iron forms a solid with a face-centered cubic unit cell. the iron atom has a radius of $1.26 \times 10^{-8}\text{ cm}$.
calculate the density of iron in $\text{g/cm}^3$. the molar mass of iron is $55.845\text{ g/mol}$.

$1.30 \times 10^{-15}\text{ g/cm}^3$
$8.20\text{ g/cm}^3$
$23.2\text{ g/cm}^3$
$2.05\text{ g/cm}^3$

Question

iron forms a solid with a face-centered cubic unit cell. the iron atom has a radius of $1.26 \times 10^{-8}\text{ cm}$.
calculate the density of iron in $\text{g/cm}^3$. the molar mass of iron is $55.845\text{ g/mol}$.

$1.30 \times 10^{-15}\text{ g/cm}^3$
$8.20\text{ g/cm}^3$
$23.2\text{ g/cm}^3$
$2.05\text{ g/cm}^3$

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": false, "concepts_used": [ "Cubic Unit Cells" ], "new_concepts": [ "Unit Cell Density", "Face-Centered Cubic" ], "current_concepts": [ "Cubic Unit Cells", "Unit Cell Density", "Face-Centered Cubic" ] } ## Determine atoms per unit cell Using the *Cubic Unit Cells* knowledge point $$ N = 4 \text{ atoms} $$ ## Calculate the unit cell edge length Using the *Cubic Unit Cells* knowledge point $$ \begin{aligned} a &= 2\sqrt{2}r \ a &= 2\sqrt{2} \times (1.26 \times 10^{-8}\text{ cm}) \approx 3.5638 \times 10^{-8}\text{ cm} \end{aligned} $$ ## Calculate the unit cell volume Using the *Cubic Unit Cells* knowledge point $$ \begin{aligned} V &= a^3 \ V &= (3.5638 \times 10^{-8}\text{ cm})^3 \approx 4.526 \times 10^{-23}\text{ cm}^3 \end{aligned} $$ ## Calculate the mass of one unit cell Using the *Cubic Unit Cells* knowledge point $$ \begin{aligned} m &= \frac{N \times M}{N_A} \ m &= \frac{4 \times 55.845\text{ g/mol}}{6.022 \times 10^{23}\text{ atoms/mol}} \approx 3.7094 \times 10^{-22}\text{ g} \end{aligned} $$ ## Calculate the density of iron Using the *Cubic Unit Cells* knowledge point $$ \begin{aligned} ho &= \frac{m}{V} \ ho &= \frac{3.7094 \times 10^{-22}\text{ g}}{4.526 \times 10^{-23}\text{ cm}^3} \approx 8.20\text{ g/cm}^3 \end{aligned} $$

(A) $1.30 \times 10^{-15}\text{ g/cm}^3$

(B) $8.20\text{ g/cm}^3$

(C) $23.2\text{ g/cm}^3$

(D) $2.05\text{ g/cm}^3$

{ "subject": "Natural Science", "question_type": "Multiple Choice", "knowledge_point": [ "Natural Science", "Chemistry", "Solid State Chemistry" ] }

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QUESTION IMAGE

Question

Explanation:

Determine atoms per unit cell

Calculate the unit cell edge length

Calculate the unit cell volume

Calculate the mass of one unit cell

Calculate the density of iron

Answer: