NCERT Solutions · Class 11 Chemistry Some Basic Concepts of Chemistry

Molarity is moles of solute per litre of SOLUTION — you need moles of sugar first, then divide by the total volume given, not by the volume of water added.Step $\displaystyle 1$ — Find the molar mass of sugar, \(\displaystyle \mathrm{C_{12}H_{22}O_{11}}\).Add up the atomic masses (C = $\displaystyle 12$ u, H = $\displaystyle 1$ u, O = $\displaystyle 16$ u) for every atom in the formula:\[M = (12 \times 12) + (22 \times 1) + (11 \times 16)\ \text{g mol}^{-1} \] \[M = 144 + 22 + 176 = 342\ \text{g mol}^{-1} \]Step $\displaystyle 2$ — Convert the given mass to moles.Use \(\displaystyle \text{moles} = \dfrac{\text{mass}}{\text{molar mass}}\), where mass is in grams and molar mass is in g mol⁻¹:\[n = \frac{20\ \text{g}}{342\ \text{g mol}^{-1}} = 0.05848\ \text{mol} \]Step $\displaystyle 3$ — Divide by the volume of solution, in litres.Molarity, \(\displaystyle C = \dfrac{n}{V}\), where \(\displaystyle n\) is moles of solute and \(\displaystyle V\) is the volume of the whole solution in litres — this is the step people slip on, because "dissolved in enough water to make a final volume up to $\displaystyle 2$ L" means the solution (sugar + water together) occupies $\displaystyle 2$ L, not that $\displaystyle 2$ L of water was added:\[C = \frac{0.05848\ \text{mol}}{2\ \text{L}} = 0.02924\ \text{mol L}^{-1} \]The data ($\displaystyle 20$ g, $\displaystyle 2$ L) supports three significant figures, so round here, at the end:\[C \approx 0.0292\ \text{mol L}^{-1} \]Answer: The concentration of sugar is \(\displaystyle 0.0292\ \text{mol L}^{-1}\) (i.e. \(\displaystyle 2.92 \times 10^{-2}\ \text{mol L}^{-1}\)).

Molarity is moles of solute per litre of SOLUTION -- so the plan is: use the molarity to find how many moles of methanol are needed, turn those moles into a mass, and then turn that mass into a volume using the density. Density is what lets you swap between "how much mass" and "how much space it occupies."Step $\displaystyle 1$ -- moles of methanol neededMolarity formula: \(\displaystyle M = \dfrac{n}{V} \), where \(\displaystyle M\) is the molarity (mol L\(\displaystyle ^{-1}\)), \(\displaystyle n\) is the number of moles of solute, and \(\displaystyle V\) is the volume of the solution in litres.Rearranging for \(\displaystyle n\): \[n = M \times V = 0.25\ \text{mol L}^{-1} \times 2.5\ \text{L} = 0.625\ \text{mol} \]Step $\displaystyle 2$ -- convert those moles to a massYou need a mass because density relates mass to volume, not moles to volume. Molar mass of methanol, \(\displaystyle \text{CH}_3\text{OH}\): \[M_{\text{molar}} = 12 + 4(1) + 16 = 32\ \text{g mol}^{-1} \]Mass needed: \[m = n \times M_{\text{molar}} = 0.625\ \text{mol} \times 32\ \text{g mol}^{-1} = 20\ \text{g} \]Step $\displaystyle 3$ -- convert that mass to a volume of pure methanol, using its densityDensity formula: \(\displaystyle \rho = \dfrac{m}{V} \), where \(\displaystyle \rho\) is density, \(\displaystyle m\) is mass, \(\displaystyle V\) is volume. Rearranged for the volume of liquid methanol you must measure out: \[V = \frac{m}{\rho} \]The density is given as \(\displaystyle 0.793\ \text{kg L}^{-1}\). The mass you just found is in grams, so the units must match before you divide -- mixing kg with g here is the step people get wrong. Convert: \[0.793\ \text{kg L}^{-1} = 793\ \text{g L}^{-1} = 0.793\ \text{g mL}^{-1} \]Now divide: \[V = \frac{20\ \text{g}}{0.793\ \text{g mL}^{-1}} = 25.22\ \text{mL} \]The density (\(\displaystyle 0.793\), $\displaystyle 3$ significant figures) is the most precise value used in the final division, so the result is rounded to $\displaystyle 3$ significant figures: \(\displaystyle 25.2\ \text{mL}\), which is \(\displaystyle 0.0252\ \text{L}\).This is the volume of liquid methanol you must measure out and then dilute up to the full $\displaystyle 2.5$ L mark with solvent -- not $\displaystyle 2.5$ L of methanol itself.Answer: $\displaystyle 25.2$ mL of methanol (≈ $\displaystyle 0.0252$ L) is needed.

Pressure is force per unit area, and force is not the same as mass — you need to bring in gravity to turn the mass of air into the weight (force) pressing down.The problem gives the mass of air per unit area, not a force per unit area, so the first job is to convert mass into force using Newton's second law, \(\displaystyle F = mg \), where \(\displaystyle m\) is mass and \(\displaystyle g\) is the acceleration due to gravity. Only after that conversion does dividing by area give you a pressure in pascals.Step $\displaystyle 1$: Get everything into SI base units first.The data is given as \(\displaystyle 1034 \text{ g cm}^{-2}\), mixing grams and square centimetres — neither is an SI unit, so both need converting before anything is substituted.Convert grams to kilograms: \[1034 \text{ g} = 1034 \times 10^{-3} \text{ kg} = 1.034 \text{ kg} \]Convert square centimetres to square metres, using \(\displaystyle 1 \text{ cm} = 10^{-2} \text{ m}\), so \(\displaystyle 1 \text{ cm}^2 = 10^{-4} \text{ m}^2\): \[1 \text{ cm}^{-2} = \frac{1}{10^{-4} \text{ m}^2} = 10^{4} \text{ m}^{-2} \]So the mass per unit area becomes \[1034 \text{ g cm}^{-2} = 1.034 \text{ kg} \times 10^{4} \text{ m}^{-2} = 1.034 \times 10^{4} \text{ kg m}^{-2} \]Step $\displaystyle 2$: Turn "mass per area" into "force per area" using \(\displaystyle F = mg\).Take \(\displaystyle g = 9.8 \text{ m s}^{-2}\) for the acceleration due to gravity. Since \(\displaystyle F = mg\), dividing both sides by area \(\displaystyle A\) gives \[\frac{F}{A} = \frac{m}{A} \times g \] which is exactly the pressure you want, because \(\displaystyle \dfrac{m}{A}\) is the quantity already found in Step 1.\[P = \frac{m}{A} \times g = \left(1.034 \times 10^{4} \text{ kg m}^{-2}\right) \times \left(9.8 \text{ m s}^{-2}\right) \]Multiplying the numbers first: \[1.034 \times 9.8 = 10.1332 \]So \[P = 10.1332 \times 10^{4} \text{ kg m}^{-1}\text{s}^{-2} = 1.01332 \times 10^{5} \text{ kg m}^{-1}\text{s}^{-2} \]Step $\displaystyle 3$: Recognize the unit \(\displaystyle \text{kg m}^{-1}\text{s}^{-2} \) as the pascal.A newton is defined as \(\displaystyle 1 \text{ N} = 1 \text{ kg m s}^{-2}\), so \[1 \text{ kg m}^{-1}\text{s}^{-2} = 1 \frac{\text{kg m s}^{-2}}{\text{m}^2} = 1 \text{ N m}^{-2} = 1 \text{ Pa} \] using the definition given in the problem, \(\displaystyle 1 \text{ Pa} = 1 \text{ N m}^{-2}\). This is the step that is easy to skip — without checking that the combined unit collapses to newtons per square metre, the number by itself means nothing.So \[P = 1.01332 \times 10^{5} \text{ Pa} \]Step $\displaystyle 4$: Round to match the precision of the data.The acceleration due to gravity was used as \(\displaystyle 9.8 \text{ m s}^{-2}\) (two significant figures), so the final pressure should not be reported with more precision than that allows. Rounding \(\displaystyle 1.01332 \times 10^{5}\) to three significant figures gives\[P \approx 1.01 \times 10^{5} \text{ Pa} \]This is, reassuringly, very close to standard atmospheric pressure — which is exactly what this calculation represents: the weight of the entire column of air above one square centimetre of sea-level surface.Answer: \(\displaystyle P \approx 1.01 \times 10^{5} \text{ Pa}\)

Mass is measured in kilograms — and since $\displaystyle 2019$, the kilogram is no longer defined by a physical object at all, but by fixing a constant of nature.Mass is the amount of matter present in a body. It must not be confused with weight, which is the force gravity exerts on that mass and which changes from place to place (the Moon, the equator, the poles) — mass itself does not.The SI unit. The SI base unit of mass is the kilogram, symbol \(\displaystyle \text{kg} \). In the laboratory, chemists usually work with the smaller sub-multiple, the gram, since \(\displaystyle 1\ \text{kg} = 1000\ \text{g} \), because the quantities of chemicals handled in an experiment are far smaller than a kilogram.How it used to be defined (retired in $\displaystyle 2019$). For over a century the kilogram was defined by an artefact, not a law of nature: the International Prototype of the Kilogram (IPK), a cylinder of platinum–iridium alloy kept under triple bell jars at the International Bureau of Weights and Measures (BIPM) in Sèvres, France. By definition, that one physical object had a mass of exactly \(\displaystyle 1\ \text{kg} \), and every balance in the world was ultimately calibrated against copies of it.The trouble with this is the mistake students make when they picture "$\displaystyle 1$ kg" as a fixed, unchangeable thing — a physical prototype is not. Cleaning, handling, and contamination over decades of comparisons showed the IPK's mass drifting by tens of micrograms relative to its official copies. A unit that can drift, however slightly, is not a safe foundation for precise science, so metrologists replaced it with something that cannot drift: a fundamental constant.How it is defined now. Since $\displaystyle 20$ May $\displaystyle 2019$, the kilogram is defined by fixing the numerical value of the Planck constant, \(\displaystyle h \), to be exactly \[h = 6.62607015 \times 10^{-34}\ \text{J s (exact, by definition)} \] Writing the joule in terms of base units, \(\displaystyle 1\ \text{J} = 1\ \text{kg m}^2\text{s}^{-2} \), so this fixes the kilogram as \[1\ \text{kg} = \frac{h}{6.62607015\times10^{-34}\ \text{m}^2\text{s}^{-1}} \] Here \(\displaystyle h \) is Planck's constant (the proportionality constant linking a photon's energy to its frequency, \(\displaystyle E = h\nu \)), and the metre and second on the right-hand side are themselves already fixed exactly by the speed of light, \(\displaystyle c \), and the frequency of a specific transition of caesium-133. So once \(\displaystyle c \), the caesium frequency, and \(\displaystyle h \) are all fixed, the kilogram is fixed too — purely by constants of nature, with no object anyone could scratch, contaminate, or lose. In practice, a device called a Kibble (watt) balance is used to realise this definition, by balancing the weight of a test mass against an electromagnetic force computed from \(\displaystyle h \).**Answer: The SI unit of mass is the kilogram (kg). It is now defined by fixing the value of the Planck constant, \(\displaystyle h = 6.62607015 \times 10^{-34}\ \text{J s} \) exactly (since $\displaystyle 20$ May $\displaystyle 2019$), replacing the earlier definition based on the platinum–iridium International Prototype Kilogram kept at Sèvres, France.

A metric prefix is just a shorthand name for a power of ten — matching it correctly means recalling the SI exponent that name is defined to mean, nothing more.Every one of these prefixes is fixed by the SI (Système International) convention, so there is nothing to calculate — only the definition to state for each, and then read off the exponent.(i) micro The prefix micro (symbol \(\displaystyle \mu \)) is defined as \[\text{micro} = 10^{-6} \] So \(\displaystyle 1\ \mu\text{g} = 10^{-6}\ \text{g} \), \(\displaystyle 1\ \mu\text{m} = 10^{-6}\ \text{m} \), and so on.Aside — this is the pair people mix up: micro and mega sound alike but sit twelve orders of magnitude apart. Micro (\(\displaystyle 10^{-6}\)) makes a unit smaller; mega (\(\displaystyle 10^{6}\)) makes it a million times bigger. Reading one for the other silently flips a millionth into a million.(ii) deca The prefix deca (symbol da) is defined as \[\text{deca} = 10^{1} \] So \(\displaystyle 1\ \text{dag} = 10\ \text{g} \). This is the smallest of all the standard prefixes above the base unit — do not confuse deca (\(\displaystyle 10^{1}\), a multiplying prefix) with the similarly-spelled deci (\(\displaystyle 10^{-1}\), a dividing prefix); they differ only by one letter but point in opposite directions.(iii) mega The prefix mega (symbol M) is defined as \[\text{mega} = 10^{6} \] So \(\displaystyle 1\ \text{MW} = 10^{6}\ \text{W} \) (one megawatt is a million watts).(iv) giga The prefix giga (symbol G) is defined as \[\text{giga} = 10^{9} \] So \(\displaystyle 1\ \text{GHz} = 10^{9}\ \text{Hz} \) — giga sits three steps of \(\displaystyle 10^3\) above mega.(v) femto The prefix femto (symbol f) is defined as \[\text{femto} = 10^{-15} \] So \(\displaystyle 1\ \text{fs} = 10^{-15}\ \text{s} \). Femto is far smaller than micro or nano; it is not to be confused with pico (\(\displaystyle 10^{-12}\)), which is three orders of magnitude larger.Collecting the five matches:

Prefix	Multiple
micro	\(\displaystyle 10^{-6}\)
deca	\(\displaystyle 10^{1}\)
mega	\(\displaystyle 10^{6}\)
giga	\(\displaystyle 10^{9}\)
femto	\(\displaystyle 10^{-15}\)

Answer: micro \(\displaystyle = 10^{-6}\), deca \(\displaystyle = 10^{1}\), mega \(\displaystyle = 10^{6}\), giga \(\displaystyle = 10^{9}\), femto \(\displaystyle = 10^{-15}\).

Significant figures are the digits in a measured (or calculated) quantity that are known with certainty, plus one digit that is estimated. They exist because no measurement is ever exact — every reading has some uncertainty in its last digit, and significant figures are how that reliability is communicated honestly. A quantity written with more digits than are actually known is just as wrong as one written with too few.To decide how many significant figures a number has, six rules are used.1. All non-zero digits are significant. \(\displaystyle 285\ \text{cm}\) has $\displaystyle 3$ significant figures; \(\displaystyle 4.32\ \text{g}\) has $\displaystyle 3$ significant figures.2. Zeros between two non-zero digits are significant. \(\displaystyle 508\) has $\displaystyle 3$ significant figures — the zero is "trapped" between the $\displaystyle 5$ and the $\displaystyle 8$, so it is a measured digit, not a placeholder.3. Leading zeros (zeros to the left of the first non-zero digit) are never significant — they only fix the position of the decimal point. \(\displaystyle 0.00508\ \text{g}\) has $\displaystyle 3$ significant figures, not 6. This is the step people get wrong most often: it is tempting to count every digit typed, but a zero whose only job is to say "this is a small number" carries no measurement information.4. Trailing zeros in a number with no decimal point are ambiguous. \(\displaystyle 100\ \text{m}\) could mean $\displaystyle 1$, $\displaystyle 2$, or $\displaystyle 3$ significant figures — there is no way to tell just from the digits. This ambiguity is exactly why scientific notation is preferred for reporting data: writing the number as \(\displaystyle N \times 10^{n}\), where \(\displaystyle 1 \le N < 10\), puts every significant digit into \(\displaystyle N\) and leaves none of them hidden. \(\displaystyle 100\ \text{m}\) written as \(\displaystyle 1 \times 10^{2}\ \text{m}\) has $\displaystyle 1$ significant figure, while \(\displaystyle 1.00 \times 10^{2}\ \text{m}\) has $\displaystyle 3$ — the notation removes the guesswork.5. Trailing zeros in a number that does have a decimal point are significant. \(\displaystyle 0.500\ \text{kg}\) has $\displaystyle 3$ significant figures, not $\displaystyle 1$ — the two trailing zeros after the $\displaystyle 5$ were deliberately written, so they assert that the measurement is trustworthy out to the third digit (unlike a leading zero, a trailing zero after a decimal point is a choice, not a placeholder).6. Exact (counted or defined) numbers have an infinite number of significant figures. The "$\displaystyle 2$" in "$\displaystyle 2$ balls" or the "$\displaystyle 60$" in "$\displaystyle 60$ minutes = $\displaystyle 1$ hour" is not a measurement at all, so it never limits the precision of a calculation.These rules also govern arithmetic, because a calculated answer cannot be more precise than the data used to get it: For example, \(\displaystyle 12.11 + 18.0 + 1.012 = 31.122\), but since \(\displaystyle 18.0\) is known only to one decimal place, the answer is reported as \(\displaystyle 31.1\) — the extra decimal places in the sum are not real information, only arithmetic residue.**Answer: Significant figures are the meaningful digits in a number — all certain digits plus one estimated (uncertain) digit — found by the rules: every non-zero digit counts; zeros between non-zero digits count; leading zeros never count; trailing zeros count only if a decimal point is present (otherwise they are ambiguous, which scientific notation \(\displaystyle N \times 10^{n}\) resolves); and exact/counted numbers have infinitely many significant figures.

Parts per million works exactly like percent, just measured against a millionth instead of a hundredth — and molality is always moles of solute per kilogram of solvent, never per kilogram of the whole solution.Converting ppm to percent by mass"$\displaystyle 15$ ppm by mass" means $\displaystyle 15$ units of mass of CHCl₃ for every \(\displaystyle 10^{6}\) units of mass of the solution (the contaminated water): \[\text{ppm (by mass)} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 10^{6} \] Percent by mass is the identical ratio, scaled by $\displaystyle 100$ instead of \(\displaystyle 10^{6}\): \[\%\text{ by mass} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100 \] So going from ppm to percent is just changing the scaling factor, \(\displaystyle \times \dfrac{100}{10^{6}}\): \[\%\text{ by mass} = 15 \times \frac{100}{10^{6}} = \frac{1500}{10^{6}} = 1.5\times10^{-3}\,\% \]Finding the molalityMolality is moles of solute per kilogram of solvent — the water by itself, not the water-plus-chloroform mixture. At $\displaystyle 15$ ppm the two masses are almost identical, but the formula still has to use the solvent's mass, not the solution's, so it is worth keeping them separate.Take a convenient sample: \(\displaystyle 10^{6}\ \text{g}\) of this water. By the definition of ppm, that sample contains: (the $\displaystyle 15$ g taken out is a $\displaystyle 0.0015$% correction to the solvent mass — too small to affect the answer at the precision we're working to, so $\displaystyle 1000$ kg is used below)Molar mass of CHCl₃, \(\displaystyle M\): \[M = 12.0\,(\text{C}) + 1.0\,(\text{H}) + 3\times35.5\,(\text{Cl}) = 12.0+1.0+106.5 = 119.5\ \text{g mol}^{-1} \]Moles of CHCl₃ in the sample, \(\displaystyle n = \dfrac{\text{mass}}{M}\): \[n = \frac{15\ \text{g}}{119.5\ \text{g mol}^{-1}} = 0.125523\ \text{mol} \]Molality, \(\displaystyle m\) = moles of solute per kilogram of solvent, \(\displaystyle w_{\text{solvent}}\): \[m = \frac{n}{w_{\text{solvent}}(\text{kg})} = \frac{0.125523\ \text{mol}}{1000\ \text{kg}} = 1.25523\times10^{-4}\ \text{mol kg}^{-1} \]The data ("$\displaystyle 15$ ppm") and the molar-mass arithmetic together justify three significant figures, so this rounds up (the digit after the third significant figure is $\displaystyle 5$, followed by more non-zero digits): \[m \approx 1.26\times10^{-4}\ \text{mol kg}^{-1} \]Answer: (i) $\displaystyle 1.5$ × $\displaystyle 10$⁻³ % by mass; (ii) molality ≈ $\displaystyle 1.26$ × $\displaystyle 10$⁻⁴ mol kg⁻¹

Scientific notation writes any number as \(\displaystyle N \times 10^{n} \), where \(\displaystyle N\) is a number with exactly one non-zero digit before the decimal point (\(\displaystyle 1 \le N < 10\)) and \(\displaystyle n\) is an integer. To find \(\displaystyle n\), count how many places the decimal point moves to land just after that first non-zero digit — moving it left (for a number bigger than $\displaystyle 10$) gives a positive \(\displaystyle n\); moving it right (for a number smaller than $\displaystyle 1$) gives a negative \(\displaystyle n\).A step people skip: significant zeros must be carried into \(\displaystyle N\), not dropped — a trailing zero after the decimal point (as in $\displaystyle 500.0$) or a zero in the middle of the digit string (as in $\displaystyle 8008$) is a measured digit, so it stays.(i) $\displaystyle 0.0048$The decimal point sits before the first non-zero digit, 4. Move it right until it is right after that $\displaystyle 4$: \[0.0048 \rightarrow 004.8 \times 10^{-3} \] The point moved $\displaystyle 3$ places right, so the exponent is \(\displaystyle -3\), and dropping the leading zeros in \(\displaystyle N\) gives \[0.0048 = 4.8 \times 10^{-3} \](ii) $\displaystyle 234,000$The first non-zero digit is 2. Move the (implied) decimal point left until it sits right after the $\displaystyle 2$ — that is $\displaystyle 5$ places: \[234000 \rightarrow 2.34000 \times 10^{5} \] The trailing zeros after "$\displaystyle 234$" only fix the point's original position; once the point has moved past them they carry no separate significance, so they need not be written in \(\displaystyle N\): \[234000 = 2.34 \times 10^{5} \](iii) $\displaystyle 8008$First non-zero digit is 8. Moving the decimal point left $\displaystyle 3$ places puts it right after that first $\displaystyle 8$: \[8008 \rightarrow 8.008 \times 10^{3} \] Here the two zeros sit between significant digits ($\displaystyle 8$ and $\displaystyle 8$), so they must stay in \(\displaystyle N\): \[8008 = 8.008 \times 10^{3} \](iv) $\displaystyle 500.0$First non-zero digit is 5. Moving the decimal point left $\displaystyle 2$ places puts it right after the $\displaystyle 5$: \[500.0 \rightarrow 5.000 \times 10^{2} \] The trailing zero after the decimal point in "$\displaystyle 500.0$" is written deliberately (it distinguishes this quantity from a bare "$\displaystyle 500$"), so it is a significant digit and must be kept in \(\displaystyle N\) as one of the trailing zeros: \[500.0 = 5.000 \times 10^{2} \](v) $\displaystyle 6.0012$The decimal point is already right after the first non-zero digit, $\displaystyle 6$, so no shift is needed and the exponent is \(\displaystyle 0\): \[6.0012 = 6.0012 \times 10^{0} \]Answer: (i) \(\displaystyle 4.8 \times 10^{-3}\) (ii) \(\displaystyle 2.34 \times 10^{5}\) (iii) \(\displaystyle 8.008 \times 10^{3}\) (iv) \(\displaystyle 5.000 \times 10^{2}\) (v) \(\displaystyle 6.0012 \times 10^{0}\)

Significant figures are the digits that carry real precision — and zeros are the trap: some of them are just placeholders that fix the position of the decimal point, and those never count.Four rules sort every digit into "counts" or "doesn't count":1. Every non-zero digit is significant. 2. A zero sitting between two non-zero digits (a "captive" zero) is significant. 3. A zero to the left of the first non-zero digit (a leading zero) is never significant — it only tells you where the decimal point is, it carries no precision. 4. A trailing zero (at the right end of the number) is significant only if the number is written with a decimal point. Without a decimal point, a trailing zero is ambiguous — it might be a measured digit or just padding to show magnitude — so by convention it is not counted.Rule $\displaystyle 3$ vs. Rule $\displaystyle 4$ is exactly where people lose marks: a zero's position relative to the first non-zero digit and the decimal point decides everything, not just "is it a zero."(i) \(\displaystyle 0.0025 \) The digits are \(\displaystyle 0,0,0,2,5\). The three zeros before the \(\displaystyle 2\) are leading zeros (Rule $\displaystyle 3$) — they only fix the decimal point's position, so they don't count. Only \(\displaystyle 2\) and \(\displaystyle 5\) are significant. \[0.0025 \;\rightarrow\; 2 \text{ significant figures} \](ii) \(\displaystyle 208 \) Digits: \(\displaystyle 2,0,8\). The \(\displaystyle 0\) sits between the \(\displaystyle 2\) and the \(\displaystyle 8\), so it is a captive zero (Rule $\displaystyle 2$) and counts. \[208 \;\rightarrow\; 3 \text{ significant figures} \](iii) \(\displaystyle 5005 \) Digits: \(\displaystyle 5,0,0,5\). Both zeros are captive — each sits between two non-zero digits (the second \(\displaystyle 0\) is between the first \(\displaystyle 0\) and the final \(\displaystyle 5\), and the whole pair sits between the two \(\displaystyle 5\)'s) — so both count. \[5005 \;\rightarrow\; 4 \text{ significant figures} \](iv) \(\displaystyle 126000 \) Digits: \(\displaystyle 1,2,6,0,0,0\). \(\displaystyle 1\), \(\displaystyle 2\), and \(\displaystyle 6\) are non-zero, so they count. The three trailing zeros come after the last non-zero digit, and the number is written with no decimal point — so by Rule $\displaystyle 4$ these trailing zeros are ambiguous and are not counted as significant. \[126000 \;\rightarrow\; 3 \text{ significant figures} \](v) \(\displaystyle 500.0 \) Digits: \(\displaystyle 5,0,0,0\). This looks like the same shape as (iv), but the decimal point after the last zero changes the rule completely: writing "\(\displaystyle .0\)" is how you tell the reader the trailing zeros were measured, not just filler. By Rule $\displaystyle 4$, because a decimal point is present, every trailing zero is significant. \[500.0 \;\rightarrow\; 4 \text{ significant figures} \](vi) \(\displaystyle 2.0034 \) Digits: \(\displaystyle 2,0,0,3,4\). The digit \(\displaystyle 2\) is non-zero (significant). The two zeros immediately after it sit between the \(\displaystyle 2\) and the \(\displaystyle 3\) — they are captive zeros (Rule $\displaystyle 2$) — so they count too. Then \(\displaystyle 3\) and \(\displaystyle 4\) are non-zero. \[2.0034 \;\rightarrow\; 5 \text{ significant figures} \]Answer: (i) $\displaystyle 2$ (ii) $\displaystyle 3$ (iii) $\displaystyle 4$ (iv) $\displaystyle 3$ (v) $\displaystyle 4$ (vi) $\displaystyle 5$ significant figures.

Rounding to a fixed number of significant figures means keeping exactly that many digits counted from the first non-zero digit, and using the very next digit only to decide whether the last kept digit stays the same or increases by one.The rule: look at the digit immediately after the last significant figure you are keeping. Leading zeros (before the first non-zero digit) are never significant, so they are never counted when you count "three figures," but they still have to stay in the answer to hold the decimal point in place.(i) $\displaystyle 34.216$The digits, in order, are \(\displaystyle 3, 4, 2, 1, 6\). The first three significant figures are \(\displaystyle 3, 4, 2\). The next digit is \(\displaystyle 1\), which is less than $\displaystyle 5$, so the third figure stays as it is.\[34.216 \rightarrow 34.2 \](ii) $\displaystyle 10.4107$The significant digits are \(\displaystyle 1, 0, 4, 1, 0, 7\) (this leading $\displaystyle 1$ and the internal zero both count — a zero between non-zero digits, or a non-zero leading digit, is always significant). Keeping the first three gives \(\displaystyle 1, 0, 4\). The next digit is \(\displaystyle 1\), less than $\displaystyle 5$, so nothing changes.\[10.4107 \rightarrow 10.4 \](iii) $\displaystyle 0.04597$Here the two zeros right after the decimal point (\(\displaystyle 0.0\ldots\)) are leading zeros — they only place the decimal point and are not significant. The significant digits start at the first non-zero digit: \(\displaystyle 4, 5, 9, 7\). Keeping three of them gives \(\displaystyle 4, 5, 9\), with the next digit \(\displaystyle 7\), which is $\displaystyle 5$ or more, so the last kept digit \(\displaystyle 9\) rounds up by 1. This is the step people miss: rounding up a $\displaystyle 9$ carries into the digit before it, \(\displaystyle 59 + 1 = 60\), so the three-figure block becomes \(\displaystyle 460\) written back in its original place:\[0.04597 \rightarrow 0.0460 \]The trailing zero here is required — it is the third significant figure, not decoration, and dropping it would silently claim only two figures of precision.(iv) $\displaystyle 2808$The digits are \(\displaystyle 2, 8, 0, 8\). The first three significant figures are \(\displaystyle 2, 8, 0\). The next digit is \(\displaystyle 8\), which is $\displaystyle 5$ or more, so the last kept digit \(\displaystyle 0\) rounds up by $\displaystyle 1$, giving \(\displaystyle 281\). Since the original number is a whole number in the thousands, this three-figure value has to be scaled back up to the right place — one more digit is needed to mark where the number actually ends, so a zero is appended to hold that place:\[2808 \rightarrow 2810 \]That final zero is a placeholder fixing the magnitude, not a fourth significant figure.Answer: (i) $\displaystyle 34.2$ (ii) $\displaystyle 10.4$ (iii) $\displaystyle 0.0460$ (iv) $\displaystyle 2810$