Consider the following reactions sequence When the product (P) is...

The given reactant is p-nitrotoluene.

Step (i): Reduction with Sn/HCl followed by $OH^-$ converts the $-NO_2$ group to an $-NH_2$ group, yielding p-toluidine.

Step (ii): Reaction with acetic anhydride, $(CH_3CO)_2O$, protects the amino group by forming N-(4-methylphenyl)acetamide.

Step (iii): Bromination with $Br_2/AlBr_3$ directs the incoming bromine to the ortho position relative to the strongly activating $-NHCOCH_3$ group (the para position is blocked by the methyl group). This yields 2-bromo-4-methylacetanilide.

Step (iv): Acid hydrolysis with $H_3O^+$ removes the acetyl group, yielding 2-bromo-4-methylaniline as the major product (P).

The molecular formula of product P (2-bromo-4-methylaniline) is $C_7H_8NBr$.
Molar mass of P = $(7 \times 12) + (8 \times 1) + 14 + 80 = 186 \text{ g mol}^{-1}$.

In Carius analysis, the bromine in the organic compound is quantitatively converted to silver bromide (AgBr).
Molar mass of AgBr = $108 + 80 = 188 \text{ g mol}^{-1}$.

Since one molecule of P contains one bromine atom, 1 mole of P produces 1 mole of AgBr.
Moles of P in $1.0 \text{ g} = \dfrac{1.0}{186} \text{ mol}$.
Moles of AgBr produced = $\dfrac{1.0}{186} \text{ mol}$.
Mass of AgBr produced = $\dfrac{1.0}{186} \times 188 \text{ g} \approx 1.01 \text{ g}$.

Rounding to the nearest integer, the mass of the AgBr precipitate is 1 g.

Answer: $1$