Question bank — Density, molar mass, molar volume

True — Avogadro's law: equal volumes hold equal numbers of molecules because gas molecules are far apart and their own size is negligible, so $V_m=RT/P$ depends only on $T,P$, not identity.

False — same molar volume, but density $=M/V_m$, so a heavier molecule (larger $M$) is denser. $O_2$ is denser than $C{H}_4$ though both occupy 22.7 L mol⁻¹.

False — 22.4/22.7 L mol⁻¹ is ONLY for ideal gases at STP. Liquid water's molar volume is $\approx 18{\text{ mL mol}}^{-1}$, about a thousand times smaller, because liquid particles touch.

False — true only for gases at fixed $T,P$. For solids/liquids, packing (number density) also matters: osmium beats lead despite similar atomic masses because its atoms pack tighter.

False — it is not coincidence, it is by design. The mole was defined so that $N_A\times (1\text{ u})\approx 1\text{ g}$; the Avogadro number is exactly the u→gram scaling factor.

True — $V_m=RT/P$, so $P\uparrow \times 2\Rightarrow V_m\downarrow \times 2$; and $\rho =PM/RT$ so $\rho$ doubles. The same moles get squeezed into half the room.

False — density is intensive: $\rho =m/V$; doubling the sample doubles both $m$ and $V$, so the ratio is unchanged.

False — $\rho =PM/RT$; heating (constant $P$) expands the gas, lowering density. Hot air rises for exactly this reason.

True — at fixed $T,P$, $\rho =PM/RT$ depends only on $M$; e.g. $N_2$ and $CO$ (both $M=28$) share the same STP density.

The number is fine because $1\text{ mL}=1{\text{ cm}}^3$, so $250\text{ mL}=250{\text{ cm}}^3$ matches ${\text{g cm}}^{-3}$. The error is mixing mL/L mid-problem — always make density units match the volume units before multiplying.

Only if mass is held fixed. If you take more of the same substance, $m$ and $V$ grow together and $\rho$ stays constant — density is a property of the material, not the amount.

$V_m$ is volume per mole. Two moles occupy $2\times 22.7=45.4$ L. $V_m$ is a ratio, not a total.

(1) Units: $44\text{ g}/22.4\text{ L}$ gives ${\text{g L}}^{-1}$, not ${\text{g cm}}^{-3}$. (2) At the modern STP (1 bar) use $22.7$: $44/22.7\approx 1.94{\text{ g L}}^{-1}$. A gas at $\sim 2{\text{ g cm}}^{-3}$ would be denser than aluminium — impossible.

It's inverted. $n=m/M$: grams divided by grams-per-mole so the grams cancel and mol remains. $M/m$ has units ${\text{mol g}}^{-1}\cdot {\text{g}}^{-1}$, nonsense.

Density $=$ (mass per atom) $\times$ (number density). Similar atomic mass is only the first factor; osmium packs atoms more tightly (higher number density), so it is denser.

Because we measure three different things — a balance reads grams, a flask reads litres, but reactions count particles (moles). Density links mass↔volume, molar mass links mass↔moles, molar volume links volume↔moles.

In a gas, molecules are so far apart their own size is irrelevant; volume is set by how much room they bounce in, which depends only on $T$ and $P$ (Avogadro's law), not molecule mass.

Same mass per particle, but far fewer particles per unit volume — the number density collapses because gas molecules are spread far apart rather than touching (see the packing figure above).

One mole is $M$ grams and occupies $V_m$ litres by definition, so grams-per-litre is just $M/V_m$. The mole absorbs the particle-counting for us.

Because density is mass-over-volume, we need $m$ in the equation — but $PV=nRT$ only has $n$. The bridge $n=m/M$ (moles = grams ÷ grams-per-mole) lets us swap $n$ for $m$: substitute to get $PV=(m/M)RT$, rearrange to isolate the ratio $m/V=PM/RT$, and that ratio $m/V$ is density $\rho$. We chose $n=m/M$ precisely because it is the only bridge carrying mass into the gas law.

At constant pressure, heating raises $T$, and $\rho =PM/RT$ shows density falls as $T$ rises; less-dense hot air floats up through denser cool air.

The mole and the atomic mass unit were defined together so that $N_A\times 1\text{ u}=1\text{ g}$; the g mol⁻¹ value inherits the u value exactly.

Zero — no mass in the volume, so $\rho =0/V=0$. The formula holds; it just returns zero, the degenerate case of "no stuff crammed in".

Density approaches zero — the gas becomes arbitrarily thin. This is also where the ideal-gas assumption is most accurate, since molecules almost never interact.

It shrinks toward zero, but real gases liquefy before reaching absolute zero, so the ideal formula stops applying — a limiting behaviour the ideal model doesn't survive.

No — at 1 atm (slightly higher pressure than 1 bar) and 273.15 K, $V_m=22.4{\text{ L mol}}^{-1}$. The number depends on which STP definition you use.

No — real gases near condensation deviate strongly; attractions and finite molecular size make $V_m$ smaller than the ideal prediction.

Yes — $V_m=V/n=M/\rho$ works for any phase; liquid water is $\approx 18{\text{ mL mol}}^{-1}$. Only the ideal-gas value 22.7 L is phase-specific.

It stays the same — $V_m=RT/P$, and doubling both leaves the ratio $T/P$ unchanged, so $V_m$ is unchanged.