Question

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

Answer 1

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.
\[{\text{Equivalent weight of compound = }}\dfrac{{{\text{Molecular weight of the compound}}}}{{{\text{no}}{\text{. of electrons lost or gained}}}}\]……..(1)
\[\mathop {SnC{l_2}}\limits_{ + 2} + C{l_2} \to \mathop {SnC{l_4}}\limits_{ + 4} \]
We can see in the reaction that in \[SnC{l_2}\], Sn is in +2 oxidation state and in \[SnC{l_4}\], 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 \[SnC{l_2}\]= Atomic weight of Sn + 2(Atomic weight of Chlorine)
Molecular weight of \[SnC{l_2}\]= 119 + 2(35.5)
Molecular weight of \[SnC{l_2}\]= 190
Putting available values into equation (1), we get
Equivalent Weight of \[SnC{l_2}\]\[ = \dfrac{{190}}{2}\]
Equivalent Weight of \[SnC{l_2}\]= 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.