Final answer:
To prepare 150 mL of a 250 mM MgCl2 solution, 3,570,375 micrograms of MgCl2 are required. This is calculated by converting the volume to liters, using molarity to find moles, and then converting moles to micrograms using the molar mass of MgCl2.
Step-by-step explanation:
The question asks how many micrograms of solid MgCl2 are required to prepare 150ml of a 250mM solution. To calculate this, we first convert the volume from milliliters to liters, then use the molarity and the molar mass of MgCl2 to find the mass in grams, and finally, convert that mass into micrograms.
- Conversion of volume: 150 mL = 0.150 L
- Decimal of Millimolarity to Molarity: 250 mM = 0.250 M
- Calculate moles of MgCl2: moles = Molarity × Volume = 0.250 M × 0.150 L = 0.0375 moles
- Molar mass of MgCl2: approximately 95.21 g/mol (exact value may vary based on isotope composition)
- Moles to mass: Mass (g) = 0.0375 moles × 95.21 g/mol = 3.570375 g
- Conversion to micrograms: 3.570375 g = 3,570,375 micrograms
Therefore, 3,570,375 micrograms of MgCl2 are required to prepare 150 mL of a 250 mM solution.