Question

The equivalent weight of ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ in acidic medium is expressed in terms of its molecular weight (M) as

• A. M/3
• B. M/4
• C. M/6
• D. M/7

Answer 1

The correct option is C

M/6

Explanation for correct option:

C. M/6

Step 1: Equivalent weight:

• “Equivalent weight is equal to the molecular/atomic mass divided by the number of electrons involved in the reaction per molecule/atom/ion.”

Step 2: Formula for equivalent weight:

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

Step 3: Reaction for the dissociation of Potassium dichromate:

• In the solution, ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ gives the following ions, the chemical reaction can be depicted as:

${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7\left(\mathrm{aq}\right)+14{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+6{\mathrm{e}}^{–}\to 2{\mathrm{K}}^{+}\left(\mathrm{aq}\right)+2{\mathrm{Cr}}^{3+}\left(\mathrm{aq}\right)+7{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

• The chemical equivalent is 6 because six electrons participate in the chemical reaction.

Step 3: Calculation for equivalent weight:

• Let's consider “M” be the atomic weight of ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$.

$\mathrm{Equivalent}\mathrm{weight}\mathrm{of}{\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7=\frac{\mathrm{M}}{6}$

Thus, the equivalent weight of ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ is M/6.

Explanation for incorrect options:

The equivalent weight of ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ is M/6. Thus, options A, B, and D are incorrect.

Hence, the correct option is C i.e. M/6.