Questions: The density of water is 1.00 g/mL at 4°C. How many water molecules are present in 2.86 mL of water at this temperature? Be sure your answer has the correct number of significant digits. Note: Reference the Fundamental constants table for additional information. molecules

The density of water is 1.00 g/mL at 4°C. How many water molecules are present in 2.86 mL of water at this temperature?
Be sure your answer has the correct number of significant digits.
Note: Reference the Fundamental constants table for additional information.
molecules

Transcript text: Textbook Problems: Chapter 3 - Part 1 Question 6 of 15 (1 point) I Question Attempt: 1 of 10 Onnah $\checkmark 1$ $\checkmark 2$ $\checkmark 3$ $\checkmark 4$ 7 8 9 10 11 12 13 The density of water is $1.00 \frac{\mathrm{~g}}{\mathrm{~mL}}$ at $4^{\circ} \mathrm{C}$. How many water molecules are present in 2.86 mL of water at this temperature? Be sure your answer has the correct number of significant digits. Note: Reference the Fundamental constants table for additional information. $\square$ molecules $\square$ Try again

Solution

Solution Steps

Step 1: Determine the Mass of Water

The density of water is given as $1.00 \, \text{g/mL}$. To find the mass of 2.86 mL of water, we use the formula:

\[ \text{mass} = \text{density} \times \text{volume} \]

Substituting the given values:

\[ \text{mass} = 1.00 \, \text{g/mL} \times 2.86 \, \text{mL} = 2.86 \, \text{g} \]

Step 2: Convert Mass to Moles

The molar mass of water ($\text{H}_2\text{O}$) is approximately $18.015 \, \text{g/mol}$. To convert the mass of water to moles, use the formula:

\[ \text{moles of water} = \frac{\text{mass}}{\text{molar mass}} \]

Substituting the values:

\[ \text{moles of water} = \frac{2.86 \, \text{g}}{18.015 \, \text{g/mol}} \approx 0.1588 \, \text{mol} \]

Step 3: Convert Moles to Molecules

To find the number of molecules, use Avogadro's number, which is $6.022 \times 10^{23} \, \text{molecules/mol}$. Multiply the number of moles by Avogadro's number:

\[ \text{molecules} = 0.1588 \, \text{mol} \times 6.022 \times 10^{23} \, \text{molecules/mol} \]

\[ \text{molecules} \approx 9.558 \times 10^{22} \, \text{molecules} \]