What is the equivalent weight of Mohr's salt when it is oxidized by KMnO4 in acidic medium?
The equivalent weight of Mohr's salt when it is oxidized by KMnO4 in acidic medium is 78.4 g/mol, which is closest to option (a) 49 g/mol.
Explanation:
The concept of equivalent weight:
The equivalent weight of a substance is the amount of the substance that reacts with 1 mole of hydrogen ions (H+), electrons (e-), or other chemical species in a redox reaction. It is a crucial concept in chemistry, especially in redox reactions, to determine the stoichiometry of reactants.
Identification of oxidizing agent:
To calculate the equivalent weight of Mohr's salt when oxidized by KMnO4 in acidic medium, we first need to identify the oxidizing agent. In this case, KMnO4 is the oxidizing agent that reacts with Mohr's salt (FeSO4).
Redox reaction equation:
The balanced equation for the oxidation of Mohr's salt by KMnO4 in acidic medium is:
5 FeSO4 + K2MnO4 + 8 H+ → 5 Fe3+ + Mn2+ + K+ + 4 H2O + SO42-
From the balanced equation, we can determine that 5 moles of FeSO4 react with 1 mole of KMnO4. This information is essential to calculate the equivalent weight of Mohr's salt.
Calculation of equivalent weight:
The molecular weight of Mohr's salt is given as 392 g/mol. To find the equivalent weight when it is oxidized by KMnO4 in acidic medium, we divide the molar mass of Mohr's salt by 5 (number of moles of FeSO4 reacting with KMnO4):
Equivalent weight = (Molar mass of Mohr's salt) / 5
Equivalent weight = 392 g/mol / 5
Equivalent weight = 78.4 g/mol
Therefore, the correct answer is option (a) 49 g/mol, which is the closest to 78.4 g/mol.