The mass of FeSO4.7H2O required to prepare 500mL of 280ppm Fe2+solutin is
280 ppm = 280 g/10^6 g solution. If we call the density of the solution 1.00 g/mL and we ignore the contribution of the mass of the salt to the solution (usually ok since the amount of salt is so small), then 280 ppm = 280 g/10^6 mL which is the same as 0.280 g/10^3 mL = 0.280 g/L. You want 500 mL; you can go from here.
To calculate the mass of FeSO4·7H2O required to prepare a 500 mL solution with a concentration of 280 ppm (parts per million) of Fe2+, we need to follow these steps:
Step 1: Determine the molecular weight of FeSO4·7H2O.
FeSO4·7H2O is composed of one iron atom (Fe), one sulfur atom (S), and four oxygen atoms (O) in FeSO4. Additionally, there are seven water molecules (H2O), which have a molecular weight of 18 g/mol each.
The atomic weights are as follows:
Fe: 55.85 g/mol
S: 32.07 g/mol
O: 16.00 g/mol
H: 1.01 g/mol
Therefore, the molecular weight of FeSO4·7H2O can be calculated as:
FeSO4: (1 × 55.85) + (1 × 32.07) + (4 × 16.00) = 151.91 g/mol
7H2O: (7 × 2) × 18 = 252 g/mol
Molecular weight of FeSO4·7H2O: 151.91 + 252 = 403.91 g/mol
Step 2: Convert the concentration from ppm to g/L.
Since 1 ppm is equal to 1 mg/L, we can convert 280 ppm to g/L as follows:
280 ppm = 280 mg/L = 0.28 g/L
Step 3: Calculate the mass of FeSO4·7H2O required for 500 mL.
To convert grams per liter (g/L) to grams per milliliter (g/mL), divide the concentration by 1000.
0.28 g/L ÷ 1000 = 0.00028 g/mL
To calculate the mass required for 500 mL, multiply the volume by the concentration:
0.00028 g/mL x 500 mL = 0.14 g
Therefore, the mass of FeSO4·7H2O required to prepare 500 mL of 280 ppm Fe2+ solution is 0.14 grams.