# Finding the volume of a cube based off the mass and density of a sphere contained inside it.

This is an extension to my last question, but since people didn't like that the problem involved atoms, I will change them to small spheres of cheese.

The mass of a sphere of cheese is $1.37\cdot 10^{-25}$ kg, and the density of the cheese is $8920 \mbox{kg/m}^3$.

Visualize a one cubic centimeter as formed by stacking up identical cubes, with one sphere of cheese at the center of each of these cubes. Determine the volume of each cube.

This seemed simple enough to me, I just solve for the volume of a sphere of cheese, then, since a cube holds exactly one sphere, the length of the sides of the cube would be equal to the diameter of the sphere.

But, when I calculate all of these things, the answer is coming up wrong on the website where I submit my homework (WebAssign).

Here are my calculations (please tell me if I'm doing something wrong!):

$$\rho = \frac{m}{V}$$ so $$V = \frac{m}{\rho}$$

$$V = \frac{1.37 \cdot 10^{-25} kg}{8920 kg/m^3}$$

$$V = 1.536 \cdot 10^{-29} m^3$$

The volume of a sphere is $V_{sphere} = \frac{4 \pi r^3}{3}$, so

$$1.536 \cdot 10^{29}m^3 = \frac{4 \pi r^3}{3}$$

$$3.667 \cdot 10^{-30}m^3 = r^3$$

$$\sqrt[3]{3.667 \cdot 10^{-30}m^3} = \sqrt[3]{r^3}$$

$$r = 1.542 \cdot 10^{-10}m$$

The diameter of a sphere is $d = 2r$, so

$$d = 1.542 \cdot 10^{-10}m \cdot 2$$

$$d = 3.084 \cdot 10^{-10}m$$

The volume of a cube is $V_{cube} = l^3$ and since a cube holds exactly one sphere, $l = d$, so

$$V = d^3$$ $$V = (3.084 \cdot 10^{-10}m)^3$$

$$V = 2.933 \cdot 10^{-29}m^3$$

$$V = 2.933 \cdot 10^{-23}cm^3$$

Where did I go wrong?

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