Question

The molar mass of ${\text{CuS}}{{\text{O}}_4} \cdot {\text{5}}{{\text{H}}_2}{\text{O}}$ is ${\text{249}}$. Its equivalent mass in the reaction (a) and (b) would be:
${\text{CuS}}{{\text{O}}_4} + {\text{KI}} \to {\text{Product}}$
Electrolysis of ${\text{CuS}}{{\text{O}}_4}$ solution.
A.$\left( {\text{a}} \right){\text{249 (b)249}}$
B.$\left( {\text{a}} \right){\text{124}}{\text{.5 (b)124}}{\text{.5}}$
C.$\left( {\text{a}} \right){\text{249 (b)124}}{\text{.5}}$
D.$\left( {\text{a}} \right){\text{124}}{\text{.5 (b)249}}$

Answer 1

Hint:The sum of the masses of the atoms of all the elements in the chemical formula of any molecule is known as the molar mass. The mass of any substance that is chemically equivalent to one gram of hydrogen is known as equivalent mass.

Complete step by step answer:
a)Write the balanced chemical equation for the reaction of ${\text{CuS}}{{\text{O}}_4}$ with ${\text{KI}}$ as follows:
${\text{CuS}}{{\text{O}}_4} \cdot {\text{5}}{{\text{H}}_2}{\text{O}} + {\text{KI}} \to {\text{C}}{{\text{u}}_2}{{\text{I}}_2} + {{\text{I}}_2} + {{\text{K}}_2}{\text{S}}{{\text{O}}_4} + {{\text{H}}_2}{\text{O}}$
Determine the oxidation state of ${\text{Cu}}$ in ${\text{CuS}}{{\text{O}}_4}$ as follows:
$\left( {{\text{1}} \times {\text{O}}{\text{.S}}{\text{. of Cu}}} \right) + \left( {1 \times {\text{O}}{\text{.S}}{\text{. of S}}{{\text{O}}_4}} \right) = {\text{Charge on CuS}}{{\text{O}}_4}$
${\text{O}}{\text{.S}}{\text{. of Cu}} + \left( {1 \times - 2} \right) = 0$
${\text{O}}{\text{.S}}{\text{. of Cu}} = + 2$
Thus, the oxidation state of ${\text{Cu}}$ in ${\text{CuS}}{{\text{O}}_4}$ is $ + {\text{2}}$.
Determine the oxidation state of ${\text{Cu}}$ in ${\text{C}}{{\text{u}}_2}{{\text{I}}_2}$ as follows:
$\left( {{\text{2}} \times {\text{O}}{\text{.S}}{\text{. of Cu}}} \right) + \left( {2 \times {\text{O}}{\text{.S}}{\text{. of I}}} \right) = {\text{Charge on C}}{{\text{u}}_2}{{\text{I}}_2}$
${\text{2}} \times {\text{O}}{\text{.S}}{\text{. of Cu}} + \left( {2 \times - 1} \right) = 0$
${\text{O}}{\text{.S}}{\text{. of Cu}} = + 1$
Thus, the oxidation state of ${\text{Cu}}$ in ${\text{C}}{{\text{u}}_2}{{\text{I}}_2}$ is $ + {\text{1}}$.
Determine the n-factor for ${\text{CuS}}{{\text{O}}_4}$ as follows:
The n-factor is the product of the two oxidation states. Thus,
$n - {\text{factor}} = 2 \times 1$
$n - {\text{factor}} = 2$
Thus, the n-factor for ${\text{CuS}}{{\text{O}}_4}$ is 2.
Determine the equivalent mass of ${\text{CuS}}{{\text{O}}_4}$ using the equation as follows:
${\text{Equivalent mass}} = \dfrac{{{\text{Molar mass}}}}{{n - {\text{factor}}}}$
Substitute ${\text{249}}$ for the molar mass of ${\text{CuS}}{{\text{O}}_4}$, 2 for the n-factor. Thus,
${\text{Equivalent mass}} = \dfrac{{{\text{249}}}}{2}$
${\text{Equivalent mass}} = 124.5$
Thus, the equivalent mass of ${\text{CuS}}{{\text{O}}_4}$ in the reaction ${\text{CuS}}{{\text{O}}_4} + {\text{KI}} \to {\text{Product}}$ is ${\text{124}}{\text{.5}}$.

b)During the electrolysis of ${\text{CuS}}{{\text{O}}_4}$ solution, the reaction at cathode is as follows:
${\text{C}}{{\text{u}}^{2 + }} + 2{e^ - } \to {\text{Cu}}$
The change in oxidation state is 2.
Thus, the n-factor for ${\text{CuS}}{{\text{O}}_4}$ is 2.
Determine the equivalent mass of ${\text{CuS}}{{\text{O}}_4}$ using the equation as follows:
${\text{Equivalent mass}} = \dfrac{{{\text{Molar mass}}}}{{n - {\text{factor}}}}$
Substitute ${\text{249}}$ for the molar mass of ${\text{CuS}}{{\text{O}}_4}$, 2 for the n-factor. Thus,
${\text{Equivalent mass}} = \dfrac{{{\text{249}}}}{2}$
${\text{Equivalent mass}} = 124.5$
Thus, the equivalent mass of ${\text{CuS}}{{\text{O}}_4}$ in the electrolysis of ${\text{CuS}}{{\text{O}}_4}$ solution is ${\text{124}}{\text{.5}}$.
Thus, the correct option is option (B).

Note:
Determine the n-factor carefully. The equivalent mass is the ratio of molar mass to the n-factor. For acids, n-factor is the number of replaceable hydrogen ions and for bases, n-factor is the number of replaceable hydroxide ions.