Question

Reaction

$$ \begin{array}{l} \text{As}_2\text{S}_3 + \text{H}^+ + \text{NO}_3^- \rightarrow \text{NO} + \text{H}_2\text{O} + \text{AsO}_4^{3-} + \text{SO}_4^{2-} \end{array} $$

The relationship between the molecular weight (M) and the equivalent weight (E) of As$_2$S$_3$ in this reaction is:

  • A $ E = \frac{M}{2} $
  • B $ E = \frac{M}{4} $
  • C $ E = \frac{M}{24} $
  • D $ E = \frac{M}{28} $

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Answer

To determine the relationship between the molecular mass ($M$) and the equivalent mass ($E$) of As$_2$S$_3$, consider the given reaction:

$$ \begin{array}{l} \mathrm{As}{2} \mathrm{S}{3}+\mathrm{H}^{+}+\mathrm{NO}{3}^{-} \rightarrow \mathrm{NO}+\mathrm{H}{2} \mathrm{O}+\mathrm{AsO}{4}^{3-} + \mathrm{SO}{4}^{2-} \end{array} $$

For As$_2$S$_3$, let's analyze the equivalence factor:

  • The factor $n$ is calculated using the number of atoms and their oxidation states.
  • For each atom in As$_2$S$_3$, the oxidation state changes need to be taken into account.

The calculation is as follows:

$$ n\text{-factor} = |2 \times 2 + 8 \times 3| = 28 $$

So, the equivalent mass ($E$) is related to the molecular mass ($M$) by the equation:

$$ E = \frac{M}{28} $$

Therefore, the correct answer is:

D. $E = \frac{M}{28}$


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