The density of 2-methyl-2-butanol is 0.806 g/mL.. How many mgs represent an aliquot of 430 µL.? How many mmols represent an aliquot of 0.650 mL. (2-methyl-2-butanol, formula weight is 88.15 g/mol)?
The Correct Answer and Explanation is :
To solve this, we need to compute the mass and the number of millimoles of 2-methyl-2-butanol for the given volumes and conditions. Here’s the breakdown:
1. Mass of 430 µL aliquot in mg:
- Convert µL to mL:
430 μL=430×10−3 mL=0.430 mL430 \, \mu\text{L} = 430 \times 10^{-3} \, \text{mL} = 0.430 \, \text{mL} - Calculate the mass (g):
Using the formula: Mass (g)=Density (g/mL)×Volume (mL)\text{Mass (g)} = \text{Density (g/mL)} \times \text{Volume (mL)} Mass=0.806 g/mL×0.430 mL=0.34658 g\text{Mass} = 0.806 \, \text{g/mL} \times 0.430 \, \text{mL} = 0.34658 \, \text{g} - Convert to mg:
0.34658 g×1000=346.58 mg0.34658 \, \text{g} \times 1000 = 346.58 \, \text{mg} Mass = 346.58 mg
2. Millimoles in 0.650 mL aliquot:
- Calculate the mass (g): Mass (g)=0.806 g/mL×0.650 mL=0.5239 g\text{Mass (g)} = 0.806 \, \text{g/mL} \times 0.650 \, \text{mL} = 0.5239 \, \text{g}
- Convert mass to moles:
Using the molar mass of 2-methyl-2-butanol (88.15 g/mol88.15 \, \text{g/mol}): Moles=Mass (g)Molar mass (g/mol)\text{Moles} = \frac{\text{Mass (g)}}{\text{Molar mass (g/mol)}} Moles=0.523988.15=0.005945 mol\text{Moles} = \frac{0.5239}{88.15} = 0.005945 \, \text{mol} - Convert to millimoles (mmol):
0.005945 mol×1000=5.945 mmol0.005945 \, \text{mol} \times 1000 = 5.945 \, \text{mmol} Millimoles = 5.945 mmol
Explanation:
- Mass Calculation: To calculate the mass from a given volume, multiply the volume by the substance’s density. Density links mass and volume, and converting units (µL to mL, g to mg) ensures consistency.
- Millimoles Calculation: To find millimoles, calculate the mass of the aliquot first using its density and volume. Then divide the mass by the molar mass of the compound to determine the moles, which can be converted to millimoles.
These calculations are crucial in chemistry for preparing solutions, performing stoichiometric calculations, and understanding the substance’s physical and chemical properties.