Avogadro's law and Avogadro's number N_A = 6.022 × 10²³: Question bank

Prerequisites worth a peek if a line stumps you: The mole concept, Ideal gas law PV=nRT, Molar mass and atomic mass unit, Boyle's law, Charles's law.

Picture the whole bank in one glance

Before the traps, look at the two images every question secretly tests.

Figure 1 — why the volume only "sees" the count. Three boxes at the same $T, P$ : helium (tiny light atoms), oxygen (medium), and $CO_{2}$ (big heavy). Each box holds the same number of molecules, and each box is the same size. The molecules differ wildly in mass, yet the boxes are identical in volume — because in $P V = n R T$ the only thing on the right that changes with "which gas" is… nothing. Look at how the amber box outlines all line up.

Figure — Avogadro's law and Avogadro's number N_A = 6.022 × 10²³

Figure 2 — the mass-vs-count fork. Take the same 8 grams of two gases. Follow the two arrows: grams → moles (divide by molar mass $M$ ) → molecule count. Because He is light and $O_{2}$ is heavy, equal grams split into very different mole counts, so very different volumes. This is the fork you must walk before ever using Avogadro's law.

True or false — justify

Equal volumes of

H_{2}

and

O_{2}

at the same

T

and

P

contain the same number of molecules.

True — that is exactly Avogadro's law; volume at fixed

T, P

depends only on molecule count, so equal volumes ⇒ equal moles ⇒ equal molecule number, regardless of which gas (see Figure 1).

Equal masses of

H_{2}

and

O_{2}

at the same

T, P

occupy the same volume.

False — mass is not "amount." 1 g of

H_{2}

0.5

mol but 1 g of

O_{2}

is only

\approx 0.031

mol, so the

H_{2}

occupies about 16× the volume. Walk the grams → moles fork of Figure 2 first.

One mole of

O_{2}

contains

N_{A}

atoms.

False — it contains

N_{A}

molecules, hence

2 N_{A}

atoms.

N_{A}

counts whatever entity you specify; here you specified molecules.

N_{A}

is a pure counting number with no unit.

Half-true — the number

6.022 \times 1 0^{23}

is dimensionless, but Avogadro's constant carries the unit

mol^{- 1}

(particles per mole). In formulas like

N = n N_{A}

the

mol^{- 1}

is what cancels the mole in

n

At fixed

T

and

P

, doubling the moles of gas doubles the volume.

True —

V / n = R T / P

is constant when

T, P

are fixed, so

V

is directly proportional to

n

; twice the moles ⇒ twice the volume.

One mole of any gas occupies exactly

22.4

L under all conditions.

False —

22.4

L is only the molar volume at the old STP (

1

atm,

273

K); modern IUPAC STP (

1

bar,

273

K) gives

22.7

L, and away from STP you must use

P V = n R T

A mole of electrons and a mole of bricks both contain

6.022 \times 1 0^{23}

objects.

True — the mole is just a count, like "a dozen"; it says nothing about size or mass, only how many. (A mole of bricks is absurdly heavy, but the count is identical.)

The molar mass of a substance in g/mol is numerically equal to its molecular mass in u.

True by design —

N_{A}

was fixed so 1 mol of C-12 weighs exactly 12 g, which forces g/mol and u to line up numerically for every substance.

Avogadro's law holds for a mixture of two different gases in the same container.

True — the total molecule count sets the volume at fixed

T, P

; the ideal gas law treats all molecules alike, so identities and proportions don't matter, only the total

n

Heating a gas at fixed pressure keeps

V / n

constant.

False —

V / n = R T / P

depends on

T

. Raising

T

raises

V / n

(this is Charles's law territory). Avogadro's constant-ratio result needs both

T

and

P

fixed.

Spot the error

"8 g of

O_{2}

has the same volume as 8 g of

He

at the same

T, P

because Avogadro's law says equal amounts give equal volumes."

The error is calling equal masses "equal amounts." Amount means moles:

8/32 = 0.25

mol

O_{2}

8/4 = 2

mol He, so the He occupies 8× the volume — exactly the fork drawn in Figure 2.

"1 mol of

CO_{2}

has

N_{A}

atoms, so

0.5

mol has

0.5 N_{A}

atoms."

Two errors chained: 1 mol

CO_{2}

has

N_{A}

molecules =

3 N_{A}

atoms (3 atoms each), so

0.5

mol has

1.5 N_{A}

atoms, not

0.5 N_{A}

"Since

P V = n R T

, and He is lighter than

O_{2}

, a mole of He takes less volume."

Nothing in

P V = n R T

mentions molecular mass; at the same

T, P

one mole of any ideal gas takes the same volume (the equal boxes of Figure 1). Lightness affects density and mass, not molar volume.

"1 u equals

1/ N_{A}

grams, so it depends on which atom we pick."

The atomic mass unit is fixed:

1 u = \frac{1}{12}

of a C-12 atom's mass

= 1/ N_{A}

\approx 1.66 \times 1 0^{- 24}

g. It is a universal constant, not something that changes per element.

"Molar volume is 22.7 L, so 44.8 L of gas is always 2 mol."

Only true at STP. At other

T

P

the molar volume differs, so you cannot read moles off volume without knowing conditions — use

n = P V / R T

"A container of

H_{2}

and one of

O_{2}

at equal volume,

T

P

must weigh the same because they have equal molecules."

Equal molecule count does not mean equal mass. Each

O_{2}

molecule is 16× heavier than each

H_{2}

, so the oxygen sample weighs 16× more despite identical counts.

Why questions

Why does gas volume at fixed

T, P

ignore the identity of the gas?

Ideal gas molecules are treated as point particles with no size and no mutual attraction, so only their number (through

P V = n R T

) sets the volume — chemical identity never enters the equation (Figure 1).

Why is

N_{A}

exactly

6.02214076 \times 1 0^{23}

and not a round number?

It was historically pinned so 1 mol of C-12 weighs precisely 12 g; since 2019 it is a defined exact value, chosen to keep that gram↔u correspondence intact.

Why must you convert to moles before applying Avogadro's law to two gas samples?

Because the law is

V \propto n

(moles), not

V \propto m

(mass). Different gases have different molar masses, so equal masses correspond to different mole counts — only moles scale linearly with volume (Figure 2).

Why does

N_{A}

act as an "exchange rate" between grams and atoms?

Atoms are far too small to count;

N_{A}

converts a weighable macroscopic amount (grams → moles via

m / M

) into an invisible microscopic count (moles → particles via

\times N_{A}

Why do both

T

and

P

have to be held constant for "

V / n

= constant"?

Because

V / n = R T / P

: if either

T

P

changes, the right side changes, so the ratio is no longer constant. Only fixing both freezes it.

Why does saying "

N_{A}

particles" invite mistakes?

"Particle" is ambiguous — it could mean atoms, molecules, ions, or electrons. Since a molecule can hold several atoms, you must state "moles of what" or your atom counts go wrong.

Edge cases

Does Avogadro's law say anything about

0

moles of gas?

Yes, consistently:

n = 0

gives

V = 0

from

V = n R T / P

. No molecules ⇒ no volume, which is the degenerate limit of

V \propto n

What happens to Avogadro's law for a real (non-ideal) gas at high pressure?

It breaks down — molecular size and attractions matter, so equal volumes no longer hold equal molecule counts exactly. The law is an ideal-gas idealisation; deviations grow as pressure rises or temperature falls.

0.25

mol a valid amount of substance even though you can't have a "quarter of a molecule"?

Yes — moles are macroscopic bulk amounts.

0.25

mol still contains

0.25 N_{A} \approx 1.5 \times 1 0^{23}

whole molecules, a perfectly integer count of real particles.

If two gases are at the same

T

and

V

but different

P

, are their molecule counts equal?

No — Avogadro's law needs equal

P

too. From

n = P V / R T

, at fixed

V, T

the higher-pressure gas has proportionally more molecules.

Does one mole of a monatomic gas (like He) contain fewer particles than one mole of a diatomic gas (like

N_{2}

Both contain

N_{A}

molecules/atoms as named. But per molecule He has 1 atom and

N_{2}

has 2, so

N_{2}

contains twice as many atoms while having the same number of molecular units.

Can two samples with equal numbers of molecules ever have different volumes at the same

T, P

For ideal gases, no — equal molecule count at equal

T, P

forces equal volume. Only non-ideal behaviour (finite molecular size, attractions) could break this, which is outside the ideal-gas model.

At the same

T, P, V

, do a helium sample and a

CO_{2}

sample have equal moles?

Yes — equal

T, P, V

forces equal

n

via

n = P V / R T

, so equal moles and equal molecule counts, even though the

CO_{2}

sample weighs much more.

Mass-based vs count-based reasoning — the master contrast

Trap phrasing	The mass-based (wrong) read	The count-based (right) read
"Equal masses ⇒ equal volume"	grams → volume	grams → moles → volume; light gas = more moles = more volume
"8 g $O_{2}$ = 8 g He in volume"	same grams, same size	0.25 mol vs 2 mol ⇒ He is 8× bigger
"Lighter gas takes less room"	mass sets volume	at fixed $T, P$ , one mole = one volume, mass irrelevant
" $N_{A}$ particles = $N_{A}$ atoms"	count the sample once	count the named entity; molecules ≠ atoms
"Equal molecules ⇒ equal mass"	count sets weight	equal count, unequal per-molecule mass ⇒ unequal weight

Connections

Parent: Avogadro's law and $N_A$
The mole concept — the mole is a pure count, the root of half these traps.
Ideal gas law PV=nRT — every gas trap resolves by returning to this equation.
Molar mass and atomic mass unit — why mass ≠ amount, and why g/mol = u.
Boyle's law and Charles's law — remind you why both $T$ and $P$ must be fixed.