular weight of SnCl2SnC{l_2}= Atomic weight of Sn + 2(Atomic weight of Chlorine)
Molecular weight of SnCl2SnC{l_2}= 119 + 2(35.5)
Molecular weight of SnCl2SnC{l_2}= 190
Putting available values into equation (1), we get
Equivalent Weight of SnC{l_2}$$$$ = \dfrac{{190}}{2}
Equivalent Weight of SnCl2SnC{l_2}= 95 gm

So, option (A) 95 is correct.

Additional Information:
- Below are the different formulas to find out the Equivalent weight of compounds in different circumstances such as if the compound is an acid or a base or an atom.
{\text{Equivalent weight of acid = }}\dfrac{{{\text{Molecular weight of the acid}}}}{{{\text{Basicity of the acid}}}}$$$${\text{Equivalent weight of base = }}\dfrac{{{\text{Molecular weight of the base}}}}{{{\text{Acidity of the base}}}}$$$${\text{Equivalent weight of an atom = }}\dfrac{{{\text{Atomic weight of the atom}}}}{{{\text{Valency}}}}

Note : - Do not divide the molecular weight by the oxidation state of the metal in the compound as it requires the change in number of oxidation states.
- Sometimes we interpret equivalent weight as molecular weight and that leads to wrong answers.