Answer
Verified
398.1k+ 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.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Summary of the poem Where the Mind is Without Fear class 8 english CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Write an application to the principal requesting five class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE