Question bank — SI units in chemistry — kg, mol, K, Pa; derived units (J, L)

True — a base unit is defined independently (one of the seven: kg, mol, K, m, s, A, cd); derived units like J and Pa are constructed from those through a defining equation.

False — pressure is important but the pascal is derived: $\text{Pa}={\text{N/m}}^2$. Substitute $\text{N}=\text{kg}\text{m}{\text{s}}^{-2}$ and divide by ${\text{m}}^2$ (subtract 2 from the m-exponent): $\text{kg}{\text{m}}^{1-2}{\text{s}}^{-2}=\text{kg}{\text{m}}^{-1}{\text{s}}^{-2}$, all base units.

False — kinetic energy tracks absolute temperature. From 20 °C to 40 °C you go 293.15 K → 313.15 K, only about a 7% rise, not double; the Celsius zero is arbitrary so ratios in °C are meaningless.

True — a litre is defined as $1{\text{dm}}^3$, a cube 10 cm on each edge, which equals $10^{-3}{\text{m}}^3$.

True — the mole counts entities, always $N_A=6.022\times 10^{23}$ of them, regardless of what the entity is; only their mass differs.

False — the coherent SI form (base units only) is kg/mol, but chemists tabulate g/mol because atomic masses come out as tidy numbers (water = 18 g/mol). Any consistent mass unit works as long as $m$ and $M$ match.

False — they are identical: $1\text{J}=1\text{N}\cdot \text{m}$ by definition, since energy is force acting over distance.

True — it is the point of zero thermal motion; for an ordinary system you cannot remove more energy than "all of it," so no temperature lies below 0 K.

Error: 0.018 is in kg/mol, but 18 is in grams. Match units first: $n=\frac{18\text{g}}{18{\text{g mol}}^{-1}}=1$ mol, or $\frac{0.018\text{kg}}{0.018{\text{kg mol}}^{-1}}=1$ mol. The grams cancel the grams; you can't cancel grams against kg.

Error: $R=8.314\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}$ carries kelvin in its units, so $T$ must be in kelvin for the K to cancel. Convert first: $T=25+273.15=298.15$ K.

The written factor is nonsense (litres to litres). The correct relation is $1\text{L}=10^{-3}{\text{m}}^3$; use it: $2\text{L}\times \frac{10^{-3}{\text{m}}^3}{1\text{L}}=2\times 10^{-3}{\text{m}}^3=0.002{\text{m}}^3$ — the litres cancel, leaving m³ (so the number was right by luck).

Error: $\text{kg}\text{m}{\text{s}}^{-2}$ is a newton (force). Energy = force × distance, so multiply by one more metre (add 1 to the m-exponent): $\text{kg}\text{m}{\text{s}}^{-2}\times \text{m}=\text{kg}{\text{m}}^2{\text{s}}^{-2}=1\text{J}$.

Work the gap: $\frac{101325-100000}{100000}=0.01325\approx 1.3\%$. So $1\text{atm}=101325\text{Pa}$ is about $1.3\%$ larger than $1\text{bar}=10^5\text{Pa}$ — small, but it matters in precise work.

Error: one mole of carbon is $N_A=6.022\times 10^{23}$ atoms, weighing about 12 g — not $6.022\times 10^{23}$ grams (which would be an astronomically huge mass). Don't confuse a count with a mass.

Atoms are far too small and numerous to count individually, yet reactions occur in fixed whole-number particle ratios — the mole packages a huge fixed count ($N_A$) into a weighable mass, bridging the particle world and the lab balance.

Because the law assumes $T$ is proportional to kinetic energy, so $T=0$ must mean zero thermal motion. Only kelvin has a physical zero; Celsius's zero (water's freezing point) is arbitrary and breaks the proportionality. The units of $R$ (…K⁻¹) also demand kelvin.

They are read off their defining equations by adding/subtracting base-unit exponents: work $=F\times d$ gives $\text{kg}{\text{m}}^2{\text{s}}^{-2}$, pressure $=F/A$ gives $\text{kg}{\text{m}}^{-1}{\text{s}}^{-2}$. Know the physics formula and the derived unit assembles itself.

A cubic metre (1000 L) is far larger than any bench-top glassware, so it's an inconvenient scale. A litre (a 10 cm cube) matches beakers and flasks, so it survives as a practical non-SI unit.

If the units don't reduce to the expected unit (e.g. joules for an energy), the formula is wrong and no number will save it. Unit analysis catches structural errors before arithmetic ever begins — see Dimensional Analysis.

It was originally chosen so that one mole of a substance weighs its atomic/molecular mass in grams, tying the count to convenient tabulated masses; $N_A$ itself is now fixed by definition.

Exactly 1 mole — $n=N/N_A=N_A/N_A=1$. This is the defining case, independent of what element it is.

Pressure $\to 0$, since $p=nRT/V$ is directly proportional to $T$. Zero absolute temperature means zero molecular motion, so no collisions push on the walls — consistent with Kinetic Theory of Gases.

Not for an ordinary gas — kelvin starts at absolute zero, so here $T\ge 0$; a negative value would predict impossible negative pressure. Always keep $T$ at or above zero when using $pV=nRT$.

Assume g/mol by convention (so CO₂ = 44 g/mol), but flag the ambiguity: without a unit the number is meaningless, and you must confirm before using it in $n=m/M$.

$0\text{L}=0{\text{m}}^3$ — the same factor $10^{-3}$ still multiplies, it just turns zero into zero. A linear conversion always carries zero to zero.

Zero moles, since $n=m/M=0/M=0$ for any nonzero molar mass. "No stuff" is a valid amount and gives a well-defined answer.