Answer:To make a 1.50 M solution of ammonium nitrate using 3.49 grams of the compound, we need to calculate the volume of the solution required and the number of moles of ammonium nitrate present in the given mass.
1. Calculate the molar mass of ammonium nitrate (NH4NO3):
- The molar mass of N = 14.01 g/mol
- The molar mass of H = 1.01 g/mol
- The molar mass of O = 16.00 g/mol
Molar mass of NH4NO3 = (1*14.01) + (4*1.01) + (3*16.00) = 80.04 g/mol
2. Determine the number of moles of NH4NO3 present in 3.49 grams:
Number of moles = Mass / Molar mass
Number of moles = 3.49 g / 80.04 g/mol ≈ 0.0436 mol
3. Calculate the volume of the solution required to make a 1.50 M solution:
Molarity (M) = Number of moles / Volume of solution (in liters)
1.50 M = 0.0436 mol / Volume (in liters)
Volume (in liters) = 0.0436 mol / 1.50 mol/L ≈ 0.029 L or 29.0 mL
Therefore, to make a 1.50 M solution of ammonium nitrate using 3.49 grams of the compound, you would need approximately 29.0 mL of the solution.
Explanation:
