Question

The equivalent weight of ${\text{CuS}}{{\text{O}}_4}$ when it is converted to ${\text{C}}{{\text{u}}_2}{{\text{I}}_2}$ {M=mol. Wt}
A. $\dfrac{{\text{M}}}{1}$
B.$\dfrac{{\text{M}}}{2}$
C.$\dfrac{{\text{M}}}{3}$
D.$2{\text{M}}$

Answer 1

Hint: Equivalent weight of a substance is different for every element or ion. Equivalent weight is the weight of the atom combined or displaced from a compound. Use formula for equivalent weight,
Equivalent Weight=$\dfrac{{{\text{Molecular weight}}}}{{{\text{No}}{\text{. of }}{{\text{e}}^ - }{\text{ gained/lost}}}}$

Complete step by step answer:
When copper sulphate reacts with potassium iodide it produces cupric iodide which is unstable. Hence cupric iodide converts readily into cuprous iodide and iodine.
The reaction is given as-
$2CuS{O_4} + 4KI \to 2Cu{I_2} + 2{K_2}S{O_4}$ -- (i)
But since cupric iodide is not stable it converts to cuprous iodide and the reaction is-
$2Cu{I_2} \to C{u_2}{I_2} + {I_2}$ --- (ii)
We have to find the equivalent weight (E)of ${\text{CuS}}{{\text{O}}_4}$ when it is converted to${\text{C}}{{\text{u}}_2}{{\text{I}}_2}$
So we can write-
\[2\mathop {Cu}\limits^{ + 2} \mathop {S{O_4}}\limits^{ - 2} \to \mathop {C{u_2}}\limits^{ + 1} \mathop {{I_2}}\limits^{ - 1} \]
From the above reaction we can find the oxidation state of copper in both compounds which will give us the number of electrons lost or gained.
In ${\text{CuS}}{{\text{O}}_4}$ , copper has oxidation state $ + 2$ and sulphur has oxidation state of $ - 2$ .Now this copper accepts an electron during reaction (i) and (ii), and thus in ${\text{C}}{{\text{u}}_2}{{\text{I}}_2}$, copper has oxidation state of $ + 1$ . This means that the number of electrons gained=$1$
And the molecular mass of ${\text{CuS}}{{\text{O}}_4}$ is given as M.
We know that the formula of equivalent weight is given as-
Equivalent Weight=$\dfrac{{{\text{Molecular weight}}}}{{{\text{No}}{\text{. of }}{{\text{e}}^ - }{\text{ gained/lost}}}}$
On putting the values in the formula we get,
$ \Rightarrow E = \dfrac{M}{1}$

Hence option A is correct.

Note:
Equivalent weight is also given by the following formulas-
For acid or base-
E=$\dfrac{M}{{Basicity/Acidity}}$
Where M is Molar mass, Basicity is the number of replaceable ${H^ + }$ ions present, Acidity is the number of replaceable \[{\text{O}}{{\text{H}}^ - }\] ions present.