Final answer:
The w/w percentage of nitrogen in an ammonium nitrate solution with the given molar concentration and density is (D) 5.6%.
Step-by-step explanation:
To calculate the w/w percentage of nitrogen in an ammonium nitrate solution, we need to use the given molar concentration of ammonium nitrate and the density of the solution.
The molar mass of ammonium nitrate (NH4NO3) is the sum of the molar masses of nitrogen (N), hydrogen (H), and oxygen (O), which is 14.01 g/mol (N) + (4 x 1.01 g/mol) (H) + (3 x 16.00 g/mol) (O) = 80.06 g/mol for ammonium nitrate.
Since there are two nitrogen atoms in ammonium nitrate, the molar mass of nitrogen in the compound is 2 x 14.01 g/mol = 28.02 g/mol.
To find the mass of ammonium nitrate in 1 dm3 (1000 cm3) of solution, we multiply the molarity by the molar mass:
2.1 mol/dm3 x 80.06 g/mol = 168.126 g/dm3.
Then we find the mass of nitrogen in the same volume: 2.1 mol/dm3 x 28.02 g/mol = 58.842 g/dm3.
Next, we calculate the w/w percentage of nitrogen using the density of the solution: Density = 1.05 g/cm3, so the mass of the solution in 1 dm3 is 1.05 g/cm3 x 1000 cm3/dm3 = 1050 g/dm3.
The w/w% = (mass of nitrogen / mass of solution) x 100% = (58.842 g / 1050 g) x 100% = 5.6%.
The correct answer is (d) 5.6% %.