MHT-CET Chemistry · Solid State

Density and Crystal Structure Calculations

One relation, ρ = nM/(a³N_A), links density, molar mass, particles per cell and the cell volume; MHT-CET hands you a lumped product such as a³N_A or ρ·a³ so that the answer is one multiplication or division.

All Solid State notes MHT-CET Chemistry strategy

Why this matters

61 PYQs, 6 HARD — the largest subtopic in the chapter. Every row is the same formula solved for a different unknown: density, molar mass, cell volume, the number of particles per cell (which then names the structure), or the number of cells or atoms in a given mass or volume. Read what is given as a LUMP (a³N_A, ρN_A, ρa³) and the question collapses to arithmetic on two numbers.

Concept 1 of 5: The Density Formula and the Lumped Constants

Mass of the cell is n atoms × (M/N_A) grams; volume is a³. Their ratio is the density. The paper rarely gives a and M separately — it gives a³·N_A (cm³ mol⁻¹), or ρ·N_A, or ρ·a³ (the mass of one cell), so the arithmetic is two numbers and n.

Definition

$\rho = \dfrac{n\,M}{a^3 N_A}$ with $n = 1, 2, 4$ for sc, bcc, fcc and $a$ in cm.
Given $a^3 N_A$ : $\rho = \dfrac{nM}{a^3 N_A}$ directly. E.g. fcc, $M = 197$ , $a^3 N_A = 40$ : $\rho = \dfrac{4 \times 197}{40} = 19.7\ \text{g cm}^{-3}$ .
Given $a$ in Å or pm: convert to cm, cube, multiply by $6.022 \times 10^{23}$ . $a = 4\ \text{Å}$ : $a^3 N_A = 64 \times 10^{-24} \times 6.022 \times 10^{23} = 38.5$ .
$\rho \cdot a^3$ is the MASS of one unit cell; $\rho \cdot a^3 / n$ is the mass of one atom.

Density of a cubic crystal

\rho = \frac{n\,M}{a^3\,N_A}

Worked example

Sodium (

M = 23

) is bcc with

a = 4.29

Å. Find its density.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 10th May Shift 2 · Q87Easy

Example 1 · Solid State · Density and Crystal Structure Calculations

Calculate the density of a metal having molar mass

197\,\text{g}\,\text{mol}^{-1}

if it forms fcc structure.

a^3 \times N_A = 40\,\text{cm}^3\,\text{mol}^{-1}

Using n = 4 for bcc

The wrong n doubles or halves the answer, and that wrong answer is always among the options. bcc is 2; fcc is 4; simple cubic is 1.

Drill 6 more on the density formula and the lumped constants

Concept 2 of 5: Molar Mass From Density

Rearrange:

M = \rho\,a^3 N_A / n

. With the lump

a^3 N_A

given, it is density × lump ÷ n. If the mass of the cell is given instead, divide by n for the mass of one atom and multiply by

N_A

Definition

$M = \dfrac{\rho \cdot a^3 N_A}{n}$ . bcc, $\rho = 5.6$ , $a^3 N_A = 75$ : $M = \dfrac{5.6 \times 75}{2} = 210\ \text{g mol}^{-1}$ .
From the cell mass: $M = \dfrac{\rho a^3}{n} \times N_A$ . fcc cell of mass $1.8 \times 10^{-22}$ g: one atom $= 4.5 \times 10^{-23}$ g, $M = 27$ .
From $a$ and $\rho$ : bcc, $\rho = 10$ , $a = 4 \times 10^{-8}$ cm: $M = \dfrac{10 \times 6.4 \times 10^{-23} \times 6.022 \times 10^{23}}{2} = 193$ .

Molar mass

M = \frac{\rho\,a^3 N_A}{n}

Worked example

An fcc metal has density

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

and

a^3 N_A = 28.4\ \text{cm}^3\ \text{mol}^{-1}

. Find its molar mass.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 11th May Shift 2 · Q97Moderate

Example 2 · Solid State · Density and Crystal Structure Calculations

A metal crystallises in bcc structure with edge length

4\times10^{-8}

cm. If density of unit cell is

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

. What is its molar mass?

Dividing by N_A when the lump already contains it

a^3 N_A

in cm³ mol⁻¹ has Avogadro's number built in. Multiply

\rho

by it and divide by n — nothing else. A second

N_A

sends the exponent off by 23.

Drill 8 more on molar mass from density

Concept 3 of 5: Unit Cell Volume From Density and Molar Mass

a^3 = nM/(\rho N_A)

. The paper gives

\rho N_A

as one lump (of order

10^{24}

), so the cell volume is n × M ÷ lump — a number of order

10^{-23}

cm³.

Definition

$a^3 = \dfrac{n\,M}{\rho\,N_A}$ . fcc, $M = 27$ , $\rho N_A = 16.0 \times 10^{23}$ : $a^3 = \dfrac{4 \times 27}{1.6 \times 10^{24}} = 6.75 \times 10^{-23}\ \text{cm}^3$ .
Without the lump: bcc Na, $M = 23$ , $\rho = 1$ : $a^3 = \dfrac{2 \times 23}{6.022 \times 10^{23}} = 7.6 \times 10^{-23}\ \text{cm}^3$ .
Sanity check: a cubic cell volume is $10^{-23}$ to $10^{-22}$ cm³ (edges 200–500 pm). An answer of $10^{-21}$ means a lost exponent.

Cell volume

a^3 = \frac{n\,M}{\rho\,N_A}

Worked example

An element of molar mass 52 forms bcc cells with

\rho N_A = 4.3 \times 10^{24}\ \text{g cm}^{-3}\ \text{mol}^{-1}

. Find the cell volume.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 9th May Shift 1 · Q54Easy

Example 3 · Solid State · Density and Crystal Structure Calculations

Calculate the volume of unit cell when metal having density

1\,\text{g\,cm}^{-3}

and molar mass

23\,\text{g\,mol}^{-1}

crystallises to form bcc structure.

Trusting a printed exponent over the order of magnitude

One 2024 paper printed

d N_A = 120 \times 10^{21}

; the working gives

6 \times 10^{-21}

, the key says

6.00 \times 10^{-23}

. Match the mantissa to the options and let the sanity range (

10^{-23}

) settle the exponent.

Drill 5 more on unit cell volume from density and molar mass

Concept 4 of 5: Which Structure? Solve for n

Solve the density relation for the particles per cell:

n = \rho a^3 N_A / M

, or, when the cell mass and atom mass are given,

n = \rho a^3 / m_{\text{atom}}

. The answer rounds to 1, 2 or 4 and names the cell: simple, body-centred, face-centred.

Definition

$n = \dfrac{\rho \cdot a^3 N_A}{M}$ . $\rho = 8.6$ , $a^3 N_A = 21.5$ , $M = 92$ : $n = 2$ → bcc.
$n = \dfrac{\text{mass of cell}}{\text{mass of one atom}} = \dfrac{\rho a^3}{m}$ . $1.792 \times 10^{-22} / 4.4 \times 10^{-23} = 4.07$ → fcc.
$n = 1$ : simple cubic. $n = 2$ : bcc. $n = 4$ : fcc/ccp. Round — the data are rarely exact.

Particles per cell

n = \frac{\rho\,a^3 N_A}{M} = \frac{\rho\,a^3}{m_{\text{atom}}}

Worked example

An element (

M = 60

) has density

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

and

a = 400

pm. Identify the cell.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 15th May Shift 2 · Q77Moderate

Example 4 · Solid State · Density and Crystal Structure Calculations

Unit cell of an element has edge length of

5

Å with density

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

, if its atomic mass is 149, identify the crystal structure.

Stopping at n and not naming the cell

Half these questions ask for the STRUCTURE, not the number. n = 2 is 'body-centred cubic'; n = 4 is 'face-centred cubic'. hcp is never the answer of a cubic-cell calculation.

Drill 5 more on which structure? solve for n

Concept 5 of 5: Counting Unit Cells and Atoms in a Mass or a Volume

Divide the sample by one cell. In a mass: cells = mass ÷ (ρ·a³), the mass of one cell; atoms = n × cells. In a volume: cells = V ÷ a³. One mole of a simple-cubic metal is

N_A

cells because each cell holds one atom.

Definition

Cells in mass $w$ : $\dfrac{w}{\rho a^3}$ . $0.9$ g, $\rho a^3 = 3 \times 10^{-22}$ : $3 \times 10^{21}$ cells.
Atoms in mass $w$ : $n \times \dfrac{w}{\rho a^3}$ . bcc, $0.3$ g, $\rho a^3 = 3 \times 10^{-22}$ : $2 \times 10^{21}$ atoms.
Cells from moles: $\dfrac{w}{M} \times \dfrac{N_A}{n}$ . $0.60$ g, $M = 60$ , fcc: $0.01 \times 6.022 \times 10^{23} / 4 = 1.5 \times 10^{21}$ .
Cells in volume $V$ : $\dfrac{V}{a^3}$ . $1\ \text{cm}^3$ , $a = 2 \times 10^{-8}$ cm: $1.25 \times 10^{23}$ .
Simple cubic, 1 mol: $6.022 \times 10^{23}$ cells (one atom per cell).

Counting cells

N_{\text{cells}} = \frac{w}{\rho\,a^3} = \frac{V}{a^3} = \frac{w}{M}\cdot\frac{N_A}{n},\qquad N_{\text{atoms}} = n\,N_{\text{cells}}

Worked example

How many atoms are in 2.0 g of an fcc metal whose unit cell weighs

2.5 \times 10^{-22}

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2022 · Shift 1 · Q92Hard

Example 5 · Solid State · Density and Crystal Structure Calculations

Calculate the number of atoms in 5-gram metal that crystallises to form simple cubic unit cell structure having edge length 336 pm. (Density of metal = 9.4 g cm

^{-3}

)

Counting cells when atoms are asked

'Number of atoms in 0.3 g of a bcc metal' is cells × 2. The cell count

10^{21}

is offered as option (A) for the student who stops one step early.

Drill 13 more on counting unit cells and atoms in a mass or a volume

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (5)

The Density Formula and the Lumped Constants
Density of a cubic crystal
$\rho = \frac{n\,M}{a^3\,N_A}$
Molar Mass From Density
Molar mass
$M = \frac{\rho\,a^3 N_A}{n}$
Unit Cell Volume From Density and Molar Mass
Cell volume
$a^3 = \frac{n\,M}{\rho\,N_A}$
Which Structure? Solve for n
Particles per cell
$n = \frac{\rho\,a^3 N_A}{M} = \frac{\rho\,a^3}{m_{\text{atom}}}$
Counting Unit Cells and Atoms in a Mass or a Volume
Counting cells
$N_{\text{cells}} = \frac{w}{\rho\,a^3} = \frac{V}{a^3} = \frac{w}{M}\cdot\frac{N_A}{n},\qquad N_{\text{atoms}} = n\,N_{\text{cells}}$

Watch out for (5)

Using n = 4 for bcc
→ The Density Formula and the Lumped Constants
Dividing by N_A when the lump already contains it
→ Molar Mass From Density
Trusting a printed exponent over the order of magnitude
→ Unit Cell Volume From Density and Molar Mass
Stopping at n and not naming the cell
→ Which Structure? Solve for n
Counting cells when atoms are asked
→ Counting Unit Cells and Atoms in a Mass or a Volume

Test yourself on Solid State

20 past MHT-CET questions from this chapter, timed at 18 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Solid State chapter test Drill all 61 questions

Related notes

Unit Cells — n and the edge that feed this formula →