An element forms a body-centered cubic crystalline substance.The edge length of the unit cell is 287 pm and the density of the crystal is 7.92 g/cm³.Calculate the atomic weight of the substance. A)45.0 amu B)48.0 amu C)56.4 amu D)63.5 amu

Question

Quizplus · Accepted Answer

The atomic weight of the substance can be calculated using the formula for the density of a crystal, $\rho = \frac{Z \times M}{a^3 \times N_A}$, where $\rho$ is the density, $Z$ is the number of atoms per unit cell (which is 2 for a body-centered cubic structure), $M$ is the molar mass (atomic weight in g/mol we are looking for), $a$ is the edge length of the unit cell in cm (287 pm = 287 x 10^-10 cm), and $N_A$ is Avogadro's number ($6.022 \times 10^{23}$ mol^-1). Rearranging the formula to solve for $M$, and substituting the given values ($\rho = 7.92$ g/cm^3, $a = 287 \times 10^{-10}$ cm), we can calculate the atomic weight $M$. The calculation yields a value close to 56.4 amu.

An Element Forms a Body-Centered Cubic Crystalline Substance