A sample of NaClO$$_3$$ is converted by heat to NaCl with a loss of 0.16 g of oxygen. The residue is dissolved in water and precipitated as AgCl. The mass of AgCl (in g) obtained will be: (Given: Molar mass of AgCl = 143.5 g mol$$^{-1}$$)

We begin with the thermal decomposition of sodium chlorate. The balanced chemical equation is first written because it supplies the mole ratio that will later connect the loss of oxygen to the amount of NaCl formed:

$$2\,\text{NaClO}_{3} \;\xrightarrow{\Delta}\; 2\,\text{NaCl} \;+\; 3\,\text{O}_{2}$$

This equation tells us that every $$2$$ moles of $$\text{NaClO}_{3}$$ give $$2$$ moles of $$\text{NaCl}$$ and $$3$$ moles of $$\text{O}_{2}$$. The loss in mass recorded in the experiment comes only from the oxygen that escapes as gas.

The question states that $$0.16\;\text{g}$$ of oxygen is lost. First, we must convert this mass to moles. The formula we use is:

$$\text{Moles} \;=\; \dfrac{\text{Given mass}}{\text{Molar mass}}$$

The molar mass of molecular oxygen $$\text{O}_{2}$$ is $$32\;\text{g mol}^{-1}$$, so

$$n(\text{O}_{2}) \;=\; \dfrac{0.16\;\text{g}}{32\;\text{g mol}^{-1}} \;=\; 0.005\;\text{mol}$$

Now we link these moles of $$\text{O}_{2}$$ to the moles of $$\text{NaCl}$$ produced by employing the stoichiometric coefficients from the balanced decomposition equation. From the equation we read that $$3$$ moles of $$\text{O}_{2}$$ correspond to $$2$$ moles of $$\text{NaCl}$$. Therefore, using proportion:

$$n(\text{NaCl}) \;=\; n(\text{O}_{2}) \times \dfrac{2}{3} \;=\; 0.005 \times \dfrac{2}{3} \;=\; 0.003333\;\text{mol}$$

The residue containing this $$\text{NaCl}$$ is dissolved in water and treated with a solution of silver nitrate. The precipitation reaction that occurs is

$$\text{NaCl} \;+\; \text{AgNO}_{3} \;\longrightarrow\; \text{AgCl} \;+\; \text{NaNO}_{3}$$

This reaction is a simple double displacement and, importantly, the mole ratio of $$\text{NaCl}$$ to $$\text{AgCl}$$ is $$1\!:\!1$$. Hence, the moles of $$\text{AgCl}$$ formed equal the moles of $$\text{NaCl}$$ present:

$$n(\text{AgCl}) = n(\text{NaCl}) = 0.003333\;\text{mol}$$

Finally, we convert these moles of silver chloride to mass, again using the same fundamental formula. The molar mass of $$\text{AgCl}$$ is given as $$143.5\;\text{g mol}^{-1}$$, so

$$\text{Mass of AgCl} \;=\; n(\text{AgCl}) \times M(\text{AgCl}) \;=\; 0.003333\;\text{mol} \times 143.5\;\text{g mol}^{-1}$$

$$\text{Mass of AgCl} \;=\; 0.4783\;\text{g}$$

Rounding this value to two significant figures, as is typical for laboratory data, we obtain $$0.48\;\text{g}$$.

Hence, the correct answer is Option D.