Calculate the mass of salt needed to prepare 100 cm3 of 0.05 M solution of:
(i) Fe(NH4)2(SO4)2.6H2O
(ii) FeNH4(SO4)2.12H2O
How many mols do you want? That's M x L = ?
Convert mols to grams. grams = mols x molar mass.
(i) The mass of salt needed to prepare a solution can be calculated using the formula:
Mass = Volume x Concentration x Molar mass
(i) Fe(NH4)2(SO4)2.6H2O:
Molar mass of Fe(NH4)2(SO4)2.6H2O = 392.14 g/mol
To calculate the mass of salt needed, we can use the formula:
Mass = Volume x Concentration x Molar mass
Mass = 100 cm^3 x 0.05 mol/cm^3 x 392.14 g/mol
Mass = 196.07 g
So, you would need approximately 196.07 grams of Fe(NH4)2(SO4)2.6H2O to prepare 100 cm^3 of 0.05 M solution.
(ii) FeNH4(SO4)2.12H2O:
Molar mass of FeNH4(SO4)2.12H2O = 392.14 g/mol
To calculate the mass of salt needed, we can use the formula:
Mass = Volume x Concentration x Molar mass
Mass = 100 cm^3 x 0.05 mol/cm^3 x 392.14 g/mol
Mass = 196.07 g
So, you would need approximately 196.07 grams of FeNH4(SO4)2.12H2O to prepare 100 cm^3 of 0.05 M solution.
However, the real question is, why do chemists like nitrates so much?
Because they're cheaper than day rates!
To calculate the mass of salt needed to prepare a solution, we first need to determine the molar mass of the salt. Then, we can use the molarity and the volume of the solution to find the mass.
(i) Fe(NH4)2(SO4)2.6H2O:
The molar mass of Fe(NH4)2(SO4)2.6H2O can be calculated by summing up the atomic masses of its constituent elements:
Fe: 55.85 g/mol
N: 14.01 g/mol (x 2)
H: 1.01 g/mol (x 4 in NH4 and x 12 in H2O)
S: 32.07 g/mol (x 2)
O: 16.00 g/mol (x 10 in SO4 and x 6 in H2O)
Now, let's calculate the total molar mass:
Molar mass = (55.85 g/mol) + (14.01 g/mol x 2) + (1.01 g/mol x 4) + (32.07 g/mol x 2) + (16.00 g/mol x 10) + (16.00 g/mol x 6)
Molar mass = 392.14 g/mol
To calculate the mass of salt needed, we can use the formula:
Mass (g) = Molar mass (g/mol) x Volume (L) x Molarity (mol/L)
The given volume is 100 cm³, which is equivalent to 100/1000 = 0.1 L.
Mass (g) = 392.14 g/mol x 0.1 L x 0.05 mol/L
Mass (g) = 1.9607 g
Therefore, you would need approximately 1.96 grams of Fe(NH4)2(SO4)2.6H2O to prepare a 100 cm³ 0.05 M solution.
(ii) FeNH4(SO4)2.12H2O:
The molar mass of FeNH4(SO4)2.12H2O can be calculated in a similar manner as before.
Fe: 55.85 g/mol
N: 14.01 g/mol
H: 1.01 g/mol (x 4 in NH4 and x 24 in H2O)
S: 32.07 g/mol (x 2)
O: 16.00 g/mol (x 10 in SO4 and x 12 in H2O)
Molar mass = (55.85 g/mol) + (14.01 g/mol) + (1.01 g/mol x 4) + (32.07 g/mol x 2) + (16.00 g/mol x 10) + (16.00 g/mol x 12)
Molar mass = 504.13 g/mol
Using the same formula as before:
Mass (g) = 504.13 g/mol x 0.1 L x 0.05 mol/L
Mass (g) = 2.5206 g
Therefore, you would need approximately 2.52 grams of FeNH4(SO4)2.12H2O to prepare a 100 cm³ 0.05 M solution.
To calculate the mass of salt needed to prepare a certain volume and concentration of a solution, you can use the formula:
Mass (g) = Volume (L) * Concentration (mol/L) * Molar mass (g/mol)
Let's calculate the mass of salt needed for each compound:
(i) Fe(NH4)2(SO4)2.6H2O:
The molar mass of Fe(NH4)2(SO4)2.6H2O can be calculated by adding up the molar masses of its constituent atoms.
Fe(NH4)2(SO4)2.6H2O = (1 * molar mass of Fe) + (2 * molar mass of N) + (8 * molar mass of H) + (2 * molar mass of S) + (24 * molar mass of O) + (6 * molar mass of H2O)
(ii) FeNH4(SO4)2.12H2O:
Similarly, the molar mass of FeNH4(SO4)2.12H2O can be calculated by adding up the molar masses of its constituent atoms.
FeNH4(SO4)2.12H2O = (1 * molar mass of Fe) + (1 * molar mass of N) + (6 * molar mass of H) + (2 * molar mass of S) + (24 * molar mass of O) + (12 * molar mass of H2O)
Once you have calculated the molar mass for each compound, you can substitute the values into the formula and solve for the mass of salt needed. Remember to convert the volume from cm3 to L by dividing by 1000.
I'm sorry, but I don't have the specific molar masses of Fe(NH4)2(SO4)2.6H2O and FeNH4(SO4)2.12H2O. You can find the molar masses from the periodic table and calculate the mass using the formula provided.