1. Write the decomposition reaction

On heating, sodium chlorate decomposes to sodium chloride and oxygen:

$2NaClO_3 \rightarrow 2NaCl + 3O_2$

2. Find moles of oxygen lost

Given mass of oxygen lost = $0.16\,g$

Since oxygen is released as $O_2$,

$n(O_2)=\frac{0.16}{32}=0.005\,mol$

3. Use stoichiometry to find moles of $NaCl$ formed

From the reaction,

$3\,mol\; O_2 \text{ are formed with } 2\,mol\; NaCl$

So,

$n(NaCl)=\frac{2}{3}\times 0.005=0.00333\,mol$

4. Relate $NaCl$ to $AgCl$ formed

When the residue is dissolved and treated with silver ion:

$NaCl + AgNO_3 \rightarrow AgCl \downarrow + NaNO_3$

Thus,

$n(AgCl)=n(NaCl)=0.00333\,mol$

5. Calculate mass of $AgCl$

Given molar mass of $AgCl = 143.5\,g\,mol^{-1}$

$m(AgCl)=0.00333\times 143.5$

$m(AgCl)\approx 0.478\,g$

So the mass of $AgCl$ obtained is approximately

$\boxed{0.48\,g}$

6. Option check

• A: $0.35$ ❌
• B: $0.41$ ❌
• C: $0.48$ ✅
• D: $0.54$ ❌

Therefore, the correct option is C.