Calculate to four decimal places the mass of purely dry MgCO3(Magnesium Carbonate) required to make 1L of a 1000 ppm Mg stock solution.
To calculate the mass of dry MgCO3 required to make a 1000 ppm (parts per million) Mg stock solution, we need to know the molar mass of MgCO3.
Molar mass of MgCO3 = (1 × Molar mass of Mg) + (1 × Molar mass of C) + (3 × Molar mass of O)
The molar mass of Mg is 24.31 g/mol, the molar mass of C is 12.01 g/mol, and the molar mass of O is 16.00 g/mol.
Molar mass of MgCO3 = (1 × 24.31 g/mol) + (1 × 12.01 g/mol) + (3 × 16.00 g/mol)
= 24.31 g/mol + 12.01 g/mol + 48.00 g/mol
= 84.32 g/mol
Now, we can calculate the mass of dry MgCO3 required to make 1L of a 1000 ppm Mg stock solution:
Mass (g) = (Desired concentration in ppm) × (Volume in L) × (Molar mass in g/mol) / 10^6
Mass (g) = 1000 ppm × 1 L × 84.32 g/mol / 10^6
= 0.08432 g
Therefore, to make a 1000 ppm Mg stock solution, you would require approximately 0.08432 grams of pure dry MgCO3.
To calculate the mass of purely dry MgCO3 required to make a 1000 ppm Mg stock solution, we need to consider the molar mass of MgCO3 and the definition of ppm.
1. Find the molar mass of MgCO3:
- The molar mass of magnesium (Mg) is approximately 24.31 g/mol.
- The molar mass of carbon (C) is approximately 12.01 g/mol.
- The molar mass of oxygen (O) is approximately 16.00 g/mol.
- Since MgCO3 contains one magnesium atom, one carbon atom, and three oxygen atoms, the molar mass of MgCO3 is calculated as follows:
Molar mass of MgCO3 = (1 × molar mass of Mg) + (1 × molar mass of C) + (3 × molar mass of O)
2. Calculate the molar mass of MgCO3:
Molar mass of MgCO3 = (1 × 24.31 g/mol) + (1 × 12.01 g/mol) + (3 × 16.00 g/mol)
3. Convert ppm to mass per volume:
Since ppm is defined as parts per million, it represents the number of parts of the solute per million parts of the solution. In this case, ppm represents µg/mL or mg/L.
4. Calculate the mass of MgCO3 required:
To calculate the mass of MgCO3 required to make a 1000 ppm Mg stock solution in 1L (1000 mL):
- Divide 1000 ppm by 1000 to convert it to mg/L.
- Multiply the result by the molar mass of MgCO3 to get the mass of MgCO3 in mg:
Mass of MgCO3 (mg) = (1000 ppm / 1000) × Molar mass of MgCO3
5. Round the answer to four decimal places:
Round the calculated mass of MgCO3 (mg) to four decimal places to get the final answer.
Following these steps, you would be able to calculate the mass of purely dry MgCO3 required to make 1L of a 1000 ppm Mg stock solution.
A good factor to remember is
1 ppm = 1 mg/L so 1000 ppm would be 1000 mg/L
So you want1000 mg (that's 1 gram) MgCO3 in a L of solution.