NCERT Solutions · Class 11 Chemistry Some Basic Concepts of Chemistry

Significant figures in a calculated result are limited by the least reliable measurement that went into it — and the rule for limiting them is different for multiplication/division than for addition/subtraction.Rule $\displaystyle 1$ — multiplication and division: the result can carry no more significant figures than the factor with the fewest significant figures.Rule $\displaystyle 2$ — addition and subtraction: the result can carry no more decimal places than the term with the fewest decimal places. Sig figs and decimal places are not the same thing, and mixing up which rule applies to which operation is exactly where this question is designed to catch you.(i) \(\displaystyle \dfrac{0.02856 \times 298.15 \times 0.112}{0.5785} \)First count significant figures in each number. A leading zero (before the first nonzero digit) is never significant — it only locates the decimal point — so: The smallest of these is $\displaystyle 3$ (from \(\displaystyle 0.112 \)), so the final answer is limited to $\displaystyle 3$ significant figures — carry the full value through the arithmetic and round only at the end.\[0.02856 \times 298.15 = 8.515164 \] \[8.515164 \times 0.112 = 0.953698368 \] \[\frac{0.953698368}{0.5785} = 1.648571\ldots \]Rounding \(\displaystyle 1.648571\ldots \) to $\displaystyle 3$ significant figures gives \(\displaystyle 1.65 \).(ii) \(\displaystyle 5 \times 5.364 \)Here the "$\displaystyle 5$" is a counting/multiplying number, not a measured quantity — it has no uncertainty at all, so it never limits the precision of the answer. (This is the trap in this part: treating every number in the expression as a measurement and reporting only $\displaystyle 1$ significant figure because "$\displaystyle 5$" looks like it has one.) The only measured value is \(\displaystyle 5.364 \), which has $\displaystyle 4$ significant figures, so the answer is reported to $\displaystyle 4$ significant figures.\[5 \times 5.364 = 26.820 \]Written to $\displaystyle 4$ significant figures: \(\displaystyle 26.82 \).(iii) \(\displaystyle 0.0125 + 0.7864 + 0.0215 \)This is addition, so Rule $\displaystyle 2$ applies — count decimal places, not significant figures. All three terms already agree at $\displaystyle 4$ decimal places, so the sum is reported to $\displaystyle 4$ decimal places with no extra rounding needed:\[0.0125 + 0.7864 + 0.0215 = 0.8204 \]Answer: (i) $\displaystyle 1.65$ ($\displaystyle 3$ significant figures) (ii) $\displaystyle 26.82$ ($\displaystyle 4$ significant figures) (iii) $\displaystyle 0.8204$ ($\displaystyle 4$ decimal places)

Average atomic mass is a weighted average — each isotope's mass is weighted by how much of it actually exists in nature, not counted equally.The idea. Argon in nature is a mixture of three isotopes, each with its own mass and its own share (abundance) of every argon sample you'd ever weigh. To get the mass you'd actually measure for "argon," you cannot just average the three isotopic masses — you must weight each one by the fraction of atoms it contributes.Formula. \[M_{\text{avg}} = \sum_i (\text{fractional abundance})_i \times (\text{isotopic molar mass})_i \] where the fractional abundance is the percentage abundance divided by $\displaystyle 100$ — this is the step people skip, using the percentage number ($\displaystyle 35.6$, $\displaystyle 0.63$, $\displaystyle 99.6$) directly instead of converting it to a fraction of $\displaystyle 1$ first.Convert the abundances to fractions. \[0.337\% = 0.00337, \qquad 0.063\% = 0.00063, \qquad 99.600\% = 0.99600 \] Check: \(\displaystyle 0.00337 + 0.00063 + 0.99600 = 1.00000\) — the fractions must add to exactly $\displaystyle 1$, since together the three isotopes make up all the argon there is.Substitute into the weighted sum.For \(\displaystyle ^{36}\text{Ar}\): \[0.00337 \times 35.96755 \text{ g mol}^{-1} = 0.1212106 \text{ g mol}^{-1} \]For \(\displaystyle ^{38}\text{Ar}\): \[0.00063 \times 37.96272 \text{ g mol}^{-1} = 0.0239165 \text{ g mol}^{-1} \]For \(\displaystyle ^{40}\text{Ar}\): \[0.99600 \times 39.9624 \text{ g mol}^{-1} = 39.8025504 \text{ g mol}^{-1} \]Add the three contributions. \[M_{\text{avg}} = 0.1212106 + 0.0239165 + 39.8025504 = 39.9476775 \text{ g mol}^{-1} \]The last two isotopes barely move the total — \(\displaystyle ^{40}\text{Ar}\) makes up $\displaystyle 99.6$% of all argon atoms, so the molar mass of argon is essentially the molar mass of \(\displaystyle ^{40}\text{Ar}\), nudged upward by a fraction of a gram per mole because of the trace amounts of the lighter isotopes.Rounding. The isotopic masses are known to six figures, but the least-precise abundance (\(\displaystyle 0.063\%\), $\displaystyle 2$ significant figures) still contributes a term of size \(\displaystyle \sim 0.02\), three orders of magnitude smaller than the total — so it does not limit the overall precision. The dominant abundance, \(\displaystyle 99.600\%\), is given to $\displaystyle 5$ significant figures, so the sum is meaningfully known to $\displaystyle 4$–$\displaystyle 5$ significant figures. Rounding to $\displaystyle 2$ decimal places ($\displaystyle 4$ significant figures): \[M_{\text{avg}} \approx 39.95 \text{ g mol}^{-1} \]This matches the accepted atomic weight of argon ($\displaystyle 39.95$ g mol⁻¹) almost exactly, which is a good check that the weighting was done correctly.Answer: The average atomic (molar) mass of naturally occurring argon is \(\displaystyle 39.9477 \text{ g mol}^{-1}\), rounding to \(\displaystyle 39.95 \text{ g mol}^{-1}\).

The mole and the atomic mass unit (u) look alike but sit at completely different scales — one counts a huge collection of atoms, the other is the mass of a single atom. Mixing them up is exactly what this question is testing.Idea used throughout: the Avogadro constant, \(\displaystyle N_A = 6.022 \times 10^{23} \text{ mol}^{-1} \), tells you how many entities (atoms, molecules, etc.) are in one mole. Whenever you have an amount in moles, multiply by \(\displaystyle N_A \) to get the number of particles. But when a mass is given in atomic mass units (u), you are already talking about individual atoms, so \(\displaystyle N_A \) does not enter at all.(i) $\displaystyle 52$ moles of ArHere the amount is already in moles, \(\displaystyle n \), so the number of atoms \(\displaystyle N \) is\[N = n \times N_A \]\[N = 52 \text{ mol} \times 6.022 \times 10^{23} \text{ mol}^{-1} = 313.144 \times 10^{23} \text{ atoms} \]The "mol" unit cancels, leaving a pure count. Since $\displaystyle 52$ has $\displaystyle 2$ significant figures, round to\[N = 3.1 \times 10^{25} \text{ atoms} \](ii) $\displaystyle 52$ u of HeThis is the step people trip on: $\displaystyle 52$ u is a mass on the atomic scale, not a bulk mass in grams — so this problem does not need Avogadro's number at all. By definition, $\displaystyle 1$ u is ($\displaystyle 1$/$\displaystyle 12$) the mass of one \(\displaystyle ^{12}\text{C}\) atom, so a mass given in u is already "per atom." One He atom has a mass of $\displaystyle 4$ u (its atomic mass). So the number of atoms is simply the total mass divided by the mass of one atom:\[N = \frac{\text{total mass}}{\text{mass of one atom}} = \frac{52 \text{ u}}{4 \text{ u atom}^{-1}} = 13 \text{ atoms} \]No conversion factor is needed here — dividing two masses in the same unit (u) leaves a pure, exact count.(iii) $\displaystyle 52$ g of HeNow the mass is in grams, a bulk/macroscopic unit, so this time you do need two steps: convert grams to moles using the molar mass, then convert moles to atoms using \(\displaystyle N_A \).Molar mass of He: \(\displaystyle M = 4 \text{ g mol}^{-1} \) (numerically the same "$\displaystyle 4$" as before, but now it is the mass of one mole, i.e. of \(\displaystyle 6.022\times10^{23}\) atoms — not of a single atom. Treating this $\displaystyle 4$ the same way as the "$\displaystyle 4$ u" in part (ii) is the trap.)Step $\displaystyle 1$ — moles of He:\[n = \frac{\text{given mass}}{\text{molar mass}} = \frac{52 \text{ g}}{4 \text{ g mol}^{-1}} = 13 \text{ mol} \]Step $\displaystyle 2$ — atoms of He:\[N = n \times N_A = 13 \text{ mol} \times 6.022 \times 10^{23} \text{ mol}^{-1} = 78.286 \times 10^{23} \text{ atoms} \]Rounding to $\displaystyle 2$ significant figures (from the $\displaystyle 52$ g):\[N = 7.8 \times 10^{24} \text{ atoms} \]Answer: (i) \(\displaystyle 3.1 \times 10^{25}\) atoms of Ar; (ii) $\displaystyle 13$ atoms of He; (iii) \(\displaystyle 7.8 \times 10^{24}\) atoms of He.

Combustion analysis works backward: the mass of CO₂ and H₂O tells you the moles of carbon and hydrogen burnt — you must go mass → moles → mass through each product's own molar mass, never treat the mass of CO₂ as if it were the mass of carbon.Finding carbon, from the CO₂.Molar mass of \(\displaystyle \text{CO}_2 \) (using \(\displaystyle C=12.0\), \(\displaystyle O=16.0\)): \[M(\text{CO}_2) = 12.0 + 2(16.0) = 44.0\ \text{g mol}^{-1} \]Moles of \(\displaystyle \text{CO}_2 \) produced: \[n(\text{CO}_2) = \frac{3.38\ \text{g}}{44.0\ \text{g mol}^{-1}} = 0.0768\ \text{mol} \]Every \(\displaystyle \text{CO}_2 \) molecule carries exactly one carbon atom, so moles of C in the original sample equal moles of \(\displaystyle \text{CO}_2 \): \[n(\text{C}) = 0.0768\ \text{mol}, \qquad m(\text{C}) = 0.0768\ \text{mol} \times 12.0\ \text{g mol}^{-1} = 0.922\ \text{g} \]Finding hydrogen, from the H₂O — this is the step people lose the factor of $\displaystyle 2$ on.Molar mass of \(\displaystyle \text{H}_2\text{O} \): \[M(\text{H}_2\text{O}) = 2(1.0) + 16.0 = 18.0\ \text{g mol}^{-1} \]Moles of \(\displaystyle \text{H}_2\text{O} \): \[n(\text{H}_2\text{O}) = \frac{0.690\ \text{g}}{18.0\ \text{g mol}^{-1}} = 0.0383\ \text{mol} \]Each water molecule carries two hydrogen atoms, so you must double this before it's moles of H — using \(\displaystyle n(\text{H}_2\text{O})\) directly as \(\displaystyle n(\text{H})\) is the single most common slip in this kind of problem: \[n(\text{H}) = 2 \times 0.0383\ \text{mol} = 0.0767\ \text{mol}, \qquad m(\text{H}) = 0.0767\ \text{mol} \times 1.0\ \text{g mol}^{-1} = 0.0767\ \text{g} \]As a check: the gas is stated to contain only C and H, so \(\displaystyle m(\text{C})+m(\text{H})\) should equal the mass of the sample that was actually burnt: \(\displaystyle 0.922 + 0.0767 = 0.999\ \text{g} \approx 1.00\ \text{g}\) — a clean number, which is reassuring, though that burnt-sample mass isn't needed for what follows.(i) Empirical formula — take the mole ratio, not the mass ratio.\[n(\text{C}) : n(\text{H}) = 0.0768 : 0.0767 \]Dividing both by the smaller number ($\displaystyle 0.0767$): \[1.00 : 1.00 \]The simplest whole-number ratio of C to H is $\displaystyle 1$:$\displaystyle 1$, so the empirical formula is CH, with empirical formula mass (EFM) \[\text{EFM} = 12.0 + 1.0 = 13.0\ \text{g mol}^{-1} \](ii) Molar mass — from how much space the gas occupies at STP.The ideal gas equation is \(\displaystyle PV = nRT\), where \(\displaystyle P\) is pressure, \(\displaystyle V\) is volume, \(\displaystyle n\) is moles, \(\displaystyle R\) is the gas constant and \(\displaystyle T\) is temperature. At STP one mole of any ideal gas occupies the molar volume \(\displaystyle V_m = 22.4\ \text{L mol}^{-1}\) (the value this chapter uses throughout), so the moles of gas in the $\displaystyle 10.0$ L sample are \[n = \frac{V}{V_m} = \frac{10.0\ \text{L}}{22.4\ \text{L mol}^{-1}} = 0.446\ \text{mol} \]Molar mass is mass divided by moles, not by volume — don't stop at "grams per litre" and call it done: \[M = \frac{m}{n} = \frac{11.6\ \text{g}}{0.446\ \text{mol}} = 26.0\ \text{g mol}^{-1} \](iii) Molecular formula — compare molar mass to the empirical formula mass.\[n = \frac{M}{\text{EFM}} = \frac{26.0\ \text{g mol}^{-1}}{13.0\ \text{g mol}^{-1}} = 2 \]This \(\displaystyle n\) multiplies every subscript in the empirical formula, not just one atom, so \[(\text{CH})_2 = \text{C}_2\text{H}_2 \]Answer: empirical formula \(\displaystyle \text{CH} \); molar mass \(\displaystyle \approx 26.0\ \text{g mol}^{-1} \); molecular formula \(\displaystyle \text{C}_2\text{H}_2 \) (acetylene, the everyday welding fuel gas).

Molarity tells you moles of solute per litre of solution — use it to get moles of HCl first, then use the equation's mole ratio to step across to CaCO₃.Step $\displaystyle 1$: Moles of HCl availableMolarity is defined as \[M = \frac{n}{V} \] where \(\displaystyle n\) is moles of solute and \(\displaystyle V\) is the volume of solution in litres.Here \(\displaystyle M = 0.75\ \text{mol L}^{-1}\) and \(\displaystyle V = 25\ \text{mL} = 0.025\ \text{L}\).\[n(\text{HCl}) = M \times V = 0.75\ \text{mol L}^{-1} \times 0.025\ \text{L} = 0.01875\ \text{mol} \]Step $\displaystyle 2$: Use the balanced equation to switch substances\[\text{CaCO}_3(s) + 2\,\text{HCl}(aq) \rightarrow \text{CaCl}_2(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(l) \]The coefficients say $\displaystyle 1$ mole of CaCO₃ reacts with $\displaystyle 2$ moles of HCl — this ratio is the whole point of balancing the equation, and skipping it (treating moles of HCl as if they equalled moles of CaCO₃) is the mistake that trips people up here.\[n(\text{CaCO}_3) = \frac{1}{2} \times n(\text{HCl}) = \frac{1}{2} \times 0.01875\ \text{mol} = 0.009375\ \text{mol} \]Step $\displaystyle 3$: Convert moles of CaCO₃ to massName the formula: mass = moles × molar mass, \[m = n \times M_{\text{molar}} \]Molar mass of CaCO₃ (Ca = $\displaystyle 40.08$, C = $\displaystyle 12.01$, O = $\displaystyle 16.00$ × $\displaystyle 3$): \[M_{\text{molar}}(\text{CaCO}_3) = 40.08 + 12.01 + 3(16.00) = 100.09\ \text{g mol}^{-1} \approx 100\ \text{g mol}^{-1} \]\[m(\text{CaCO}_3) = 0.009375\ \text{mol} \times 100\ \text{g mol}^{-1} = 0.9375\ \text{g} \]Step $\displaystyle 4$: Round to the precision the data supportsThe given data ($\displaystyle 25$ mL, $\displaystyle 0.75$ M) each carry two significant figures, so the final mass is rounded to two significant figures at the very end — not partway through the calculation.\[m(\text{CaCO}_3) \approx 0.94\ \text{g} \]Answer: $\displaystyle 0.94$ g of CaCO₃ (≈ $\displaystyle 0.009375$ mol) is required to react completely with $\displaystyle 25$ mL of $\displaystyle 0.75$ M HCl.

The balanced equation tells you the mole ratio in which substances react — you always convert grams → moles → moles (using that ratio) → grams, never grams straight to grams.The reaction is \[4\text{HCl (aq)} + \text{MnO}_2\text{(s)} \rightarrow 2\text{H}_2\text{O (l)} + \text{MnCl}_2\text{(aq)} + \text{Cl}_2\text{(g)} \]Read the coefficients as a mole statement: every $\displaystyle 1$ mole of \(\displaystyle \text{MnO}_2\) that reacts uses up $\displaystyle 4$ moles of HCl. That "$\displaystyle 4$" is the number people forget — treating this as a $\displaystyle 1$:$\displaystyle 1$ reaction is the mistake that quietly halves the answer to a quarter of its true value.Step $\displaystyle 1$ — Moles of \(\displaystyle \text{MnO}_2\) available.Use \(\displaystyle n = \dfrac{m}{M} \), where \(\displaystyle n\) is the amount in moles, \(\displaystyle m\) is the given mass, and \(\displaystyle M\) is the molar mass.Molar mass of \(\displaystyle \text{MnO}_2\): \[M(\text{MnO}_2) = 54.94 + 2(16.00) = 86.94\ \text{g mol}^{-1} \]\[n(\text{MnO}_2) = \frac{5.0\ \text{g}}{86.94\ \text{g mol}^{-1}} = 0.05751\ \text{mol} \]Step $\displaystyle 2$ — Moles of HCl that react, from the mole ratio.From the balanced equation, \(\displaystyle \dfrac{n(\text{HCl})}{n(\text{MnO}_2)} = \dfrac{4}{1} \), so\[n(\text{HCl}) = 4 \times n(\text{MnO}_2) = 4 \times 0.05751\ \text{mol} = 0.2300\ \text{mol} \]This is the step that is easy to skip past — you are not scaling the $\displaystyle 5.0$ g by $\displaystyle 4$, you are scaling the moles by $\displaystyle 4$, because the coefficients in a balanced equation are a mole ratio, not a mass ratio.Step $\displaystyle 3$ — Convert moles of HCl back to mass.Rearrange \(\displaystyle n = \dfrac{m}{M} \) to \(\displaystyle m = n \times M \).Molar mass of HCl: \[M(\text{HCl}) = 1.008 + 35.45 = 36.46\ \text{g mol}^{-1} \]\[m(\text{HCl}) = 0.2300\ \text{mol} \times 36.46\ \text{g mol}^{-1} = 8.39\ \text{g} \]Step $\displaystyle 4$ — Round to the precision the data justifies.The given mass, $\displaystyle 5.0$ g, carries $\displaystyle 2$ significant figures, so the answer is reported to $\displaystyle 2$ significant figures:\[m(\text{HCl}) \approx 8.4\ \text{g} \]Answer: $\displaystyle 8.4$ g of HCl reacts with $\displaystyle 5.0$ g of \(\displaystyle \mathrm{MnO_{2}}\).