Question

Equivalent mass of $N{a_2}C{O_3}$ in the above equation will be.
A)$106$
B)$53$
C)$26.5$
D)$13.25$

Answer 1

Hint: We characterize the Equivalent mass of the substance as the proportion of the sub-atomic weight or mass of the compound to the n-factor or the causticity or basicity. We can figure the n-factor by deciding the adjustment in oxidation state. We realize that the leading intensity of the apparent multitude of particles delivered by one gram likeness electrolyte is named as equal conductance.

Complete step by step solution:
We can calculate the Equal mass of sodium carbonate as,
We know that the atomic mass of sodium carbonate \[ = 106g/mol\]
We realize that \[N{a_2}C{O_3} \rightleftharpoons 2N{a^ + } + C{O_3}^{2 - }\]
Hence, it’s’ n factor or valence factor can be calculated as,
${\text{n factor}} = {\text{Change in oxidation number}}$
\[{\text{n factor}} = 2\]
Thus the equivalent mass can be given as,
The equivalent weight of $N{a_2}C{O_3} = \dfrac{{{\text{Molecular weight}}}}{2}$
The equivalent weight of $N{a_2}C{O_3} = \dfrac{{106\,g/mol}}{2}$
The equivalent weight of $N{a_2}C{O_3} = 53$
The equivalent weight of $N{a_2}C{O_3}$ is estimated as $53.$
Therefore, the option B is correct.

Note: We characterize the Equivalent mass of the substance as the proportion of the sub-atomic weight or mass of the compound to the n-factor or the causticity or basicity. We can figure the n-factor by deciding the adjustment in oxidation state. The 'n' factor of an acid is that the quantity of particles supplanted by one mole of corrosive. The n-factor for acid isn't the quantity of moles of usable Hydrogen iotas present in one mole of acid. The oxidation condition of the single component is zero yet in the event that it has any charge present on it, at that point it is considered as n-factor while basicity for the acidic substance and the acidity is characterized for the essential substance.