Question

In the standardization of $\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3$ using $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$ by iodometry, the equivalent weight of $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$ is

• A. molecular weight/2
• B. molecular weight/6
• C. molecular weight/3
• D. same as molecular weight

Hint:

Equivalent weight = $\frac{\text{Molecular weight}}{n\text{-factor}}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Equivalent weight = $\frac{\text{Molecular weight}}{n\text{-factor}}$.
Step 2: Detailed Explanation:
$\mathrm{Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O}$. $n$-factor = 6, so equivalent weight = $\frac{M}{6}$.
Step 3: Final Answer:
Equivalent weight is molecular weight/6.