WBJEE · Chemistry · d and f Block Elements
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 & Solution
Correct Answer
(C) \(M / 6\)
Step-by-step Solution
Detailed explanation
In acidic medium, \[ \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \longrightarrow \mathrm{Cr}^{3+} \] Equivalent weight of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}=\frac{\text { Molecular weight }}{\text { Change in oxidation state of } \mathrm{c}}\)…
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