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# The ratio between Gram molecular weight and gram equivalent weight for $Fe{C_2}{O_4}$, when it is oxidized to $F{e^{ + 3}}$ and $C{O_2}$with $MnO_4^ -$ in acidic medium.A.1B.3C.4D.2

Last updated date: 18th Jun 2024
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Hint: To answer this question, you must recall the concept of equivalent mass and n- factor of a substance. We know that in an acidic medium, $MnO_4^ -$is a very strong oxidizing agent.
Formula used:
$E = \dfrac{M}{n}$
Where, $E$ is the gram equivalent weight of $Fe{C_2}{O_4}$
$M$ is the gram molecular weight of $Fe{C_2}{O_4}$
And $n$ is the n- factor of $Fe{C_2}{O_4}$ in the reaction occurring.

Gram molecular weight is nothing but the molecular mass of a substance expressed in the unit of grams. Gram molecular weight is given by the sum of the atomic weights of all the atoms present in a molecule of a certain compound expressed in grams.
Gram equivalent weight of a substance is the mass of the given substance, which releases or reacts with one mole of hydrogen ions during an acid– base neutralization reaction, or releases or accepts one mole of electrons during a redox reaction, expressed in grams.
In the question, $Fe{C_2}{O_4}$ is oxidized by permanganate ions.
The redox reaction occurring is given by:
$F{e^{2 + }} + {C_2}O_4^{2 - } \to 2C{O_2} + F{e^{3 + }} + 3{e^ - }$
The number of electrons transferred by one mole of ferrous oxalate is 3.
So the ratio between the Gram molecular weight and gram equivalent weight is given as:
$\dfrac{M}{E} = \dfrac{M}{{M/3}} = 3$

Thus, the correct answer is option B.

Note:
The n factor of a compound is also known as the valence factor and its value depends upon the compound that is being considered. For instance, for an acid, the basicity of the acid is its n factor and for a base, the n factor is its acidity. In a redox reaction, the n- factor is equal to the number of electrons accepted or released by the compound.