Final answer:
The percent composition of nitrogen in silver nitrate (AgNO3) is approximately 8.25%, which doesn't match any of the provided options. The question appears to have an error in its answer choices.
Step-by-step explanation:
To find the percent composition of nitrogen in silver nitrate, AgNO3, you first need to calculate the molar mass of AgNO3. Silver (Ag) has an atomic mass of 107.87 g/mol, nitrogen (N) has an atomic mass of 14.01 g/mol, and oxygen (O) has an atomic mass of 16.00 g/mol. Since there's one nitrogen atom in the compound, the mass of nitrogen in the formula is simply its atomic mass.
The molar mass of AgNO3 is calculated as follows:
107.87 g/mol (Ag) + 14.01 g/mol (N) + 3 x 16.00 g/mol (O) = 169.87 g/mol
Then, to find the percent composition of nitrogen, you divide the mass of nitrogen by the total molar mass and multiply by 100%:
(14.01 g/mol) / (169.87 g/mol) × 100% = 8.248%, which is approximately 8.25%.
None of the given options match this calculation. Hence, the question as stated does not have a correct answer listed among the provided choices a) 36.5%, b) 22.7%, c) 63.5%, d) 14.3%.