Question

calculate the number of carbon atoms in 0.75 l of acetone (ch₃coch₃). (density = 1.3 g/ml and boiling point = 48.5 degrees celsius) 1. 3.0×10²² atoms 2. 1.0×10²⁵ atoms 3. 1.8×10²² atoms 4. 3.0×10²⁵ atoms 5. 1.0×10²² atoms 6. 1.8×10²⁵ atoms

Explanation:

Step1: Convert volume to mass

First, convert the volume of acetone from liters to milliliters: \( 0.75 \, \text{L} = 0.75 \times 1000 = 750 \, \text{mL} \).
Using the density formula \( \text{mass} = \text{density} \times \text{volume} \), the mass of acetone is \( 1.3 \, \text{g/mL} \times 750 \, \text{mL} = 975 \, \text{g} \).

Step2: Calculate molar mass of acetone

The formula for acetone is \( \text{CH}_3\text{COCH}_3 \) (or \( \text{C}_3\text{H}_6\text{O} \)).
Molar mass of \( \text{C} = 12.01 \, \text{g/mol} \), \( \text{H} = 1.008 \, \text{g/mol} \), \( \text{O} = 16.00 \, \text{g/mol} \).
Molar mass of acetone: \( 3(12.01) + 6(1.008) + 16.00 = 36.03 + 6.048 + 16.00 = 58.078 \, \text{g/mol} \).

Step3: Find moles of acetone

Moles of acetone: \( \frac{\text{mass}}{\text{molar mass}} = \frac{975 \, \text{g}}{58.078 \, \text{g/mol}} \approx 16.79 \, \text{mol} \).

Step4: Moles of carbon atoms

Each mole of acetone (\( \text{C}_3\text{H}_6\text{O} \)) contains 3 moles of C.
Moles of C: \( 16.79 \, \text{mol} \times 3 = 50.37 \, \text{mol} \).

Step5: Number of carbon atoms

Using Avogadro’s number (\( 6.022 \times 10^{23} \, \text{atoms/mol} \)):
Number of C atoms: \( 50.37 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} \approx 3.03 \times 10^{25} \, \text{atoms} \), which rounds to \( 3.0 \times 10^{25} \, \text{atoms} \) (matching option 4).

Answer:

1. \( 3.0 \times 10^{25} \) atoms