NCERT Solutions · Class 11 Chemistry Some Basic Concepts of Chemistry

Distance covered = speed × time, and every unit must be in the same system (metres and seconds) before you multiply.The relevant relation is \[\text{distance} = \text{speed} \times \text{time} \] where speed is how far light travels per second, and time is how long it travels for.Step $\displaystyle 1$ — write down the data. \[\text{speed of light} = 3.0 \times 10^{8}\ \text{m s}^{-1} \] \[\text{time} = 2.00\ \text{ns} \]Step $\displaystyle 2$ — convert nanoseconds to seconds.A nanosecond is \(\displaystyle 10^{-9}\) s (n = nano = \(\displaystyle 10^{-9}\)). This is the step people get wrong — plugging in "$\displaystyle 2.00$" directly as if it were in seconds would give an answer a billion times too large. \[2.00\ \text{ns} = 2.00 \times 10^{-9}\ \text{s} \]Step $\displaystyle 3$ — substitute into distance = speed × time. \[\text{distance} = \left(3.0 \times 10^{8}\ \text{m s}^{-1}\right) \times \left(2.00 \times 10^{-9}\ \text{s}\right) \]Multiply the numbers and the powers of ten separately: \[3.0 \times 2.00 = 6.00 \] \[10^{8} \times 10^{-9} = 10^{-1} \]So \[\text{distance} = 6.00 \times 10^{-1}\ \text{m} \]Step $\displaystyle 4$ — round to the correct precision.The speed of light is given to $\displaystyle 2$ significant figures ($\displaystyle 3.0$), so the answer can only be trusted to $\displaystyle 2$ significant figures, even though the time was given to 3. \[\text{distance} = 6.0 \times 10^{-1}\ \text{m} = 0.60\ \text{m} \]Answer: The distance covered by light in $\displaystyle 2.00$ ns is \(\displaystyle 6.0 \times 10^{-1}\ \text{m} = 0.60\ \text{m}\) (to $\displaystyle 2$ significant figures).

The limiting reagent is whichever species runs out first when you compare the amounts actually supplied to the mole ratio the equation demands — not whichever number happens to look smaller.The reaction is\[A + B_2 \rightarrow AB_2 \]The coefficients are both $\displaystyle 1$, so the equation demands exactly one particle (or one mole) of \(\displaystyle B_2\) for every one particle (or one mole) of \(\displaystyle A\) — a $\displaystyle 1$:$\displaystyle 1$ ratio. To find the limiting reagent in each mixture, take the amount of A given, work out how much \(\displaystyle B_2\) it would need at this $\displaystyle 1$:$\displaystyle 1$ ratio, and compare that to the \(\displaystyle B_2\) actually supplied. Whichever reactant falls short of what the ratio requires is the limiting reagent.(i) $\displaystyle 300$ atoms of A + $\displaystyle 200$ molecules of \(\displaystyle B_2\)At a $\displaystyle 1$:$\displaystyle 1$ ratio, $\displaystyle 300$ atoms of A require $\displaystyle 300$ molecules of \(\displaystyle B_2\): \[300 \text{ atoms A} \times \frac{1 \text{ molecule } B_2}{1 \text{ atom A}} = 300 \text{ molecules } B_2 \text{ needed} \] Only $\displaystyle 200$ molecules of \(\displaystyle B_2\) are available, so \(\displaystyle B_2\) is used up before all the A reacts ($\displaystyle 100$ atoms of A are left over).Limiting reagent: \(\displaystyle B_2\)(ii) $\displaystyle 2$ mol A + $\displaystyle 3$ mol \(\displaystyle B_2\)$\displaystyle 2$ mol A requires $\displaystyle 2$ mol \(\displaystyle B_2\) at the $\displaystyle 1$:$\displaystyle 1$ ratio. $\displaystyle 3$ mol \(\displaystyle B_2\) is supplied — $\displaystyle 1$ mol \(\displaystyle B_2\) is left over once all the A is consumed.Limiting reagent: A(iii) $\displaystyle 100$ atoms of A + $\displaystyle 100$ molecules of \(\displaystyle B_2\)$\displaystyle 100$ atoms of A require exactly $\displaystyle 100$ molecules of \(\displaystyle B_2\), and exactly $\displaystyle 100$ molecules of \(\displaystyle B_2\) are supplied. The mixture is in the exact stoichiometric ratio, so both reactants are used up at the same instant — neither is in excess.Limiting reagent: none (the mixture is exactly stoichiometric)(iv) $\displaystyle 5$ mol A + $\displaystyle 2.5$ mol \(\displaystyle B_2\)$\displaystyle 5$ mol A requires $\displaystyle 5$ mol \(\displaystyle B_2\). Only $\displaystyle 2.5$ mol \(\displaystyle B_2\) is available — half of what is needed — so the reaction stops once the \(\displaystyle B_2\) is gone, leaving $\displaystyle 2.5$ mol A unreacted.This is the case people get wrong: $\displaystyle 5$ mol looks like "more," but a larger mole number is not automatically in excess — it is in excess only relative to what the ratio actually asks for.Limiting reagent: \(\displaystyle B_2\)(v) $\displaystyle 2.5$ mol A + $\displaystyle 5$ mol \(\displaystyle B_2\)$\displaystyle 2.5$ mol A requires only $\displaystyle 2.5$ mol \(\displaystyle B_2\). $\displaystyle 5$ mol \(\displaystyle B_2\) is supplied, so $\displaystyle 2.5$ mol \(\displaystyle B_2\) remains once all the A is consumed.Limiting reagent: AAnswer: (i) \(\displaystyle B_2\) is limiting; (ii) A is limiting; (iii) no limiting reagent — the mixture is exactly stoichiometric; (iv) \(\displaystyle B_2\) is limiting; (v) A is limiting.

Before you can compare two reactants, put both on the same footing — moles, not grams — because $\displaystyle 1$ g of \(\displaystyle N_2 \) and $\displaystyle 1$ g of \(\displaystyle H_2 \) are not the same number of particles.The balanced equation for this reaction is\[N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \]This says $\displaystyle 1$ mole of dinitrogen reacts with exactly $\displaystyle 3$ moles of dihydrogen to give $\displaystyle 2$ moles of ammonia. Masses can't be read off this equation directly — moles can.Step $\displaystyle 1$: Convert both given masses to moles.Using \(\displaystyle n = \dfrac{\text{given mass}}{\text{molar mass}} \), with \(\displaystyle M(N_2) = 28.0\ \text{g mol}^{-1} \) and \(\displaystyle M(H_2) = 2.0\ \text{g mol}^{-1} \):\[n(N_2) = \frac{2.00\times10^{3}\ \text{g}}{28.0\ \text{g mol}^{-1}} = 71.43\ \text{mol} \]\[n(H_2) = \frac{1.00\times10^{3}\ \text{g}}{2.0\ \text{g mol}^{-1}} = 500.0\ \text{mol} \]Step $\displaystyle 2$: Find which reactant runs out first.The trap here is eyeballing the two masses ($\displaystyle 2000$ g vs. $\displaystyle 1000$ g) and assuming dinitrogen, the bigger mass, must be in excess. That compares grams, but the reaction runs on the $\displaystyle 1$ : $\displaystyle 3$ mole ratio, not on grams — so you must check moles against that ratio instead.The equation needs $\displaystyle 3$ mol \(\displaystyle H_2 \) for every $\displaystyle 1$ mol \(\displaystyle N_2 \). For all $\displaystyle 71.43$ mol of \(\displaystyle N_2 \) to react, the dihydrogen required is\[n(H_2)_{\text{required}} = 3 \times 71.43\ \text{mol} = 214.3\ \text{mol} \]Only $\displaystyle 214.3$ mol of \(\displaystyle H_2 \) is needed, but $\displaystyle 500.0$ mol is actually available. Since there is more \(\displaystyle H_2 \) present than the reaction can use, dihydrogen is in excess and dinitrogen is the limiting reagent — it is \(\displaystyle N_2 \) that decides how much \(\displaystyle NH_3 \) can form.Step $\displaystyle 3$: Get the mass of ammonia from the limiting reagent, \(\displaystyle N_2 \).From the equation, $\displaystyle 1$ mol \(\displaystyle N_2 \) gives $\displaystyle 2$ mol \(\displaystyle NH_3 \), so\[n(NH_3) = 2 \times n(N_2) = 2 \times 71.43\ \text{mol} = 142.9\ \text{mol} \]With \(\displaystyle M(NH_3) = 17.0\ \text{g mol}^{-1} \):\[\text{mass}(NH_3) = 142.9\ \text{mol} \times 17.0\ \text{g mol}^{-1} = 2428.6\ \text{g} \]Rounded to three significant figures, matching the precision of the data given (\(\displaystyle 2.00\times10^{3}\), \(\displaystyle 1.00\times10^{3}\)):\[\text{mass}(NH_3) \approx 2.43\times10^{3}\ \text{g} \]Step $\displaystyle 4$: Yes — the excess reactant, \(\displaystyle H_2 \), is left over. Find how much.The dihydrogen actually consumed is tied to the \(\displaystyle N_2 \) that reacted, using the $\displaystyle 3$ : $\displaystyle 1$ ratio again:\[n(H_2)_{\text{used}} = 3 \times n(N_2) = 3 \times 71.43\ \text{mol} = 214.3\ \text{mol} \]\[\text{mass}(H_2)_{\text{used}} = 214.3\ \text{mol} \times 2.0\ \text{g mol}^{-1} = 428.6\ \text{g} \]The leftover dihydrogen is what was supplied minus what got used — not a separate calculation from scratch:\[\text{mass}(H_2)_{\text{left}} = 1000\ \text{g} - 428.6\ \text{g} = 571.4\ \text{g} \]Rounded to three significant figures:\[\text{mass}(H_2)_{\text{left}} \approx 5.71\times10^{2}\ \text{g} \]Answer: (i) \(\displaystyle 2.43\times10^{3}\ \text{g}\) of \(\displaystyle NH_3 \) is produced; (ii) yes, dihydrogen is left unreacted, since dinitrogen is the limiting reagent; (iii) about \(\displaystyle 5.71\times10^{2}\ \text{g}\) ($\displaystyle 571$ g) of \(\displaystyle H_2 \) remains unreacted.

"mol" names a fixed amount of substance; "M" (mol L⁻¹) names a concentration, and how much substance that gives you depends on how much solution you take.$\displaystyle 0.50$ mol Na₂CO₃ — a fixed amount, full stopThis is simply a quantity of matter: half a mole of sodium carbonate, with nothing said about it being in solution at all. Using the molar mass of \(\displaystyle \text{Na}_2\text{CO}_3 \),\[M(\text{Na}_2\text{CO}_3) = 2(23) + 12 + 3(16) = 46 + 12 + 48 = 106\ \text{g mol}^{-1} \]the mass this corresponds to is fixed by \(\displaystyle \text{mass} = n \times M \):\[\text{mass} = 0.50\ \text{mol} \times 106\ \text{g mol}^{-1} = 53\ \text{g} \]That $\displaystyle 53$ g of \(\displaystyle \text{Na}_2\text{CO}_3 \) is the same $\displaystyle 53$ g whether it sits as a dry solid on a balance pan or is later dissolved in some water — "mol" makes no reference to a solvent or a volume.$\displaystyle 0.50$ M Na₂CO₃ — a concentration, so the amount depends on volume"$\displaystyle 0.50$ M" is shorthand for a molarity of \(\displaystyle 0.50\ \text{mol L}^{-1} \): the solution is made by dissolving $\displaystyle 0.50$ mol of \(\displaystyle \text{Na}_2\text{CO}_3 \) in enough water that the total volume of solution comes to exactly $\displaystyle 1$ L. Molarity is defined as\[C = \frac{n}{V} \]where \(\displaystyle C \) is molarity (mol L⁻¹), \(\displaystyle n \) is moles of solute, and \(\displaystyle V \) is the volume of the solution in litres — not the volume of water added before mixing. This is the step people get wrong: molarity is per litre of final solution, so if you dissolve the solute and then top up with more water, the molarity drops even though \(\displaystyle n \) hasn't changed.Because \(\displaystyle C \) is fixed at $\displaystyle 0.50$ mol L⁻¹, the moles of \(\displaystyle \text{Na}_2\text{CO}_3 \) you actually have depends on how much of the solution you take, via \(\displaystyle n = C \times V \):\[V = 1\ \text{L}: \quad n = 0.50\ \text{mol L}^{-1} \times 1\ \text{L} = 0.50\ \text{mol} \;\; (53\ \text{g in that litre}) \]\[V = 0.500\ \text{L}: \quad n = 0.50\ \text{mol L}^{-1} \times 0.500\ \text{L} = 0.25\ \text{mol} \;\; (13.25\ \text{g in that half-litre}) \]\[V = 0.250\ \text{L}: \quad n = 0.50\ \text{mol L}^{-1} \times 0.250\ \text{L} = 0.125\ \text{mol} \;\; (6.625\ \text{g in that quarter-litre}) \]So "$\displaystyle 0.50$ M Na₂CO�$\displaystyle 3$" describes a bottle you could pour out in any volume, and each portion carries proportionally less \(\displaystyle \text{Na}_2\text{CO}_3 \) — the label describes the solution's strength, not one fixed lump of substance.The core distinction"$\displaystyle 0.50$ mol Na₂CO₃" is an absolute amount of substance ($\displaystyle 53$ g, fixed, no water implied). "$\displaystyle 0.50$ M Na₂CO₃" is a concentration — $\displaystyle 0.50$ mol of solute per litre of solution — and the actual moles or mass of \(\displaystyle \text{Na}_2\text{CO}_3 \) present scales with the volume of that solution you take, via \(\displaystyle n = C \times V \).Answer: $\displaystyle 0.50$ mol Na₂CO₃ is a fixed amount of substance ($\displaystyle 53$ g, independent of any solvent); $\displaystyle 0.50$ M Na₂CO₃ is a concentration of $\displaystyle 0.50$ mol L⁻¹ of solution, so the actual moles/mass of Na₂CO₃ present depend on the volume of solution taken (e.g. $\displaystyle 0.50$ mol in $\displaystyle 1$ L, but only $\displaystyle 0.25$ mol in $\displaystyle 500$ mL).

Gay-Lussac's law of gaseous volumes says that when gases react, the volumes of reactants and products (all measured at the same temperature and pressure) are in a simple whole-number ratio — the same ratio as the moles in the balanced equation. That means you can read this problem straight off the balanced equation without ever converting to moles.The balanced equation for the formation of water vapour is\[2\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(g) \]By Avogadro's law, equal volumes of gases (at the same temperature and pressure) contain equal numbers of moles. So the coefficients \(\displaystyle 2 : 1 : 2\) in the equation are also the ratio of volumes of \(\displaystyle \text{H}_2\), \(\displaystyle \text{O}_2\), and \(\displaystyle \text{H}_2\text{O}\) that react and form.Check the volumes actually given against this ratio first — this is the step people skip. You are given $\displaystyle 10$ volumes of \(\displaystyle \text{H}_2\) and $\displaystyle 5$ volumes of \(\displaystyle \text{O}_2\). Their ratio is\[\frac{V_{\text{H}_2}}{V_{\text{O}_2}} = \frac{10}{5} = \frac{2}{1} \]which is exactly the \(\displaystyle 2:1\) ratio required by the equation. So both gases are used up completely — there is no leftover (excess) reagent to worry about, and every volume of \(\displaystyle \text{H}_2\) supplied goes on to form water vapour.Now use the equation ratio directly:\[2 \text{ volumes } \text{H}_2 \rightarrow 2 \text{ volumes } \text{H}_2\text{O} \]so $\displaystyle 1$ volume of \(\displaystyle \text{H}_2\) produces $\displaystyle 1$ volume of \(\displaystyle \text{H}_2\text{O}\) vapour (a $\displaystyle 1$:$\displaystyle 1$ correspondence between \(\displaystyle \text{H}_2\) consumed and \(\displaystyle \text{H}_2\text{O}\) formed). Scaling up to the $\displaystyle 10$ volumes of \(\displaystyle \text{H}_2\) actually supplied,\[V_{\text{H}_2\text{O}} = 10 \text{ volumes } \text{H}_2 \times \frac{2 \text{ volumes } \text{H}_2\text{O}}{2 \text{ volumes } \text{H}_2} = 10 \text{ volumes} \]As a cross-check, using the \(\displaystyle \text{O}_2\) side of the equation instead:\[V_{\text{H}_2\text{O}} = 5 \text{ volumes } \text{O}_2 \times \frac{2 \text{ volumes } \text{H}_2\text{O}}{1 \text{ volume } \text{O}_2} = 10 \text{ volumes} \]Both routes agree, confirming the answer.Answer: $\displaystyle 10$ volumes of water vapour are produced.

The SI base unit of length is the metre (m), and the SI base unit of mass is the kilogram (kg) — "convert to basic units" means rewriting the number in metres or kilograms, not just in grams or as a bare number.The prefix "pico" (p) means \(\displaystyle 10^{-12} \), so \[1\ \text{pm} = 1 \times 10^{-12}\ \text{m} \]The prefix "milli" (m) means \(\displaystyle 10^{-3} \), so \[1\ \text{mg} = 1 \times 10^{-3}\ \text{g} \]But grams are not the SI base unit of mass — kilograms are. A common slip here is stopping at grams; you still need one more step, using \[1\ \text{g} = 1 \times 10^{-3}\ \text{kg} \](i) $\displaystyle 28.7$ pm to metres\[28.7\ \text{pm} = 28.7 \times 10^{-12}\ \text{m} \]Writing the coefficient between $\displaystyle 1$ and $\displaystyle 10$ (proper scientific notation): \[28.7 \times 10^{-12}\ \text{m} = 2.87 \times 10^{-11}\ \text{m} \](ii) $\displaystyle 15.15$ pm to metres\[15.15\ \text{pm} = 15.15 \times 10^{-12}\ \text{m} = 1.515 \times 10^{-11}\ \text{m} \](iii) $\displaystyle 25365$ mg to kilogramsFirst convert milligrams to grams: \[25365\ \text{mg} = 25365 \times 10^{-3}\ \text{g} = 25.365\ \text{g} \]This is the step that is easy to stop at by mistake — grams are not a base unit, so the conversion is not finished. Now convert grams to kilograms: \[25.365\ \text{g} = 25.365 \times 10^{-3}\ \text{kg} \]Writing this in scientific notation: \[25.365 \times 10^{-3}\ \text{kg} = 2.5365 \times 10^{-2}\ \text{kg} \]Each result is left with the same number of significant figures as the data given, since these are exact unit-conversion factors ( \(\displaystyle 10^{-12}, 10^{-3} \) ) and do not limit precision.Answer: (i) \(\displaystyle 2.87 \times 10^{-11}\ \text{m} \) (ii) \(\displaystyle 1.515 \times 10^{-11}\ \text{m} \) (iii) \(\displaystyle 2.5365 \times 10^{-2}\ \text{kg} \)

Fewer grams per atom means more atoms — the number of atoms in a fixed mass depends only on how light each atom is.The number of atoms in a mass \(\displaystyle m\) of an element is\[n_{\text{atoms}} = \frac{m}{M} \times N_A \]where \(\displaystyle M\) is the molar (atomic) mass in \(\displaystyle \text{g mol}^{-1} \) and \(\displaystyle N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}\) is Avogadro's number. Since \(\displaystyle m = 1\ \text{g}\) is the same for all four samples, the one with the smallest molar mass packs in the most atoms — you don't even need to finish the arithmetic to see the trend, but let's carry it through with units so the comparison is exact.(i) $\displaystyle 1$ g Au(s) — atomic mass of Au \(\displaystyle = 197\ \text{g mol}^{-1}\)\[n_{\text{Au}} = \frac{1\ \text{g}}{197\ \text{g mol}^{-1}} \times 6.022 \times 10^{23}\ \text{mol}^{-1} = 3.06 \times 10^{21}\ \text{atoms} \](ii) $\displaystyle 1$ g Na(s) — atomic mass of Na \(\displaystyle = 23\ \text{g mol}^{-1}\)\[n_{\text{Na}} = \frac{1\ \text{g}}{23\ \text{g mol}^{-1}} \times 6.022 \times 10^{23}\ \text{mol}^{-1} = 2.62 \times 10^{22}\ \text{atoms} \](iii) $\displaystyle 1$ g Li(s) — atomic mass of Li \(\displaystyle = 7\ \text{g mol}^{-1}\)\[n_{\text{Li}} = \frac{1\ \text{g}}{7\ \text{g mol}^{-1}} \times 6.022 \times 10^{23}\ \text{mol}^{-1} = 8.60 \times 10^{22}\ \text{atoms} \](iv) $\displaystyle 1$ g Cl₂(g) — here is the step people miss: \(\displaystyle M(\text{Cl}_2) = 71\ \text{g mol}^{-1}\) gives you moles of molecules, and each Cl₂ molecule contributes two Cl atoms, so the atom count needs an extra factor of $\displaystyle 2$ (equivalently, divide by the atomic mass of Cl, \(\displaystyle 35.5\ \text{g mol}^{-1}\), directly).\[n_{\text{Cl}_2\text{ molecules}} = \frac{1\ \text{g}}{71\ \text{g mol}^{-1}} \times 6.022 \times 10^{23}\ \text{mol}^{-1} = 8.48 \times 10^{21}\ \text{molecules} \]\[n_{\text{Cl atoms}} = 2 \times 8.48 \times 10^{21} = 1.70 \times 10^{22}\ \text{atoms} \]Now line up all four counts (to $\displaystyle 3$ significant figures, matching the precision of the atomic masses used):\[\text{Au}: 3.06 \times 10^{21} \quad \text{Na}: 2.62 \times 10^{22} \quad \text{Li}: 8.60 \times 10^{22} \quad \text{Cl}_2: 1.70 \times 10^{22} \]Lithium's atoms are the lightest of the four (even lighter, per atom, than the chlorine in Cl₂), so $\displaystyle 1$ g of Li fits in the most of them.Answer: $\displaystyle 1$ g of Li(s) contains the largest number of atoms, about \(\displaystyle 8.60 \times 10^{22}\) atoms — more than Na (\(\displaystyle 2.62 \times 10^{22}\)), Cl₂ (\(\displaystyle 1.70 \times 10^{22}\)), or Au (\(\displaystyle 3.06 \times 10^{21}\)).

A mole fraction tells you the ratio of moles, not volumes — so the first job is to convert it into an amount of substance you can weigh out, before molarity (moles per litre of solution) can be found.Step $\displaystyle 1$: Fix a convenient amount of solution.Mole fraction of ethanol, \(\displaystyle x_{\text{ethanol}} = 0.040 \). Since mole fractions of the two components must add to $\displaystyle 1$,\[x_{\text{water}} = 1 - 0.040 = 0.960 \]Pick the total amount of solution to be exactly $\displaystyle 1$ mol (any basis works — the ratio is what matters). Then\[n_{\text{ethanol}} = 0.040 \text{ mol}, \qquad n_{\text{water}} = 0.960 \text{ mol} \]Step $\displaystyle 2$: Turn the water's moles into a volume — this is the step people skip.Molarity needs the volume of the solution, not moles of water, so convert \(\displaystyle n_{\text{water}} \) to mass, then to volume using the given density.Mass of water: \[m_{\text{water}} = n_{\text{water}} \times M_{\text{water}} = 0.960 \text{ mol} \times 18 \text{ g mol}^{-1} = 17.28 \text{ g} \]Density of water is given as \(\displaystyle 1 \text{ g mL}^{-1} \), so\[V_{\text{water}} = \frac{m_{\text{water}}}{\rho_{\text{water}}} = \frac{17.28 \text{ g}}{1 \text{ g mL}^{-1}} = 17.28 \text{ mL} \]Because the solution is dilute (only $\displaystyle 4$ mol % ethanol), the volume of the solution is taken as essentially equal to the volume of the water present — this is the approximation the problem is built around, and it's exactly why it hands you the density of water rather than of the solution.\[V_{\text{solution}} \approx 17.28 \text{ mL} = 0.01728 \text{ L} \]Step $\displaystyle 3$: Apply the definition of molarity.Molarity \(\displaystyle M \) is moles of solute per litre of solution:\[M = \frac{n_{\text{ethanol}}}{V_{\text{solution}}(\text{L})} \]where \(\displaystyle n_{\text{ethanol}} \) is the moles of ethanol (the solute) and \(\displaystyle V_{\text{solution}} \) is the volume of the whole solution in litres — not the volume of water alone, and not the mass of anything.\[M = \frac{0.040 \text{ mol}}{0.01728 \text{ L}} = 2.3148 \text{ mol L}^{-1} \]The data (mole fraction to $\displaystyle 2$ significant figures, molar mass to $\displaystyle 2$) justifies rounding to three significant figures at the end, not before:\[M \approx 2.31 \text{ mol L}^{-1} \]Answer: The molarity of the ethanol solution is approximately \(\displaystyle 2.31 \text{ mol L}^{-1} \) ($\displaystyle 2.31$ M).

The mole is defined so that exactly $\displaystyle 12$ g of ¹²C contains exactly \(\displaystyle 6.022\times10^{23}\) atoms — that number (Avogadro's constant) is the bridge between a single atom's mass and a molar mass in grams.Because the whole gram/mole scale is pinned to ¹²C, the molar mass of ¹²C itself is exactly\[M(^{12}\text{C}) = 12 \text{ g mol}^{-1} \]A slip to watch for: do not use \(\displaystyle 12.011\) g mol⁻¹ here. That is the average atomic mass of naturally occurring carbon (a mix of ¹²C, ¹³C, and trace ¹⁴C). The question asks specifically about the isotope ¹²C, whose mass is fixed at exactly \(\displaystyle 12\) by definition — mixing the two gives a slightly wrong, and conceptually wrong, answer.Formula. The mass of one atom is the molar mass shared out over one mole of atoms:\[m_{\text{atom}} = \dfrac{M}{N_A} \]where Substituting:\[m_{\text{atom}} = \dfrac{12 \text{ g mol}^{-1}}{6.022\times10^{23} \text{ mol}^{-1}} \]The mol⁻¹ units cancel, leaving grams, as they must for a single-atom mass:\[m_{\text{atom}} = \dfrac{12}{6.022\times10^{23}} \text{ g} = 1.99269\ldots \times 10^{-23} \text{ g} \]The data (\(\displaystyle N_A = 6.022\times10^{23}\), four significant figures) justifies rounding to four significant figures at the very end, not before:\[m_{\text{atom}} \approx 1.9927 \times 10^{-23} \text{ g} \]Answer: The mass of one ¹²C atom is \(\displaystyle 1.9927 \times 10^{-23}\) g.