If a cube has sides of 3 nm and a mass of 81 ng, what is its density in g/cm^3?

• A. 10 g/cm^3
• B. 0.01 g/cm^3
• C. 0.1 g/cm^3
• D. 1.0 g/cm^3

Answer: (D)

Verified step by step guidance

1. Convert the side length of the cube from nanometers (nm) to centimeters (cm). Since 1 nm = 1 \times 10^{-7} cm, multiply 3 nm by this conversion factor: \$3 \ \text{nm} \times 1 \times 10^{-7} \ \text{cm/nm}$.
2. Calculate the volume of the cube using the formula for the volume of a cube: \(V = s^3\), where \(s\) is the side length in centimeters. Use the converted side length from step 1.
3. Convert the mass from nanograms (ng) to grams (g). Since 1 ng = 1 \times 10^{-9} g, multiply 81 ng by this conversion factor: \$81 \ \text{ng} \times 1 \times 10^{-9} \ \text{g/ng}$.
4. Calculate the density using the formula \(\rho = \frac{m}{V}\), where \(m\) is the mass in grams and \(V\) is the volume in cubic centimeters.
5. Compare the calculated density value to the given options to identify the correct answer.