What mass of aluminium nitrate (Al(NO3)3) would be required to prepare 2.000 L of a 0.0160 M aqueous solution of this salt?
To determine the mass of aluminum nitrate required, we need to use the formula:
Molarity (M) = (moles of solute) / (volume of solution in liters)
Rearranging the formula, we can solve for the moles of solute:
(moles of solute) = Molarity (M) * (volume of solution in liters)
Now, let's substitute the given values into the equation:
Molarity (M) = 0.0160 M
Volume of solution = 2.000 L
(moles of solute) = 0.0160 M * 2.000 L
(moles of solute) = 0.0320 moles
Since aluminum nitrate (Al(NO3)3) is a compound made up of aluminum (Al) and three nitrate (NO3) ions, we need to determine the molar mass of aluminum nitrate to convert moles to grams.
The molar mass of aluminum (Al) is 26.98 g/mol, and the molar mass of nitrate (NO3) is 62.01 g/mol.
The molar mass of aluminum nitrate (Al(NO3)3) can be calculated by adding the molar masses of aluminum and three nitrate ions:
Molar mass of Al(NO3)3 = (1 * molar mass of Al) + (3 * molar mass of NO3)
Molar mass of Al(NO3)3 = (1 * 26.98 g/mol) + (3 * 62.01 g/mol)
Molar mass of Al(NO3)3 = 26.98 g/mol + 186.03 g/mol
Molar mass of Al(NO3)3 = 213.01 g/mol
Finally, we can determine the mass of aluminum nitrate required:
Mass = (moles of solute) * (molar mass of Al(NO3)3)
Mass = 0.0320 moles * 213.01 g/mol
Mass = 6.82 grams
Therefore, to prepare a 2.000 L 0.0160 M aqueous solution of aluminum nitrate, you would need 6.82 grams of aluminum nitrate.
How many mols of Al(NO3)^2 do you need? That's mols - M x L = ?
Now convert mols Al(NO3)2 to grams. That's g = mols x molar mass = ?