Calculate the mass of salt needed to prepare 100 cm3 of 0.05 M solution of:
(i) Fe(NH4)2(SO4)2.6H2O
To calculate the mass of salt needed to prepare a solution, you need to know the molar mass of the salt and the desired concentration of the solution.
(i) Fe(NH4)2(SO4)2.6H2O is a complex compound that contains iron, ammonium, and sulfate ions, as well as 6 water molecules. To find the molar mass of this compound, you need to add up the atomic masses of each element.
Atomic masses:
Fe: 55.845 g/mol
N: 14.007 g/mol
H: 1.008 g/mol
S: 32.06 g/mol
O: 16.00 g/mol
First, let's calculate the molar mass of Fe(NH4)2(SO4)2.6H2O:
Fe(NH4)2(SO4)2: (55.845 g/mol) + 2[(14.007 g/mol) + 4(1.008 g/mol)] + 2[(32.06 g/mol) + 4(16.00 g/mol)] = 278.04 g/mol
Then, let's calculate the molar mass of water (H2O):
H2O: 2[(1.008 g/mol) + 16.00 g/mol] = 18.02 g/mol
Now, we can calculate the molar mass of the entire compound:
Molar mass = (278.04 g/mol) + 6(18.02 g/mol) = 392.32 g/mol
To prepare a 0.05 M solution, you need 0.05 moles of Fe(NH4)2(SO4)2.6H2O per liter of solution. Since you want to prepare 100 cm3 of solution (0.1 L), you need to use the following formula:
Mass of salt (g) = molar mass (g/mol) x concentration (mol/L) x volume (L)
Mass of salt = 392.32 g/mol x 0.05 mol/L x 0.1 L = 1.96 g
Therefore, you will need 1.96 grams of Fe(NH4)2(SO4)2.6H2O to prepare 100 cm3 of a 0.05 M solution.