I have to prepare 1ppm of Fe+2 solution in 1litre. How many grams of ferrous ammonium sulphate will be dissolved in 1litre to make above solution.. plzz guide me
You want 1 ppm and that is 1 mg/L.
Convert 1 mg Fe^2+ to ferrous ammonium sulfate.
1 mg Fe^2+ x [molar mass Fe(NH4)2(SO4)2] = ?
To calculate the number of grams of ferrous ammonium sulfate required to prepare a 1 ppm (parts per million) Fe+2 solution in 1 liter, you need to consider the molar mass of the compound and the molarity of the iron solution.
Here's a step-by-step guide on how to do it:
Step 1: Determine the molar mass of ferrous ammonium sulfate (Fe(NH4)2(SO4)2).
The molar mass of Fe(NH4)2(SO4)2 can be calculated by looking up the atomic masses of each element involved and adding them up. The molar masses are as follows:
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
Molar mass of Fe(NH4)2(SO4)2 = (2 * (2 * (1.008) + 14.007) + 55.845 + 2 * (32.06 + 4 * 16.00)) g/mol
Step 2: Calculate the number of moles of Fe+2 required for a 1 ppm solution in 1 liter.
1 ppm means 1 part per million, which can be expressed as 1 g/L (since 1 ppm = 1 mg/L = 1 μg/mL). Therefore, you need to find the number of moles for 1 g of Fe+2:
Number of moles of Fe+2 = (1 g) / (molar mass of Fe+2)
Step 3: Calculate the number of moles of ferrous ammonium sulfate required.
Since the stoichiometry of the compound is 1:1, the number of moles of ferrous ammonium sulfate required is equal to the number of moles of Fe+2.
Number of moles of ferrous ammonium sulfate = Number of moles of Fe+2
Step 4: Convert moles to grams.
To convert the number of moles to grams, you can use the formula:
Mass (g) = Number of moles * Molar mass
Therefore,
Mass of ferrous ammonium sulfate (g) = Number of moles of ferrous ammonium sulfate * Molar mass of ferrous ammonium sulfate
By following these steps, you will be able to calculate the number of grams of ferrous ammonium sulfate needed to prepare a 1 ppm Fe+2 solution in 1 liter.