Chapter 3: Problem 48
The density of water is \(1.00 \mathrm{~g} / \mathrm{mL}\) at \(4^{\circ} \mathrm{C}\). How many water molecules are present in \(15.78 \mathrm{~mL}\) of water at this temperature?
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Mass of Water
- Mass \(= \text{Density} \times \text{Volume} = 1.00 \text{ g/mL} \times 15.78 \text{ mL} = 15.78 \text{ g}\).
Moles of Water
- Moles \(= \frac{\text{Mass}}{\text{Molar Mass}}\).
- Moles of Water = \(\frac{15.78 \text{ g}}{18.02 \text{ g/mol}}\approx 0.8759 \text{ moles}\).
Avogadro's Number
- Molecules = \(0.8759 \text{ moles} \times 6.022 \times 10^{23} \text{ molecules/mole} = 5.277 \times 10^{23} \text{ molecules}\).