NCERT Solutions for Class 11 Chemistry Chapter 1: Some Basic Concepts of Chemistry

Concept. The molar mass is the mass of one mole (6.022 × 10²³ entities) of a substance in grams. It is numerically equal to the molecular mass in atomic mass units (u): add up the atomic masses of all the atoms in the molecule, multiplying each by the number of atoms of that element. We use the NCERT atomic masses H = 1.008 u, C = 12.011 u, O = 16.00 u.

Molar mass (g mol⁻¹) = Σ (number of atoms × atomic mass)

1. (i) H₂O: 2 × 1.008 + 1 × 16.00 = 2.016 + 16.00 = 18.016 ≈ 18.02 g mol⁻¹.
2. (ii) CO₂: 1 × 12.011 + 2 × 16.00 = 12.011 + 32.00 = 44.011 ≈ 44.01 g mol⁻¹.
3. (iii) CH₄: 1 × 12.011 + 4 × 1.008 = 12.011 + 4.032 = 16.043 ≈ 16.04 g mol⁻¹.

Extra depth. For ionic or network solids that do not contain discrete molecules (e.g. NaCl), the same sum is called the formula mass. Molar masses are the bridge between the mass that is weighed in the laboratory and the number of particles that react, so they appear in every stoichiometric calculation. The values agree with the chapter’s own examples: water 18.02 u and methane 16.043 u.

Conclusion: The molar masses are H₂O = 18.02 g mol⁻¹, CO₂ = 44.01 g mol⁻¹ and CH₄ = 16.04 g mol⁻¹.

Concept. Mass per cent (percentage composition) of an element in a compound is the mass of that element present in one mole of the compound divided by the molar mass of the compound, multiplied by 100. The percentages of all the elements must add up to 100%. It tells us how much of each element a pure sample contains and so helps to check the purity of a sample.

Mass % of an element = (mass of the element in 1 mol of compound / molar mass of compound) × 100

Step 1 — molar mass of Na₂SO₄. With Na = 22.99, S = 32.06, O = 16.00: M = 2(22.99) + 32.06 + 4(16.00) = 45.98 + 32.06 + 64.00 = 142.04 g mol⁻¹.

1. Sodium: (45.98 / 142.04) × 100 = 32.37%.
2. Sulphur: (32.06 / 142.04) × 100 = 22.57%.
3. Oxygen: (64.00 / 142.04) × 100 = 45.06%.

Extra depth. Check: 32.37 + 22.57 + 45.06 = 100.00%. The same method gives the composition of water (H 11.18%, O 88.79%) and ethanol (C 52.14%, H 13.13%, O 34.73%) shown in the chapter. If the percentages are known, the empirical formula follows by the reverse process (Ex 1.3).

Conclusion: Sodium sulphate contains 32.37% sodium, 22.57% sulphur and 45.06% oxygen by mass.

Concept. The empirical formula gives the simplest whole-number ratio of the atoms in a compound. It is found from the percentage composition by converting the mass of each element in a 100 g sample into moles, dividing by the smallest number of moles and, if necessary, multiplying to get whole numbers.

1. Step 1 — masses in 100 g: Fe = 69.9 g, O = 30.1 g.
2. Step 2 — moles: Fe = 69.9 / 55.85 = 1.25 mol; O = 30.1 / 16.00 = 1.88 mol.
3. Step 3 — divide by the smallest (1.25): Fe = 1.00; O = 1.88 / 1.25 = 1.50.
4. Step 4 — convert to whole numbers: Fe : O = 1 : 1.5. Multiply both by 2 → Fe : O = 2 : 3.
5. Step 5 — empirical formula: Fe₂O₃.

Extra depth. A ratio ending in .5 means that every quantity must be doubled; ratios ending in .33 or .67 must be tripled, and .25 or .75 quadrupled. The oxide is iron(III) oxide (haematite). Its molecular formula is also Fe₂O₃ as shown in Ex 1.8.

Conclusion: The empirical formula of the iron oxide is Fe₂O₃.

Concept. Carbon burns in oxygen according to C(s) + O₂(g) → CO₂(g), i.e. 1 mol C (12 g) + 1 mol O₂ (32 g) → 1 mol CO₂ (44 g). When the oxygen supply is limited, the reagent present in the smaller stoichiometric amount (the limiting reagent) decides how much CO₂ forms.

1. (i) 1 mole of carbon burnt in air. Air provides excess O₂, so carbon is the limiting reagent. 1 mol C → 1 mol CO₂ = 44 g (1 mol).
2. (ii) 1 mol C in 16 g O₂. 16 g O₂ = 16/32 = 0.5 mol. For 1 mol C, 1 mol O₂ is needed; only 0.5 mol is available, so O₂ is the limiting reagent. CO₂ formed = 0.5 mol = 0.5 × 44 = 22 g.
3. (iii) 2 mol C in 16 g O₂. Again 0.5 mol O₂ is available while 2 mol O₂ would be needed; O₂ is limiting, giving 0.5 mol CO₂ = 22 g; 1.5 mol C remains unreacted.

Extra depth. In (iii) doubling the carbon does not increase the product because the product is controlled by the limiting reagent, not by the excess one. (In real life, insufficient oxygen would also produce some CO, but the NCERT question assumes complete conversion to CO₂.)

Conclusion: The CO₂ produced is (i) 44 g (1 mol), (ii) 22 g (0.5 mol) and (iii) 22 g (0.5 mol), because O₂ is the limiting reagent in (ii) and (iii).

Concept. Molarity (M) is the number of moles of solute dissolved in one litre of solution: M = n / V(L). So the amount (moles) of solute required is M × V, and the mass is moles × molar mass.

1. Moles needed: n = M × V = 0.375 mol L⁻¹ × 0.500 L = 0.1875 mol.
2. Mass: m = n × M_r = 0.1875 mol × 82.0245 g mol⁻¹ = 15.38 g.

Mass = M × V(L) × molar mass = 0.375 × 0.500 × 82.0245 = 15.38 g

Extra depth. The solution is prepared by dissolving 15.38 g of sodium acetate in some water and making the volume up to exactly 500 mL in a volumetric flask (not by adding 500 mL of water, since the solute also occupies volume). Because molarity is based on solution volume, it changes slightly with temperature.

Conclusion: 15.38 g of sodium acetate is required to make 500 mL of 0.375 M solution.

Concept. To convert mass per cent into molarity, take a convenient volume (1 L of solution), use the density to find the mass of the solution, use the mass per cent to find the mass of solute in it, and convert that mass to moles using the molar mass.

1. Mass of 1 L (1000 mL) solution = 1000 mL × 1.41 g mL⁻¹ = 1410 g.
2. Mass of HNO₃ in it = 1410 g × 69/100 = 972.9 g.
3. Molar mass of HNO₃ = 1.008 + 14.01 + 3(16.00) = 63.01 g mol⁻¹.
4. Moles of HNO₃ = 972.9 g / 63.01 g mol⁻¹ = 15.44 mol.
5. Since this amount is in 1 L, the molarity = 15.44 mol L⁻¹.

M = (w/w % × 10 × d) / M_r = (69 × 10 × 1.41) / 63.01 = 15.44 mol L⁻¹

Extra depth. This is the usual concentrated nitric acid of the laboratory (about 15.4 M). The shortcut formula M = (per cent × 10 × density)/molar mass is obtained from exactly the steps above. The molality of the same acid would be much higher (about 35 m) because it is moles per kg of solvent.

Conclusion: The concentration of the nitric acid sample is 15.44 mol L⁻¹.

Concept. One mole of CuSO₄ contains one mole of copper. Therefore, the mass of copper that can be obtained from a given mass of copper sulphate equals the mass per cent of copper in the compound multiplied by that mass — or, in mole terms, 1 mol CuSO₄ → 1 mol Cu.

1. Molar mass of CuSO₄ = 63.55 + 32.07 + 4(16.00) = 159.62 g mol⁻¹.
2. Moles in 100 g = 100 / 159.62 = 0.6265 mol.
3. Moles of Cu obtainable = 0.6265 mol (1 : 1).
4. Mass of Cu = 0.6265 mol × 63.55 g mol⁻¹ = 39.81 g.

Mass of Cu = (63.55 / 159.62) × 100 g = 39.81 g

Extra depth. The answer assumes that all the copper is recovered (100% yield). In practice, copper is obtained from copper sulphate solution by displacement with iron or by electrolysis, and the actual yield is a little less. The result again uses the idea of mass percentage composition.

Conclusion: About 39.8 g of copper can be obtained from 100 g of CuSO₄.

Concept. The molecular formula is a whole-number multiple (n) of the empirical formula: molecular formula = (empirical formula)_n with n = molar mass / empirical formula mass. In Ex 1.3 the empirical formula of this oxide was found to be Fe₂O₃. The printed question does not give the molar mass of the oxide; the standard textbook solution takes the molar mass of the oxide as 159.69 g mol⁻¹, which is what is used below.

1. Empirical formula (Ex 1.3): Fe₂O₃.
2. Empirical formula mass = 2(55.85) + 3(16.00) = 111.70 + 48.00 = 159.70 g mol⁻¹.
3. n = molar mass / empirical formula mass = 159.69 / 159.70 ≈ 1.
4. Molecular formula = (Fe₂O₃)₁ = Fe₂O₃.

Extra depth. Without the molar mass, only the empirical formula can be obtained from percentage composition; a different molar mass (e.g. 319.4 g mol⁻¹) would give a different molecular formula (Fe₄O₆). Iron(III) oxide is an ionic solid whose formula unit is Fe₂O₃, so n = 1 is also the chemically sensible answer.

Conclusion: The molecular formula of the oxide is Fe₂O₃ (same as its empirical formula, n = 1).

Concept. Naturally occurring elements are mixtures of isotopes. The average atomic mass is the weighted average of the isotopic masses, each multiplied by its fractional abundance. This is the value printed in the periodic table.

Average atomic mass = Σ (fractional abundance × isotopic mass)

1. Contribution of ³⁵Cl = 0.7577 × 34.9689 = 26.4959.
2. Contribution of ³⁷Cl = 0.2423 × 36.9659 = 8.9568.
3. Average atomic mass = 26.4959 + 8.9568 = 35.4527 ≈ 35.45 u (g mol⁻¹).

Extra depth. The value 35.45 lies closer to 35 than to 37 because ³⁵Cl is about three times as abundant as ³⁷Cl. The same method gave 12.011 u for carbon in the chapter and is used for argon in Ex 1.32. No individual chlorine atom has a mass of 35.45 u; this is only an average.

Conclusion: The average atomic mass of chlorine is 35.45 u.

Concept. The subscripts in a formula give the number of atoms of each element per molecule, and hence the number of moles of atoms per mole of molecules. One mole of any substance contains N_A = 6.022 × 10²³ entities.

1. (i) Moles of C atoms: 1 mol C₂H₆ contains 2 mol C; so 3 mol contain 3 × 2 = 6 mol of carbon atoms.
2. (ii) Moles of H atoms: 1 mol C₂H₆ contains 6 mol H; so 3 mol contain 3 × 6 = 18 mol of hydrogen atoms.
3. (iii) Number of molecules: 3 mol × 6.022 × 10²³ molecules mol⁻¹ = 1.807 × 10²⁴ molecules (3 × 6.022 × 10²³ = 18.066 × 10²³).

Extra depth. The number of C atoms is 6 × 6.022 × 10²³ = 3.613 × 10²⁴ and of H atoms 18 × 6.022 × 10²³ = 1.084 × 10²⁵. Mass of 3 mol ethane = 3 × 30.07 = 90.2 g.

Conclusion: Three moles of ethane contain 6 mol of C atoms, 18 mol of H atoms and 1.807 × 10²⁴ molecules.

Concept. Concentration in mol L⁻¹ is the molarity: moles of solute divided by the volume of solution in litres. First convert the 20 g of sugar to moles using its molar mass.

1. Molar mass of C₁₂H₂₂O₁₁ = 12(12.011) + 22(1.008) + 11(16.00) = 144.13 + 22.18 + 176.00 = 342.3 g mol⁻¹.
2. Moles of sugar = 20 g / 342.3 g mol⁻¹ = 0.0584 mol.
3. Molarity = 0.0584 mol / 2 L = 0.0292 mol L⁻¹ (2.92 × 10⁻² M).

M = (20 g / 342.3 g mol⁻¹) / 2 L = 0.0292 mol L⁻¹

Extra depth. The 2 L is the total volume of the solution, not the volume of water added; that is why the problem says “enough water to make a final volume up to 2 L”. A solution this dilute (0.029 M) contains about 10 g of sugar per litre (1%).

Conclusion: The concentration of the sugar solution is 0.0292 mol L⁻¹.

Concept. The volume of a pure liquid solute is obtained from its mass and density (volume = mass/density). The mass in turn comes from the number of moles needed for the required molarity and volume of solution.

1. Moles needed = M × V = 0.25 mol L⁻¹ × 2.5 L = 0.625 mol.
2. Molar mass of CH₃OH = 12.011 + 4(1.008) + 16.00 = 32.04 g mol⁻¹.
3. Mass of methanol = 0.625 mol × 32.04 g mol⁻¹ = 20.03 g.
4. Density = 0.793 kg L⁻¹ = 0.793 g mL⁻¹.
5. Volume = mass / density = 20.03 g / 0.793 g mL⁻¹ = 25.3 mL (25.25 mL).

V = (M × V_solution × M_r) / d = (0.25 × 2.5 × 32.04) / 0.793 = 25.3 mL

Extra depth. Note that 0.793 kg L⁻¹ and 0.793 g mL⁻¹ are numerically equal. In practice 25.3 mL of methanol is placed in a 2.5 L volumetric flask and made up to the mark with water.

Conclusion: About 25.3 mL of methanol is needed to prepare 2.5 L of 0.25 M solution.

Concept. Pressure = force / area. The “mass of air at sea level is 1034 g per cm²” means that a column of air above each square centimetre has this mass; its weight (force) = mass × g, where g = 9.8 m s⁻². The pressure in pascal needs the force in newton and the area in m².

1. Mass per cm² = 1034 g = 1.034 kg.
2. Force per cm² = m g = 1.034 kg × 9.8 m s⁻² = 10.13 N.
3. Area = 1 cm² = (10⁻² m)² = 10⁻⁴ m².
4. Pressure = 10.13 N / 10⁻⁴ m² = 1.013 × 10⁵ N m⁻² = 1.01 × 10⁵ Pa.

P = (1.034 kg × 9.8 m s⁻²) / 10⁻⁴ m² = 1.013 × 10⁵ Pa

Extra depth. This is the standard atmospheric pressure: 1 atm = 1.01325 × 10⁵ Pa (about 1.013 bar). The standard pressure for reporting gas properties (STP) is now taken as exactly 10⁵ Pa (1 bar).

Conclusion: The pressure of the air at sea level is about 1.01 × 10⁵ Pa.

Concept. Mass is the amount of matter in a body and is independent of location. The SI base unit of mass is the kilogram (kg); its fraction, the gram (1 kg = 1000 g), is used in the laboratory for smaller amounts.

Definition. The kilogram, symbol kg, is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015 × 10⁻³⁴ when expressed in the unit J s, which is equal to kg m² s⁻¹, where the metre and the second are defined in terms of the speed of light c and the caesium frequency ν_Cs.

• Earlier definition: since 1889 the kilogram was the mass of a platinum–iridium (Pt–Ir) cylinder kept in an airtight jar at the International Bureau of Weights and Measures at Sèvres, France; Pt–Ir was chosen because it is highly resistant to chemical attack and its mass does not change.
• Why change? A definition in terms of a physical constant is universal, reproducible and does not depend on an artefact that can lose or gain mass.

Extra depth. Mass (constant) differs from weight (the force of gravity on the mass, which varies from place to place). Mass in the laboratory is determined with an analytical balance. National standards are kept by the National Physical Laboratory (NPL), New Delhi.

Conclusion: The SI unit of mass is the kilogram, defined by fixing the Planck constant at 6.62607015 × 10⁻³⁴ J s (formerly the mass of a Pt–Ir cylinder kept at Sèvres).

Concept. SI prefixes are used with a unit to denote multiples or submultiples by powers of ten (Table 1.3). Remember the order of the commonly used ones: femto (10⁻¹⁵), pico (10⁻¹²), nano (10⁻⁹), micro (10⁻⁶), milli (10⁻³), centi (10⁻²), deci (10⁻¹), deca (10), hecto (10²), kilo (10³), mega (10⁶), giga (10⁹), tera (10¹²).

• micro (μ) — one millionth: 1 μm = 10⁻⁶ m.
• deca (da) — ten times: 1 dam = 10 m.
• mega (M) — one million: 1 MHz = 10⁶ Hz.
• giga (G) — one billion (10⁹): 1 GHz = 10⁹ Hz.
• femto (f) — 10⁻¹⁵: 1 fs = 10⁻¹⁵ s.

Extra depth. Prefixes make chemical quantities convenient: bond lengths in pm (10⁻¹² m), wavelengths in nm (10⁻⁹ m), enthalpies in kJ (10³ J).

Conclusion: micro = 10⁻⁶, deca = 10, mega = 10⁶, giga = 10⁹, femto = 10⁻¹⁵.

Concept. Every experimental measurement has some uncertainty because of the limitations of the measuring instrument and the skill of the observer. For example, a platform balance may give a mass of 9.4 g while an analytical balance gives 9.4213 g for the same object: the extra digits are known more reliably by the better instrument. The uncertainty is indicated by the number of significant figures.

Definition. Significant figures are the meaningful digits in a measured or calculated quantity that are known with certainty, plus one digit that is estimated or uncertain. If a result is reported as 11.2 mL, the digits 1 and 1 are certain and the last 2 is uncertain (an uncertainty of ±1 in the last digit is understood unless stated).

Rules for counting significant figures

1. All non-zero digits are significant (285 cm → 3; 0.25 mL → 2).
2. Zeros preceding the first non-zero digit are not significant; they only locate the decimal point (0.03 → 1; 0.0052 → 2).
3. Zeros between two non-zero digits are significant (2.005 → 4).
4. Terminal zeros are significant only if they are to the right of the decimal point (0.200 g → 3; 100.0 → 4; 100 → 1). Such numbers are best written in scientific notation (1 × 10², 1.0 × 10², 1.00 × 10²).
5. Exact numbers (counting, e.g. 20 eggs) have an infinite number of significant figures; in scientific notation all digits shown are significant (4.01 × 10² → 3).

Extra depth. Significant figures are linked with precision (agreement among repeated measurements) and accuracy (agreement with the true value), which are illustrated by students A, B and C in Table 1.4.

Conclusion: Significant figures are the certain digits of a measurement plus one estimated digit; they indicate the reliability of the measurement and must be correctly carried through calculations.

Concept. ppm (parts per million) by mass is the number of grams of solute per 10⁶ g of solution: ppm = (mass of solute / mass of solution) × 10⁶. Percent by mass uses a basis of 100 g. Molality needs moles of solute per kilogram of solvent.

1. (i) Per cent by mass. 15 ppm = 15 g of CHCl₃ in 10⁶ g of solution. Mass % = (15 / 10⁶) × 100 = 1.5 × 10⁻³ %.
2. (ii) Molality. Molar mass of CHCl₃ = 12.011 + 1.008 + 3(35.45) = 119.37 g mol⁻¹. Moles of CHCl₃ in 10⁶ g of solution = 15 / 119.4 = 0.1256 mol. The mass of solvent (water) ≈ 10⁶ g − 15 g ≈ 10⁶ g = 1000 kg (the solute mass is negligible). Molality = 0.1256 mol / 1000 kg = 1.26 × 10⁻⁴ m.

Extra depth. For very dilute aqueous solutions molality and molarity are almost equal (here both are about 1.26 × 10⁻⁴) because 1 kg of solution is about 1 L. ppm is the usual unit for traces of pollutants in water and air.

Conclusion: 15 ppm chloroform is 1.5 × 10⁻³ % by mass and corresponds to a molality of about 1.26 × 10⁻⁴ m.

Concept. In scientific (exponential) notation a number is written as N × 10ⁿ, where 1 ≤ N < 10 and n is a positive or negative integer. To write a number, move the decimal point until only one non-zero digit lies to its left; the number of places moved gives n (positive if moved to the left, negative if moved to the right). All digits retained in N are significant, so the notation shows the correct number of significant figures.

• (i) 0.0048: the decimal point is moved three places to the right, so the exponent is −3.
• (ii) 234,000: moving the point five places to the left gives 2.34 and exponent +5; the trailing zeros are not significant here.
• (iii) 8008: zeros between non-zero digits are significant, so all four digits remain: 8.008 × 10³.
• (iv) 500.0: the zeros after the decimal point are significant, so we write 5.000 × 10².
• (v) 6.0012 is already between 1 and 10, so n = 0.

Extra depth. Scientific notation is essential in chemistry where quantities range from 10⁻³¹ (mass of an electron in kg, about 9.1 × 10⁻³¹) to 10²³ (Avogadro constant). Multiplication and division use the laws of exponents; addition and subtraction require equal exponents first.

Conclusion: (i) 4.8 × 10⁻³; (ii) 2.34 × 10⁵; (iii) 8.008 × 10³; (iv) 5.000 × 10²; (v) 6.0012.

Concept. Use the rules: non-zero digits are significant; zeros between non-zero digits are significant; leading zeros are not; trailing zeros are significant only after a decimal point (or if a decimal point is shown).

Extra depth. The number 126,000 is ambiguous without scientific notation: 1.26 × 10⁵ (3 s.f.), 1.260 × 10⁵ (4 s.f.) or 1.2600 × 10⁵ (5 s.f.). In scientific notation every digit shown is significant, which removes the ambiguity.

Conclusion: The numbers contain (i) 2, (ii) 3, (iii) 4, (iv) 3, (v) 4 and (vi) 5 significant figures.

Concept. To round a number to three significant figures, keep the first three significant digits and look at the digit that follows: if it is more than 5, increase the last retained digit by one; if less than 5, leave it unchanged; if exactly 5, make the last retained digit even (increase it by one if it is odd, leave it if it is even).

• (i) 34.216 → digit after 34.2 is 1, less than 5, so the number is unchanged: 34.2.
• (ii) 10.4107 → after 10.4 comes 1; result 10.4.
• (iii) 0.04597 → three significant figures are 4, 5, 9; the next digit 7 is greater than 5, so 0.0459 becomes 0.0460 (the trailing zero is retained to show three significant figures).
• (iv) 2808 → the first three digits 280 followed by 8 (greater than 5) become 281; written as 2.81 × 10³ (i.e. 2810) so that only three digits are significant.

Extra depth. The “round-to-even” rule for a lone 5 (e.g. 6.35 → 6.4 but 6.25 → 6.2) avoids a systematic upward bias when many numbers are rounded.

Conclusion: To three significant figures: (i) 34.2, (ii) 10.4, (iii) 0.0460, (iv) 2.81 × 10³.

Concept. When two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other can be compared. Here the fixed mass is chosen as 28 g of dinitrogen.

(a) Law obeyed. Convert each pair to the mass of dioxygen that combines with 28 g of dinitrogen:

The masses of dioxygen combining with 28 g of dinitrogen are 32, 64, 32 and 80 g, i.e. in the ratio 32 : 64 : 32 : 80 = 2 : 4 : 2 : 5 — a ratio of small whole numbers. Thus the data obey the Law of Multiple Proportions.

Statement (Dalton, 1803). If two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element are in the ratio of small whole numbers. (Note: (i) and (iii) have the same composition, 14 : 16 = 28 : 32 = 7 : 8 by mass, i.e. NO; (ii) corresponds to NO₂ and (iv) to N₂O₅.)

(b) Unit conversions

Extra depth. Dalton used the law of multiple proportions to support his atomic theory: oxygen atoms combine with nitrogen atoms in simple whole-number ratios because atoms are indivisible. Compare the other example in the chapter: H₂O (2 g H : 16 g O) and H₂O₂ (2 g H : 32 g O), ratio 1 : 2.

Conclusion: The data obey the law of multiple proportions (32 : 64 : 32 : 80 = 2 : 4 : 2 : 5); and 1 km = 10⁶ mm = 10¹⁵ pm, 1 mg = 10⁻⁶ kg = 10⁶ ng, 1 mL = 10⁻³ L = 10⁻³ dm³.

Concept. Distance = speed × time. The prefix nano (n) stands for 10⁻⁹, so 2.00 ns = 2.00 × 10⁻⁹ s.

1. Time t = 2.00 ns = 2.00 × 10⁻⁹ s.
2. Speed c = 3.0 × 10⁸ m s⁻¹.
3. Distance d = c × t = (3.0 × 10⁸ m s⁻¹)(2.00 × 10⁻⁹ s) = 6.0 × 10⁻¹ m = 0.60 m.

d = (3.0 × 10⁸)(2.00 × 10⁻⁹) m = 0.60 m

Extra depth. The result has two significant figures because the speed of light (3.0) has two. Light travels about 30 cm in 1 ns — useful for judging the sizes of laboratory set-ups and the precision of modern time measurements.

Conclusion: Light covers 0.60 m (60 cm) in 2.00 ns.

Concept. From the balanced equation 1 atom (mole) of A reacts with 1 molecule (mole) of B₂ to form 1 formula unit of AB₂. The limiting reagent is the reactant present in the smaller stoichiometric amount; the one in excess is left over. In (ii), (iv) and (v) “B” means B₂ molecules, as the reaction involves B₂.

1. (i) Each A needs one B₂. With only 200 B₂, only 200 A can react; 100 A remain. B₂ is the limiting reagent and 200 AB₂ form.
2. (ii) 2 mol A need 2 mol B₂, 3 mol are present; 1 mol B₂ is in excess, so A is the limiting reagent (2 mol AB₂).
3. (iii) The reactants are present in the exact 1 : 1 ratio; both are completely consumed, so there is no limiting reagent (100 AB₂).
4. (iv) 5 mol A need 5 mol B₂ but only 2.5 mol B₂ are present; B₂ is limiting (2.5 mol AB₂); 2.5 mol A remain.
5. (v) 2.5 mol A need only 2.5 mol B₂ out of 5 mol; A is limiting (2.5 mol AB₂); 2.5 mol B₂ remain.

Extra depth. Always compare the ratio of available amounts to the stoichiometric ratio: n(A)/1 versus n(B₂)/1. The smaller quotient identifies the limiting reagent, and the product amount is determined by it.

Conclusion: The limiting reagents are (i) B₂, (ii) A, (iii) none, (iv) B₂ and (v) A.

Concept. In the balanced equation 1 mol N₂ (28.01 g) reacts with 3 mol H₂ (6.048 g) to give 2 mol NH₃ (34.06 g). Convert both reactant masses to moles, find the limiting reagent, and use it to find the product and the amount of the excess reagent left.

1. Moles of N₂ = 2.00 × 10³ g / 28.01 g mol⁻¹ = 71.4 mol.
2. Moles of H₂ = 1.00 × 10³ g / 2.016 g mol⁻¹ = 496 mol.
3. H₂ required for 71.4 mol N₂ = 3 × 71.4 = 214.2 mol, which is less than 496 mol available. Hence N₂ is the limiting reagent.
4. NH₃ formed = 2 × 71.4 = 142.8 mol × 17.03 g mol⁻¹ = 2.43 × 10³ g (2.43 kg).
5. (ii) Yes, dihydrogen remains unreacted.
6. (iii) H₂ left = 496 − 214.2 = 281.8 mol × 2.016 g mol⁻¹ = 5.68 × 10² g (about 568 g).

Extra depth. Check by conservation of mass: 2000 g N₂ + 1000 g H₂ = 3000 g = 2432 g NH₃ + 568 g H₂ left. This is the same pattern as Problem 1.5 of the chapter, where H₂ was limiting; here N₂ is limiting because the amounts differ.

Conclusion: 2.43 × 10³ g of NH₃ is formed; N₂ is the limiting reagent, and about 5.68 × 10² g of H₂ remains unreacted.

Concept. “mol” is a unit of the amount of a substance, whereas “M” (mol L⁻¹) is a unit of concentration (molarity). The two expressions look similar but describe different things.

• 0.50 mol = 53 g of Na₂CO₃ (molar mass 106 g mol⁻¹). It may be a solid, a pure substance or part of any solution.
• 0.50 M is a solution in which each litre contains 0.50 mol (53 g). A 250 mL sample of 0.50 M solution contains only 0.125 mol (13.25 g); 2 L contain 1.0 mol (106 g).

Extra depth. Amount of substance n = molarity × volume (in litres). The same amount (0.50 mol) can thus be present in 1 L of 0.50 M solution, in 0.5 L of 1.0 M solution, or in 100 mL of 5.0 M solution — the concentration changes but the amount does not.

Conclusion: 0.50 mol Na₂CO₃ is an amount (53 g) of the compound, whereas 0.50 M Na₂CO₃ is a concentration: 0.50 mol (53 g) in every litre of solution.

Concept. According to Gay-Lussac’s law of gaseous volumes and Avogadro’s law, gases at the same temperature and pressure react in simple ratios of volumes which equal the ratio of their moles in the balanced equation: 2 H₂(g) + O₂(g) → 2 H₂O(g), i.e. 2 volumes : 1 volume : 2 volumes.

1. The required ratio is H₂ : O₂ = 2 : 1; the given volumes 10 : 5 are in exactly this ratio, so both are completely consumed (no limiting reagent).
2. 2 volumes of H₂ give 2 volumes of water vapour, so 10 volumes of H₂ give 10 volumes of water vapour.

Extra depth. Gay-Lussac’s discovery of integer ratios was properly explained by Avogadro (1811): equal volumes of gases contain equal numbers of molecules, and H₂ and O₂ are diatomic; each oxygen molecule splits to supply the O atom of two water molecules. The law applies only when all gases are measured at the same temperature and pressure (water vapour, not liquid water).

Conclusion: 10 volumes of water vapour are produced (2 H₂ + O₂ → 2 H₂O, all reactants consumed).

Concept. The basic (SI) units here are the metre for length and the kilogram for mass. Use the prefix definitions: pico (p) = 10⁻¹²; milli (m) = 10⁻³; and 1 g = 10⁻³ kg.

1. (i) 28.7 pm = 28.7 × 10⁻¹² m = 2.87 × 10⁻¹¹ m.
2. (ii) 15.15 pm = 15.15 × 10⁻¹² m = 1.515 × 10⁻¹¹ m.
3. (iii) 25365 mg = 25365 × 10⁻³ g = 25.365 g = 25.365 × 10⁻³ kg = 2.5365 × 10⁻² kg.

Extra depth. 1 mg = 10⁻³ g = 10⁻⁶ kg, so 25365 mg = 25365 × 10⁻⁶ kg. The picometre is the typical unit for atomic radii and bond lengths (28.7 pm = 0.287 Å). Unit conversion by the factor-label method is explained in Section 1.4.3.

Conclusion: 28.7 pm = 2.87 × 10⁻¹¹ m; 15.15 pm = 1.515 × 10⁻¹¹ m; 25365 mg = 2.5365 × 10⁻² kg.

Concept. The number of atoms in a given mass is (mass / molar mass) × N_A × (atoms per formula). For equal masses, the substance with the smallest atomic (molar) mass contains the largest number of particles.

• Au: 1 g / 197 g mol⁻¹ = 0.00508 mol = 3.06 × 10²¹ atoms.
• Na: 1 / 22.99 = 0.0435 mol = 2.62 × 10²² atoms.
• Li: 1 / 6.94 = 0.1441 mol = 0.1441 × 6.022 × 10²³ = 8.68 × 10²² atoms (largest).
• Cl₂: 1 / 70.9 = 0.0141 mol of molecules × 2 = 0.0282 mol of atoms = 1.70 × 10²² atoms.

Extra depth. The order is Li > Na > Cl₂ (atoms) > Au. Notice that chlorine is diatomic, so each molecule contributes two atoms, but even then the lighter lithium wins. Li has the smallest atomic mass among the four, so it has the largest number of atoms per gram.

Conclusion: 1 g of Li(s) contains the largest number of atoms (about 8.68 × 10²²).

Concept. Mole fraction x_A = n_A/(n_A + n_B). Take a sample containing 1 mol of the solution, find the mass (and volume) of water in it, and use the assumption that the volume of the solution is approximately the volume of the water (density of water = 1 g mL⁻¹), as the question directs.

1. Mole fraction of ethanol = 0.040, so in 1 mol of solution: n(C₂H₅OH) = 0.040 mol and n(H₂O) = 1 − 0.040 = 0.960 mol.
2. Mass of water = 0.960 mol × 18.02 g mol⁻¹ = 17.3 g.
3. Volume of water = 17.3 g / (1 g mL⁻¹) = 17.3 mL = 0.0173 L, taken as the volume of the solution.
4. Molarity = 0.040 mol / 0.0173 L = 2.31 mol L⁻¹.

M = n(ethanol) / V(solution) = 0.040 mol / 0.0173 L = 2.31 M

Extra depth. The calculation neglects the contribution of ethanol to the volume (0.04 mol of ethanol is about 2.3 mL), so it is an approximation, as the question states by giving the density of water. It shows how mole fraction, a temperature-independent quantity, can be converted into molarity.

Conclusion: The molarity of the ethanol solution is about 2.31 mol L⁻¹.

Concept. One mole of ¹²C atoms has a mass of exactly 12 g and contains N_A = 6.022 × 10²³ atoms. The mass of a single atom is therefore the molar mass divided by Avogadro’s number.

Mass of one ¹²C atom = 12 g mol⁻¹ / 6.022 × 10²³ atoms mol⁻¹ = 1.993 × 10⁻²³ g

1. Molar mass of ¹²C = 12 g mol⁻¹.
2. Number of atoms in 1 mol = 6.022 × 10²³.
3. Mass of one atom = 12 / (6.022 × 10²³) = 1.993 × 10⁻²³ g (= 1.993 × 10⁻²⁶ kg).

Extra depth. This value, 1.992648 × 10⁻²³ g, was determined by mass spectrometry and was used in the chapter to find Avogadro’s number (12 / 1.992648 × 10⁻²³ = 6.0221367 × 10²³). It also gives the atomic mass unit: 1 u = (1/12) × 1.9926 × 10⁻²³ g = 1.66 × 10⁻²⁴ g.

Conclusion: The mass of one ¹²C atom is 1.993 × 10⁻²³ g.

Concept. In multiplication and division the result carries as many significant figures as the quantity with the fewest significant figures; in addition and subtraction the result carries as many decimal places as the quantity with the fewest decimal places. Exact (counted) numbers have infinite significant figures.

1. (i) Significant figures of the factors: 0.02856 → 4; 298.15 → 5; 0.112 → 3; 0.5785 → 4. The least is 3 (0.112), so the answer has 3 significant figures (value 1.65).
2. (ii) 5 is a counted/exact number (infinite significant figures) and 5.364 has 4; hence the answer has 4 significant figures (value 26.82).
3. (iii) 0.0125, 0.7864 and 0.0215 each have 4 decimal places, so the sum has 4 decimal places: 0.0125 + 0.7864 + 0.0215 = 0.8204, which has 4 significant figures.

Extra depth. If 5 in (ii) were a measured quantity with one significant figure, the answer would have to be rounded to 1 s.f. (30). The question treats it as a counted number. In (iii) the number of significant figures in the answer (4) happens to equal the number of decimal places, but the rule is about decimal places.

Conclusion: The answers should have (i) 3, (ii) 4 and (iii) 4 significant figures.

Concept. The molar mass of a naturally occurring element is the abundance-weighted average of the molar masses of its isotopes: M = Σ (fractional abundance × isotopic molar mass).

1. ³⁶Ar: 0.00337 × 35.96755 = 0.12121.
2. ³⁸Ar: 0.00063 × 37.96272 = 0.02392.
3. ⁴⁰Ar: 0.99600 × 39.9624 = 39.80255.
4. Sum = 0.12121 + 0.02392 + 39.80255 = 39.9477 ≈ 39.95 g mol⁻¹.

Extra depth. The abundances add up to 100.000%. Because ⁴⁰Ar makes up almost 99.6% of natural argon (formed by the radioactive decay of ⁴⁰K in the Earth’s crust), the average mass is very close to 40. The accepted atomic mass of argon, 39.95 u, matches this calculation, and is the reason why argon (39.95) appears before potassium (39.10) in the periodic table despite its higher mass.

Conclusion: The molar mass of naturally occurring argon is about 39.95 g mol⁻¹.

Concept. One mole of any element contains N_A = 6.022 × 10²³ atoms and has a mass equal to its atomic mass in grams. A mass in u refers to the mass of individual atoms (1 atom of He has an atomic mass of about 4 u).

1. (i) 52 mol of Ar (monoatomic): 52 × 6.022 × 10²³ = 3.13 × 10²⁵ atoms.
2. (ii) 52 u of He: one He atom has a mass of 4 u (atomic mass of He = 4.00 u). Number of atoms = 52 u / 4 u = 13 atoms.
3. (iii) 52 g of He: moles = 52 g / 4 g mol⁻¹ = 13 mol; number of atoms = 13 × 6.022 × 10²³ = 7.83 × 10²⁴ atoms.

Extra depth. Noble gases exist as single atoms, so atoms = molecules here. The contrast between (ii) and (iii) shows the meaning of atomic mass units versus grams: 1 g = 6.022 × 10²³ u, so 52 g of He is 6.022 × 10²³ times more helium than 52 u.

Conclusion: (i) 3.13 × 10²⁵ atoms; (ii) 13 atoms; (iii) 7.83 × 10²⁴ atoms.

Concept. Combustion analysis: all the carbon of the hydrocarbon ends up as CO₂ and all the hydrogen as H₂O. The moles of C and H in the sample are obtained from the masses of CO₂ and H₂O; their ratio gives the empirical formula. The molar mass comes from the gas density at STP (1 mol of gas occupies 22.4 L at 273 K and 1 atm), and n = molar mass / empirical formula mass gives the molecular formula.

1. Moles of C: n(CO₂) = 3.38 g / 44.01 g mol⁻¹ = 0.0768 mol, so n(C) = 0.0768 mol (mass 0.922 g).
2. Moles of H: n(H₂O) = 0.690 g / 18.02 g mol⁻¹ = 0.0383 mol; each water molecule has 2 H, so n(H) = 2 × 0.0383 = 0.0766 mol (mass 0.0772 g).
3. Ratio C : H = 0.0768 : 0.0766 ≈ 1 : 1. (i) Empirical formula = CH (empirical formula mass 13.02 g mol⁻¹).
4. (ii) Molar mass: 10.0 L at STP weighs 11.6 g, so the mass of 22.4 L = (11.6 / 10.0) × 22.4 = 25.98 ≈ 26 g mol⁻¹.
5. (iii) n = 26 / 13 = 2, so the molecular formula = (CH)₂ = C₂H₂ (ethyne / acetylene).

Extra depth. Acetylene (ethyne) burns with a very hot flame (about 3000 °C) when mixed with oxygen, which is why it is used as a welding fuel. Check: C₂H₂ has molar mass 2(12.01) + 2(1.008) = 26.04 g mol⁻¹. If STP is taken as 1 bar (22.7 L mol⁻¹) the molar mass is 26.3 g mol⁻¹, still giving n = 2.

Conclusion: The empirical formula is CH, the molar mass is about 26 g mol⁻¹ and the molecular formula is C₂H₂.

Concept. Find the moles of HCl from molarity and volume; the equation shows that 1 mol CaCO₃ reacts with 2 mol HCl; then convert the moles of CaCO₃ to mass.

1. Moles of HCl = M × V = 0.75 mol L⁻¹ × 0.025 L = 0.01875 mol.
2. Moles of CaCO₃ = 0.01875 mol HCl × (1 mol CaCO₃ / 2 mol HCl) = 0.009375 mol.
3. Molar mass of CaCO₃ = 40.08 + 12.01 + 3(16.00) = 100.09 g mol⁻¹.
4. Mass = 0.009375 mol × 100.09 g mol⁻¹ = 0.938 g.

m(CaCO₃) = ½ × (0.75 × 0.025) × 100.09 = 0.938 g

Extra depth. The CO₂ evolved is 0.009375 mol = 0.206 g (0.2 L at 25 °C and 1 atm approximately). The same stoichiometry underlies the use of limestone or chalk to neutralise acids.

Conclusion: 0.938 g of CaCO₃ is required to react completely with 25 mL of 0.75 M HCl.

Concept. From the balanced equation, 1 mol MnO₂ (86.94 g) reacts with 4 mol HCl (4 × 36.46 = 145.8 g). Use the mole ratio to find the mass of HCl needed for 5.0 g of MnO₂.

1. Molar mass of MnO₂ = 54.94 + 2(16.00) = 86.94 g mol⁻¹.
2. Moles of MnO₂ = 5.0 g / 86.94 g mol⁻¹ = 0.0575 mol.
3. Moles of HCl = 4 × 0.0575 = 0.230 mol.
4. Molar mass of HCl = 1.008 + 35.45 = 36.46 g mol⁻¹.
5. Mass of HCl = 0.230 mol × 36.46 g mol⁻¹ = 8.4 g (8.39 g).

m(HCl) = (4 × 36.46 / 86.94) × 5.0 g = 8.39 g

Extra depth. This is a redox reaction in which MnO₂ oxidises chloride ion to chlorine (Chapter 7). In the laboratory preparation concentrated HCl is used and the mixture is heated, so the actual amount of acid used is greater than the stoichiometric 8.4 g because only part of the HCl acts as the reductant and a part also forms MnCl₂.

Conclusion: 5.0 g of MnO₂ reacts with about 8.4 g of HCl.