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Some Basic Concepts in Chemistry: Chemistry | JEE Main

Iron oxide

\text{FeO}

, crystallises in a cubic lattice with a unit cell edge length of

5.0\text{Å}

. If density of the

\text{FeO}

in the crystal is

4.0\text{ g cm}^{-3}

, then the number of

\text{FeO}

units present per unit cell is

\underline{\quad\quad\quad}

(Nearest integer) Given: Molar mass of

\text{Fe}

and

\text{O}

56

and

16\text{ g mol}^{-1}

respectively.

\text{N}_A = 6.0 \times 10^{23}\text{ mol}^{-1}

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Step-by-step solutionView

Correct answer

The number of FeO formula units per unit cell is 4.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

Edge length $a = 5.0\text{ Å} = 5.0 \times 10^{-8}\text{ cm}$ , density $d = 4.0\text{ g cm}^{-3}$ , molar mass of $\text{FeO} = 56 + 16 = 72\text{ g mol}^{-1}$ , $N_A = 6.0 \times 10^{23}\text{ mol}^{-1}$ .

02find

Number of $\text{FeO}$ formula units present per unit cell ( $z$ ).

03visualise

A cubic unit cell of volume $V = a^3 = (5.0 \times 10^{-8}\text{ cm})^3 = 1.25 \times 10^{-22}\text{ cm}^3$ containing $z$ formula units of mass $z \times M / N_A$ .

04strategise

Apply the cubic crystal density formula: $d = \frac{z \times M}{N_A \times a^3} \implies z = \frac{d \times N_A \times a^3}{M}$ .

05execute

Substitute the values: $z = \frac{4.0 \times (6.0 \times 10^{23}) \times (5.0 \times 10^{-8})^3}{72} = \frac{4.0 \times 6.0 \times 10^{23} \times 125 \times 10^{-24}}{72} = \frac{24 \times 12.5}{72} = \frac{300}{72} \approx 4.17 \approx 4$ .

✓verify

For a rock-salt-like structure (NaCl-type) typical of FeO, $z = 4$ is the standard theoretical number of formula units per unit cell. The experimental density gives $z \approx 4.17$ , which rounds cleanly to 4.

Hints that build this answer step by step

What formula relates the crystal density ( $\rho$ ), number of formula units per unit cell ( $Z$ ), molar mass ( $M$ ), unit cell edge length ( $a$ ), and Avogadro's number ( $N_A$ )?
$\rho = \frac{Z \times M}{a^3 \times N_A}$
What are the molar mass $M$ and unit cell volume $V = a^3$ in appropriate CGS units?
$M = 72\text{ g mol}^{-1}$ , $a^3 = 1.25 \times 10^{-22}\text{ cm}^3$
Solving for $Z = \frac{\rho \times a^3 \times N_A}{M}$ , what is the value of $Z$ to the nearest integer?
$4$

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Chemistry
Concepts assessed: Chemistry
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why is

z

not exactly an integer like 4.00?

Real FeO crystals often exhibit non-stoichiometry (metal deficiency defects like $\text{Fe}_{0.95}\text{O}$ ), or experimental rounding in the problem's provided density leads to an approximate value (~4.17). The question specifically asks for the nearest integer, which is 4.

Answer

The number of FeO formula units per unit cell is 4.

Why each option works or fails

Step-by-step solution

given: Edge length $a = 5.0\text{ Å} = 5.0 \times 10^{-8}\text{ cm}$ , density $d = 4.0\text{ g cm}^{-3}$ , molar mass of $\text{FeO} = 56 + 16 = 72\text{ g mol}^{-1}$ , $N_A = 6.0 \times 10^{23}\text{ mol}^{-1}$ .
find: Number of $\text{FeO}$ formula units present per unit cell ( $z$ ).
visualise: A cubic unit cell of volume $V = a^3 = (5.0 \times 10^{-8}\text{ cm})^3 = 1.25 \times 10^{-22}\text{ cm}^3$ containing $z$ formula units of mass $z \times M / N_A$ .
strategise: Apply the cubic crystal density formula: $d = \frac{z \times M}{N_A \times a^3} \implies z = \frac{d \times N_A \times a^3}{M}$ .
execute: Substitute the values: $z = \frac{4.0 \times (6.0 \times 10^{23}) \times (5.0 \times 10^{-8})^3}{72} = \frac{4.0 \times 6.0 \times 10^{23} \times 125 \times 10^{-24}}{72} = \frac{24 \times 12.5}{72} = \frac{300}{72} \approx 4.17 \approx 4$ .
verify: For a rock-salt-like structure (NaCl-type) typical of FeO, $z = 4$ is the standard theoretical number of formula units per unit cell. The experimental density gives $z \approx 4.17$ , which rounds cleanly to 4.

Some Basic Concepts in Chemistry: Chemistry | JEE Main

What feels right?

The number of FeO formula units per unit cell is 4.

Why each option works or fails

Solution

01given

02find

03visualise

04strategise

05execute

✓verify

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