Question

The equivalent weight of ${\mathrm{K}}_2{\mathrm{CrO}}_4$, when it reacts with ${\mathrm{AgNO}}_3$ to give ${\mathrm{Ag}}_2{\mathrm{CrO}}_4$ is _______.

• A. $\mathrm{M}$
• B. $\frac{\mathrm{M}}{2}$
• C. $\frac{\mathrm{M}}{3}$
• D. Infinite

Answer 1

The correct option is B

$\frac{\mathrm{M}}{2}$

Explanation for correct option:

B. $\frac{\mathrm{M}}{2}$

Step 1: Equivalent mass:

• Equivalent mass is equal to the molecular/atomic mass divided by the number of electrons involved in the reaction per molecule/atom/ion.
• It is expressed as:

$\mathrm{Equivalent}\mathrm{weight}=\frac{\mathrm{Atomic}\mathrm{weight}}{\mathrm{Valency}}$

Step 2: Reaction of Potassium chromate and Silver nitrate:

• When Potassium chromate$\left({\mathrm{K}}_2{\mathrm{CrO}}_4\right)$ reacts with Silver nitrate$\left({\mathrm{AgNO}}_3\right)$, it produces Silver chromate$\left({\mathrm{Ag}}_2{\mathrm{CrO}}_4\right)$ and Potassium nitrate$\left({\mathrm{KNO}}_3\right)$.
• The chemical reaction can be depicted as:

${\mathrm{K}}_2{\mathrm{CrO}}_4\left(\mathrm{aq}\right)+2{\mathrm{AgNO}}_3\left(\mathrm{aq}\right)\to {\mathrm{Ag}}_2{\mathrm{CrO}}_4\left(\mathrm{aq}\right)+2{\mathrm{KNO}}_3\left(\mathrm{aq}\right)$

Step 3: Calculation of Equivalent mass:

The valency of Potassium in Potassium chromate is $2$.

If M is the molecular weight of Potassium chromate, thus the equivalent weight will be:

$\mathrm{Equivalent}\mathrm{weight}=\frac{\mathrm{M}}{2}$

The explanation for incorrect options:

The equivalent weight of ${\mathrm{K}}_2{\mathrm{CrO}}_4$ is $\frac{\mathrm{M}}{2}$. Thus, options A, C and D are incorrect.

Hence, the correct option is B i.e. the equivalent weight of ${\mathrm{K}}_2{\mathrm{CrO}}_4$, when it reacts with ${\mathrm{AgNO}}_3$ to give ${\mathrm{Ag}}_2{\mathrm{CrO}}_4$ is $\frac{\mathrm{M}}{2}$ .