Animated Solution for Chemistry - States of Matter: An element with molar mass $2.7\times 10^{-2}{\text{ kg mol}}^{-1}$ forms a cubic unit cell with edge length $405\text{ pm}$. If its density is $2.7\times 10^3{\text{ kg m}}^{-3}$, the radius of the element is approximately ............ $\times 10^{-12}\text{ m}$ (to the nearest integer).

Imagine you are handed a microscopic building block of a crystal, but you have no idea how the atoms are arranged inside it. Is it a simple cube? Is it body-centered? Or is it face-centered? This is the exact puzzle we are facing. We are given the density, the edge length, and the molar mass of an element, and our ultimate mission is to find the radius of a single atom.

To embark on this mission, we need to unlock the identity of the lattice. The key to this identity is $Z$, the number of atoms per unit cell.

We can find $Z$ using the master density formula:

Before we plug in the numbers, we must be extremely careful with our units. Mixing meters, centimeters, and picometers is a recipe for disaster. Let's convert everything into the standard CGS system. The molar mass $M$ becomes $27{\text{ g mol}}^{-1}$. The edge length $a$ is $405\text{ pm}$, which translates to $405\times 10^{-10}\text{ cm}$. Finally, the density $d$ is $2.7{\text{ g cm}}^{-3}$.

Now, we rearrange our formula to solve for $Z$:

Substituting our carefully converted values into this equation, we get:

After crunching the numbers, $Z$ comes out to be approximately $4$. This is a massive breakthrough! A $Z$ value of $4$ definitively tells us that our mystery crystal is a Face-Centered Cubic (FCC) lattice.

Now that we know we are dealing with an FCC lattice, we can visualize its geometry. Picture the face of this cubic unit cell. In an FCC structure, the atoms touch each other along the face diagonal.

If we draw a line across this diagonal, its length is $\sqrt{2}a$ according to the Pythagorean theorem. This diagonal cuts through one full atom in the center (contributing $2r$) and two half-atoms at the corners (contributing $r$ each). Therefore, the total length of the diagonal is $4r$. This gives us our crucial geometric relationship:

We are now in the home stretch. We need to isolate the radius $r$:

We can now substitute our original edge length, $a=405\text{ pm}$, back into the equation:

Calculating this yields a radius of approximately $143.16\text{ pm}$. The question asks for the answer in the format of $...\times 10^{-12}\text{ m}$. Since $10^{-12}\text{ m}$ is exactly one picometer, our value is already in the correct format.

Rounding to the nearest integer, we arrive at our final answer: $143$.