Chromium metal crystallizes as a body-centered cubic lattice.If the atomic radius of Cr is 1.25 angstroms,what is the density of Cr metal in g/cm³? A)5.52 B)7.18 C)14.4 D)2.76 E)3.59

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Quizplus · Accepted Answer

The density of a crystal in a body-centered cubic (bcc) lattice can be calculated using the formula: $$ \text{Density} = \frac{2 \times \text{Atomic Mass}}{\text{Volume of the unit cell}} \times N_A $$ where $N_A$ is Avogadro's number ($6.022 \times 10^{23} \text{ atoms/mol}$), and the volume of the unit cell for a bcc lattice is given by $a^3$, with $a = 4r/\sqrt{3}$ (where $r$ is the atomic radius). Given that the atomic radius of Cr is 1.25 Å (or $1.25 \times 10^{-8}$ cm), and the atomic mass of Cr is approximately 52 g/mol, we can calculate the density as follows:1. Calculate the edge length of the unit cell ($a$):$$ a = \frac{4r}{\sqrt{3}} = \frac{4 \times 1.25 \times 10^{-8} \text{ cm}}{\sqrt{3}} \approx 2.89 \times 10^{-8} \text{ cm} $$2. Calculate the volume of the unit cell ($V = a^3$):$$ V = (2.89 \times 10^{-8} \text{ cm})^3 \approx 2.41 \times 10^{-23} \text{ cm}^3 $$3. Calculate the density:$$ \text{Density} = \frac{2 \times 52 \text{ g/mol}}{2.41 \times 10^{-23} \text{ cm}^3} \times 6.022 \times 10^{23} \text{ mol}^{-1} \approx 7.18 \text{ g/cm}^3 $$Therefore, the correct answer is 7.18 g/cm^3, which corresponds to choice B.

Chromium Metal Crystallizes as a Body-Centered Cubic Lattice