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# Equivalent weight of ${{{K}}_2}{{Cr}}{{{O}}_4}$, when it reacts with ${{AgN}}{{{O}}_3}$ to give ${{A}}{{{g}}_2}{{Cr}}{{{O}}_4}$ is:A. infiniteB. ${{M}}$C. $\dfrac{{{M}}}{2}$D. $\dfrac{{{M}}}{3}$

Last updated date: 13th Jun 2024
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Hint: Equivalent weight is calculated by dividing molecular weight by its valency. Valency depends on the total charge of the ions. It may be different in the acid-base concept. Equivalent weight is the weight of a molecule which has one equivalent of proton in acids and hydroxide in bases.

Complete step by step answer:
Before calculating the equivalent weight of ${{{K}}_2}{{Cr}}{{{O}}_4}$, let’s know about the molecular weight and equivalent weight.
Molecular weight is calculated using chemical formula. The chemical formula of potassium chromate is ${{{K}}_2}{{Cr}}{{{O}}_4}$. It is the sum of atomic weights of each element in a compound.
Now let’s calculate the molecular weight of potassium chromate. Atomic weight of potassium is $39$, chromium is $52$ and oxygen is $16$. There are two potassium atoms and four oxygen atoms. Thus we have to multiply the atomic weight of potassium with two and that of oxygen with four.
It can be calculated as $\left( {39 \times 2} \right) + 52 + \left( {16 \times 4} \right) = 194$
Molecular weight is represented by ${{M}}$.
Equivalent weight is the ratio of molecular weight to the valency where valency is the total charge of cation. Thus valency is $2$.
Thus equivalent weight is $\dfrac{{{M}}}{2}$.

Hence, the correct option is C.