If the equivalent mass of S in SO, is 8, then equivalent weight of S in SO; is :

🔎 Key Concepts You Need

1. Oxidation Number (O.N.)

   • Shows how many electrons an atom effectively loses (positive) or gains (negative) in a compound.

2. Equivalent Mass (E.M.) of an Element in a Redox Change
   $\text{Equivalent Mass} = \frac{\text{Atomic Mass}}{\lvert\Delta\text{O.N.}\rvert}$

   • $\lvert\Delta\text{O.N.}\rvert$ is the magnitude of change in oxidation number per atom of that element.

3. Given Information as a Consistency Check

   • In $SO_2$, sulfur’s oxidation number changes from $0 \to +4$ ((\Delta \text{O.N.}=4)).
   • Atomic mass of sulfur $=32\,$g.
   • Plug in: $\frac{32}{4}=8\,\text{g}$ — matches the given value, confirming the formula we’ll use.

Logical Chain of Steps to Solve

1. Determine oxidation number of S in $SO_3$.
   Oxygen is $-2$ each; three oxygens give $-6$. For the molecule to be neutral, sulfur must be +6.

2. Calculate change in oxidation number.
   From elemental S $(0)$ to $+6$ is a change of $6$.

3. Apply the equivalent-mass formula.
   $\text{Equivalent mass of S in }SO_3 = \frac{32}{6}$

4. Simplify/Express answer.
   $= \frac{16}{3}\,\text{g} \approx 5.33\,\text{g}$

🧒🏻 Imagine sulfur (S) is climbing a ladder of charges

1. At the bottom of the ladder, sulfur has no charge (0).
2. In $SO_2$, sulfur climbs 4 steps up (charge +4).
   • We are told that this 4-step climb makes its equivalent mass 8 g.
3. In $SO_3$, sulfur climbs 6 steps up (charge +6).

Because the climb in $SO_3$ is bigger (6 steps instead of 4), each "step" of mass must get smaller so the total stays the same atomic weight.

So we simply share the 32 g atomic weight of sulfur over 6 steps instead of 4:

$\text{Equivalent mass in }SO_3 = \frac{32\,\text{g}}{6} = 5.33\,\text{g}$

That’s all!

Step-by-Step Solution

1. Oxidation state of S in $SO_2$ (check) $\text{S} + 2(-2) = 0 \;\Rightarrow\; \text{S} = +4$ Given E.M. $=\frac{32}{4}=8\,\text{g}$ — matches the statement.

2. Oxidation state of S in $SO_3$ $\text{S} + 3(-2) = 0 \;\Rightarrow\; \text{S} = +6$

3. Change in oxidation number per S atom $\Delta \text{O.N.} = 0 \to +6 = 6$

4. Compute equivalent mass

5. Simplify

Final Answer: $\boxed{\dfrac{16}{3}\;\text{g} \;\text{(}\approx 5.33\,\text{g}\text{)}}$