×
a 25cm³ aliquot of ammonium iron(ii) tetraoxomanganate(vii) containing 32.9g of its salt in 250cm³ solution, needed 25cm³ of potassium tetraoxomanganate(vii) solution for a colour change to occur. what is the concentration of the potassium tetraoxomanganate(vii) solution jiskha
To find the concentration of the potassium tetraoxomanganate(VII) solution, you need to use the information provided and perform some calculations.
First, let's calculate the moles of ammonium iron(II) tetraoxomanganate(VII) in the 25 cm³ aliquot.
1. Calculate the molar mass of the ammonium iron(II) tetraoxomanganate(VII):
- Ammonium (NH₄⁺): 1 nitrogen (N) * 1 atom + 4 hydrogens (H) * 4 atoms = 1 * 14.01 g/mol + 4 * 1.01 g/mol = 18.05 g/mol
- Iron(II) (Fe²⁺): 1 iron (Fe) * 1 atom = 1 * 55.85 g/mol = 55.85 g/mol
- Tetraoxomanganate(VII) (MnO₄⁻): 1 manganese (Mn) * 1 atom + 4 oxygens (O) * 4 atoms = 1 * 54.94 g/mol + 4 * 16.00 g/mol = 74.94 g/mol
- Total molar mass = 18.05 g/mol + 55.85 g/mol + 74.94 g/mol = 148.84 g/mol
2. Calculate the number of moles of ammonium iron(II) tetraoxomanganate(VII) using its molar mass and mass:
- Moles = Mass / Molar Mass = 32.9 g / 148.84 g/mol
Next, let's find the concentration of the potassium tetraoxomanganate(VII) solution.
3. Calculate the molarity of the ammonium iron(II) tetraoxomanganate(VII) solution using moles and volume:
- Molarity = Moles / Volume (in liters) = Moles / (250 cm³ / 1000) = (32.9 g / 148.84 g/mol) / (250 cm³ / 1000)
4. Since the same volume (25 cm³) of the potassium tetraoxomanganate(VII) solution is required for a color change, we can equate the molarity of both solutions.
- Molarity of potassium tetraoxomanganate(VII) = Molarity of ammonium iron(II) tetraoxomanganate(VII)
Now, you can substitute the calculated value of the molarity of the ammonium iron(II) tetraoxomanganate(VII) solution into the equation to find the concentration of the potassium tetraoxomanganate(VII) solution.