One drop of water from a medicine dropper has a volume of approximately 0.050 mL (1 mL = 1 cm3) a.) Determine the number of water molecules in a drop of water. (The density of water is 1.0 g/cm3). b.) How many H atoms would be present in these molecules?
The Correct Answer and Explanation is:
To determine the number of water molecules and the number of hydrogen atoms in a drop of water, let’s break this down step by step.
a) Determining the number of water molecules in a drop of water:
- Volume of the drop: We are given that the volume of a drop of water is 0.050 mL. Since 1 mL = 1 cm³, the volume of the drop is also 0.050 cm³.
- Mass of the water in the drop: The density of water is given as 1.0 g/cm³. So, the mass of the water in the drop is: Mass=Density×Volume\text{Mass} = \text{Density} \times \text{Volume}Mass=Density×Volume Mass=1.0 g/cm3×0.050 cm3=0.050 g\text{Mass} = 1.0 \, \text{g/cm}^3 \times 0.050 \, \text{cm}^3 = 0.050 \, \text{g}Mass=1.0g/cm3×0.050cm3=0.050g
- Moles of water: The molar mass of water (H₂O) is approximately 18.015 g/mol. To find the number of moles in the drop, we use the formula: Moles of H₂O=MassMolar mass\text{Moles of H₂O} = \frac{\text{Mass}}{\text{Molar mass}}Moles of H₂O=Molar massMass Moles of H₂O=0.050 g18.015 g/mol≈0.00277 mol\text{Moles of H₂O} = \frac{0.050 \, \text{g}}{18.015 \, \text{g/mol}} \approx 0.00277 \, \text{mol}Moles of H₂O=18.015g/mol0.050g≈0.00277mol
- Number of water molecules: To find the number of water molecules, we use Avogadro’s number, which is 6.022×10236.022 \times 10^{23}6.022×1023 molecules per mole: Number of molecules=Moles of H₂O×Avogadro’s number\text{Number of molecules} = \text{Moles of H₂O} \times \text{Avogadro’s number}Number of molecules=Moles of H₂O×Avogadro’s number Number of molecules=0.00277 mol×6.022×1023 molecules/mol\text{Number of molecules} = 0.00277 \, \text{mol} \times 6.022 \times 10^{23} \, \text{molecules/mol}Number of molecules=0.00277mol×6.022×1023molecules/mol Number of molecules≈1.67×1021 molecules\text{Number of molecules} \approx 1.67 \times 10^{21} \, \text{molecules}Number of molecules≈1.67×1021molecules
b) Determining the number of H atoms:
- Each water molecule (H₂O) contains 2 hydrogen atoms.
- The total number of hydrogen atoms is: Number of H atoms=2×Number of water molecules\text{Number of H atoms} = 2 \times \text{Number of water molecules}Number of H atoms=2×Number of water molecules Number of H atoms=2×1.67×1021=3.34×1021 H atoms\text{Number of H atoms} = 2 \times 1.67 \times 10^{21} = 3.34 \times 10^{21} \, \text{H atoms}Number of H atoms=2×1.67×1021=3.34×1021H atoms
Final Answers:
- a) The number of water molecules in a drop of water is approximately 1.67×10211.67 \times 10^{21}1.67×1021.
- b) The number of hydrogen atoms present in these molecules is approximately 3.34×10213.34 \times 10^{21}3.34×1021.
This calculation assumes that the drop is pure water and that all water molecules in the drop are identical.
