Answer
425.4k+ views
Hint:
The basicity/ acidity/ n-factor used to calculate the equivalent mass can be determined by the difference in the oxidation state. The Mohr's salt has a molecular formula of $\text{FeS}{{\text{O}}_{4}}{{\left( \text{N}{{\text{H}}_{4}} \right)}_{2}}\text{S}{{\text{O}}_{4}}\centerdot 6{{\text{H}}_{2}}\text{O}$
Complete step by step answer:
-We know that the formula of the Mohr's salt is $\text{FeS}{{\text{O}}_{4}}{{\left( \text{N}{{\text{H}}_{4}} \right)}_{2}}\text{S}{{\text{O}}_{4}}\centerdot 6{{\text{H}}_{2}}\text{O}$
-The iron that is combined with the sulphate group gets oxidised by the potassium permanganate.
-The balanced chemical reaction is:
$\text{5F}{{\text{e}}^{2+}}\text{ + MnO}_{4}^{-}\text{ + 8}{{\text{H}}^{+}}\to \text{ 5F}{{\text{e}}^{3+}}\text{ + M}{{\text{n}}^{2+}}\text{ + 4}{{\text{H}}_{2}}\text{O}$
-As we can see that the oxidation number of iron changes from +2 to +3 which clearly shows that the iron is oxidised.
-So, the n-factor will be the difference of the oxidation states i.e.
+3 - (+2) = 1
-Now, we will calculate the molar mass of Mohr's salt:
$\begin{align}
& \text{FeS}{{\text{O}}_{4}}{{\left( \text{N}{{\text{H}}_{4}} \right)}_{2}}\text{S}{{\text{O}}_{4}}\text{ }\centerdot \text{ 6}{{\text{H}}_{2}}\text{O} \\
& 56+32+\left( 16\text{ }\cdot \text{ 4} \right)+2\left( 14+\left( 1\text{ }\cdot \text{ 4} \right) \right)\ \text{+ 32 + 16 }\cdot \text{ 4 + 6 }\cdot \text{ 18} \\
& \text{88 + 64 + 36 + 32 + 64 + 108} \\
& \text{392} \\
\end{align}$
-Now, by applying the formula of equivalent mass we will get:
Equivalent mass = $\dfrac{392}{1}\text{ = 392}$
Therefore, the equivalent mass of the Mohr's salt is 392.
Hence option A is correct.
Note:
The equivalent mass is defined as the elements whose mass is fixed, which are combined to form a compound in terms of hydrogen, chlorine and oxygen. It means that the fixed mass of an element can displace 1 P/m of hydrogen, 35.5 P/m of chlorine and 8 P/m oxygen. In the titration, Mohr's salt acts as a reducing agent and it is also used in analytical chemistry.
The basicity/ acidity/ n-factor used to calculate the equivalent mass can be determined by the difference in the oxidation state. The Mohr's salt has a molecular formula of $\text{FeS}{{\text{O}}_{4}}{{\left( \text{N}{{\text{H}}_{4}} \right)}_{2}}\text{S}{{\text{O}}_{4}}\centerdot 6{{\text{H}}_{2}}\text{O}$
Complete step by step answer:
-We know that the formula of the Mohr's salt is $\text{FeS}{{\text{O}}_{4}}{{\left( \text{N}{{\text{H}}_{4}} \right)}_{2}}\text{S}{{\text{O}}_{4}}\centerdot 6{{\text{H}}_{2}}\text{O}$
-The iron that is combined with the sulphate group gets oxidised by the potassium permanganate.
-The balanced chemical reaction is:
$\text{5F}{{\text{e}}^{2+}}\text{ + MnO}_{4}^{-}\text{ + 8}{{\text{H}}^{+}}\to \text{ 5F}{{\text{e}}^{3+}}\text{ + M}{{\text{n}}^{2+}}\text{ + 4}{{\text{H}}_{2}}\text{O}$
-As we can see that the oxidation number of iron changes from +2 to +3 which clearly shows that the iron is oxidised.
-So, the n-factor will be the difference of the oxidation states i.e.
+3 - (+2) = 1
-Now, we will calculate the molar mass of Mohr's salt:
$\begin{align}
& \text{FeS}{{\text{O}}_{4}}{{\left( \text{N}{{\text{H}}_{4}} \right)}_{2}}\text{S}{{\text{O}}_{4}}\text{ }\centerdot \text{ 6}{{\text{H}}_{2}}\text{O} \\
& 56+32+\left( 16\text{ }\cdot \text{ 4} \right)+2\left( 14+\left( 1\text{ }\cdot \text{ 4} \right) \right)\ \text{+ 32 + 16 }\cdot \text{ 4 + 6 }\cdot \text{ 18} \\
& \text{88 + 64 + 36 + 32 + 64 + 108} \\
& \text{392} \\
\end{align}$
-Now, by applying the formula of equivalent mass we will get:
Equivalent mass = $\dfrac{392}{1}\text{ = 392}$
Therefore, the equivalent mass of the Mohr's salt is 392.
Hence option A is correct.
Note:
The equivalent mass is defined as the elements whose mass is fixed, which are combined to form a compound in terms of hydrogen, chlorine and oxygen. It means that the fixed mass of an element can displace 1 P/m of hydrogen, 35.5 P/m of chlorine and 8 P/m oxygen. In the titration, Mohr's salt acts as a reducing agent and it is also used in analytical chemistry.
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