To determine the mass of Silver chloride ($$\mathrm{AgCl}$$) that gets precipitated, we need to first understand the reaction between Silver nitrate ($$\mathrm{AgNO_3}$$) and Sodium chloride ($$\mathrm{NaCl}$$) to form Silver chloride and Sodium nitrate ($$\mathrm{NaNO_3}$$).

The balanced chemical equation is:

$$\mathrm{AgNO_3} + \mathrm{NaCl} \rightarrow \mathrm{AgCl} + \mathrm{NaNO_3}$$

Given:

Volume of $$\mathrm{AgNO_3}$$ solution = $$150 \mathrm{~ml}$$
Concentration of $$\mathrm{AgNO_3}$$ solution = $$32 \%$$
Volume of $$\mathrm{NaCl}$$ solution = $$150 \mathrm{ml}$$
Concentration of $$\mathrm{NaCl}$$ solution = $$11 \%$$

We will start by calculating the masses of both $$\mathrm{AgNO_3}$$ and $$\mathrm{NaCl}$$ in their respective solutions.

Step 1: Calculate the mass of $$\mathrm{AgNO_3}$$

The mass percent concentration (w/v) gives the mass of the solute in grams per 100 mL of solution. Hence:

$$\text{Mass of } \mathrm{AgNO_3} = \left( \frac{32}{100} \right) \times 150 \mathrm{~ml} = 48 \mathrm{~g}$$

The molar mass of $$\mathrm{AgNO_3}$$ is calculated as:

$$108 (\mathrm{Ag}) + 14 (\mathrm{N}) + 16 \times 3 (\mathrm{O}) = 170 \mathrm{~g/mol}$$

Thus, the number of moles of $$\mathrm{AgNO_3}$$ is:

$$\text{Moles of } \mathrm{AgNO_3} = \frac{48 \mathrm{~g}}{170 \mathrm{~g/mol}} = 0.282 \mathrm{~mol}$$

Step 2: Calculate the mass of $$\mathrm{NaCl}$$

$$\text{Mass of } \mathrm{NaCl} = \left( \frac{11}{100} \right) \times 150 \mathrm{~ml} = 16.5 \mathrm{~g}$$

The molar mass of $$\mathrm{NaCl}$$ is calculated as:

$$23 (\mathrm{Na}) + 35.5 (\mathrm{Cl}) = 58.5 \mathrm{~g/mol}$$

Thus, the number of moles of $$\mathrm{NaCl}$$ is:

$$\text{Moles of } \mathrm{NaCl} = \frac{16.5 \mathrm{~g}}{58.5 \mathrm{~g/mol}} = 0.282 \mathrm{~mol}$$

Step 3: Determine the limiting reagent

Since both $$\mathrm{AgNO_3}$$ and $$\mathrm{NaCl}$$ have the same number of moles (0.282 mol) and they react in a 1:1 molar ratio, neither reagent is in excess. Therefore, both are completely consumed.

Step 4: Calculate the mass of $$\mathrm{AgCl}$$ precipitate

The number of moles of $$\mathrm{AgCl}$$ formed is the same as the moles of the limiting reagent, which is 0.282 mol.

The molar mass of $$\mathrm{AgCl}$$ is calculated as:

$$108 (\mathrm{Ag}) + 35.5 (\mathrm{Cl}) = 143.5 \mathrm{~g/mol}$$

Thus, the mass of $$\mathrm{AgCl}$$ precipitated is:

$$\text{Mass of } \mathrm{AgCl} = 0.282 \mathrm{~mol} \times 143.5 \mathrm{~g/mol} = 40.497 \mathrm{~g}$$

Conclusion:

Therefore, the mass of Silver chloride that gets precipitated is approximately 40.47 grams.

Correct Answer: Option A - 40.47

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