Some Basic Concepts of Chemistry Class 11 Chemistry NCERT Solutions

Step-by-step solution

1. 1(i) H₂O: the molecular mass of water equals the sum of the atomic masses of its constituent atoms.
2. 2
   $$M_{{\mathrm{H}}_2\mathrm{O}}=(2\times \text{atomic mass of H})+(1\times \text{atomic mass of O})$$
3. 3
   $$=[2(1.008)+1(16.00)]\mathrm{u}=2.016\mathrm{u}+16.00\mathrm{u}=18.016\mathrm{u}$$
4. 4Rounded to two decimals: 18.02 u.
5. 5(ii) CO₂.
6. 6
   $$M_{\mathrm{C}{\mathrm{O}}_2}=(1\times 12.011)+(2\times 16.00)=12.011\mathrm{u}+32.00\mathrm{u}=44.01\mathrm{u}$$
7. 7(iii) CH₄.
8. 8
   $$M_{\mathrm{C}{\mathrm{H}}_4}=(1\times 12.011)+(4\times 1.008)=12.011\mathrm{u}+4.032\mathrm{u}=16.043\mathrm{u}$$

Step-by-step solution

1. 1First find the molar mass of sodium sulphate.
2. 2
   $$M_{\mathrm{N}{\mathrm{a}}_2\mathrm{S}{\mathrm{O}}_4}=(2\times 23.0)+32.066+4(16.00)=46.0+32.066+64.0=142.066\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$$
3. 3Mass percent of an element is given by:
4. 4
   $$\text{Mass}\%=\frac{\text{mass of that element in the compound}}{\text{molar mass of the compound}}\times 100$$
5. 5Mass percent of sodium:
6. 6
   $$\%\mathrm{Na}=\frac{46.0\mathrm{g}}{142.066\mathrm{g}}\times 100=32.379\%\approx 32.4\%$$
7. 7Mass percent of sulphur:
8. 8
   $$\%\mathrm{S}=\frac{32.066\mathrm{g}}{142.066\mathrm{g}}\times 100=22.57\%\approx 22.6\%$$
9. 9Mass percent of oxygen (present as 4 atoms, mass 4 × 16.00 = 64.0 g):
10. 10
    $$\%\mathrm{O}=\frac{64.0\mathrm{g}}{142.066\mathrm{g}}\times 100=45.049\%\approx 45.05\%$$

Step-by-step solution

1. 1Per cent of iron by mass = 69.9%; per cent of oxygen by mass = 30.1% (given).
2. 2Relative moles of each element = mass per cent ÷ atomic mass.
3. 3
   $$\text{Moles of Fe}=\frac{69.9}{55.85}=1.25$$
4. 4
   $$\text{Moles of O}=\frac{30.1}{16.00}=1.88$$
5. 5Simplest molar ratio of iron to oxygen:
6. 6
   $$\mathrm{Fe}:\mathrm{O}=1.25:1.88=1:1.5\approx 2:3$$
7. 7The empirical formula is therefore Fe₂O₃.

Step-by-step solution

1. 1Write the balanced combustion reaction:
2. 2
   $${\mathrm{C}}_{(\mathrm{s})}+{\mathrm{O}}_{2(\mathrm{g})}⟶\mathrm{C}{\mathrm{O}}_{2(\mathrm{g})}$$
3. 31 mole C (12 g) needs 1 mole O₂ (32 g) and gives 1 mole CO₂ (44 g).
4. 4(i) In air, 1 mole of carbon burns completely to give 1 mole of CO₂ = 44 g.
5. 5(ii) Only 16 g of dioxygen is available, i.e. 0.5 mole of O₂. It can combine with only 0.5 mole of carbon, so dioxygen is the limiting reactant.
6. 6
   $$0.5{\text{mol CO}}_2\times 44\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}=22\mathrm{g}$$
7. 7(iii) Again only 16 g of dioxygen is available; it is the limiting reactant and can combine with only 0.5 mole of carbon, producing the same 22 g of CO₂.

Step-by-step solution

1. 1A 0.375 M aqueous solution of sodium acetate contains 0.375 moles of sodium acetate in 1000 mL of solution.
2. 2Number of moles of sodium acetate in 500 mL:
3. 3
   $$\text{moles}=\frac{0.375}{1000}\times 500=0.1875\text{mol}$$
4. 4Molar mass of sodium acetate = 82.0245 g mol⁻¹ (given).
5. 5Required mass of sodium acetate:
6. 6
   $$\text{mass}=(82.0245\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1})(0.1875\text{mol})=15.38\mathrm{g}$$

Step-by-step solution

1. 1Mass percent of nitric acid in the sample = 69% (given): 100 g of solution contains 69 g of HNO₃ by mass.
2. 2Molar mass of nitric acid, HNO₃:
3. 3
   $$M_{\mathrm{HN}{\mathrm{O}}_3}=1+14+3(16)=63\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$$
4. 4Number of moles in 69 g of HNO₃:
5. 5
   $$n=\frac{69\mathrm{g}}{63\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}}=1.095\text{mol}$$
6. 6Volume of 100 g of solution using its density:
7. 7
   $$V=\frac{100\mathrm{g}}{1.41\mathrm{g}\mathrm{m}{\mathrm{L}}^{-1}}=70.92\mathrm{mL}=70.92\times 10^{-3}\mathrm{L}$$
8. 8Concentration (molarity) of nitric acid:
9. 9
   $$\text{M}=\frac{1.095\text{mol}}{70.92\times 10^{-3}\mathrm{L}}=15.44\mathrm{mol}{\mathrm{L}}^{-1}$$

Step-by-step solution

1. 11 mole of CuSO₄ contains 1 mole (i.e. 1 g-atom) of copper.
2. 2Molar mass of CuSO₄:
3. 3
   $$M_{\mathrm{CuS}{\mathrm{O}}_4}=63.5+32.00+4(16.00)=159.5\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$$
4. 4Thus 159.5 g of CuSO₄ contains 63.5 g of copper.
5. 5Copper obtainable from 100 g of CuSO₄:
6. 6
   $$\text{mass of Cu}=\frac{63.5\times 100}{159.5}=39.81\mathrm{g}$$

Step-by-step solution

1. 1Mass percent of iron = 69.9% and of oxygen = 30.1% (given).
2. 2
   $$\text{Moles of Fe}=\frac{69.9}{55.85}=1.25$$
3. 3
   $$\text{Moles of O}=\frac{30.1}{16.0}=1.88$$
4. 4Ratio of iron to oxygen: 1.25 : 1.88 = 1 : 1.5 = 2 : 3, so the empirical formula is Fe₂O₃.
5. 5Empirical formula mass of Fe₂O₃:
6. 6
   $$M_e=2(55.85)+3(16.00)=159.7\mathrm{g}$$
7. 7The factor n relating molecular and empirical formula mass:
8. 8
   $$n=\frac{\text{molar mass}}{\text{empirical formula mass}}=\frac{159.69\mathrm{g}}{159.7\mathrm{g}}\approx 1$$
9. 9Since n = 1, the molecular formula equals the empirical formula, Fe₂O₃.

Step-by-step solution

1. 1The average atomic mass is the abundance-weighted mean of the isotopic masses.
2. 2
   $$\bar{M}=\left(\frac{75.77}{100}\right)(34.9689\mathrm{u})+\left(\frac{24.23}{100}\right)(36.9659\mathrm{u})$$
3. 3
   $$=26.4959+8.9568=35.4527\mathrm{u}$$

Step-by-step solution

1. 1(i) 1 mole of C₂H₆ contains 2 moles of carbon atoms.
2. 2
   $$\text{moles of C atoms}=2\times 3=6$$
3. 3(ii) 1 mole of C₂H₆ contains 6 moles of hydrogen atoms.
4. 4
   $$\text{moles of H atoms}=3\times 6=18$$
5. 5(iii) 1 mole of C₂H₆ contains 6.023 × 10²³ molecules.
6. 6
   $$\text{molecules}=3\times 6.023\times 10^{23}=18.07\times 10^{23}$$

Step-by-step solution

1. 1Molarity (M) of a solution is the number of moles of solute per litre of solution:
2. 2
   $$\text{M}=\frac{\text{number of moles of solute}}{\text{volume of solution in litres}}=\frac{\text{mass of sugar}/\text{molar mass of sugar}}{2\mathrm{L}}$$
3. 3Molar mass of sugar C₁₂H₂₂O₁₁:
4. 4
   $$M=(12\times 12)+(1\times 22)+(11\times 16)=144+22+176=342\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$$
5. 5Moles of sugar = 20 g / 342 g mol⁻¹ = 0.0585 mol.
6. 6
   $$\text{M}=\frac{0.0585\text{mol}}{2\mathrm{L}}=0.02925\mathrm{mol}{\mathrm{L}}^{-1}$$

Step-by-step solution

1. 1Molar mass of methanol, CH₃OH:
2. 2
   $$M=(1\times 12)+(4\times 1)+(1\times 16)=32\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}=0.032\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$$
3. 3Molarity of the stock methanol, since density gives mass per unit volume:
4. 4
   $$M_1=\frac{0.793\mathrm{kg}{\mathrm{L}}^{-1}}{0.032\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}}=24.78\mathrm{mol}{\mathrm{L}}^{-1}$$
5. 5Apply the dilution relation M₁V₁ = M₂V₂ (stock solution and solution to be prepared):
6. 6
   $$(24.78\mathrm{mol}{\mathrm{L}}^{-1})V_1=(2.5\mathrm{L})(0.25\mathrm{mol}{\mathrm{L}}^{-1})$$
7. 7
   $$V_1=\frac{0.625}{24.78}\mathrm{L}=0.0252\mathrm{L}=25.2\mathrm{mL}$$

Step-by-step solution

1. 1Pressure is force acting per unit area: P = F/A. Taking g = 9.8 m s⁻², the force per unit area due to the air column is (mass per unit area) × g.
2. 2
   $$P=\frac{F}{A}=\frac{1034\mathrm{g}\times 9.8\mathrm{m}{\mathrm{s}}^{-2}}{\mathrm{c}{\mathrm{m}}^2}$$
3. 3Convert g to kg and cm² to m²:
4. 4
   $$P=1034\times 9.8\times \frac{1\mathrm{kg}}{1000\mathrm{g}}\times \frac{(100{)}^2\mathrm{c}{\mathrm{m}}^2}{1{\mathrm{m}}^2}\mathrm{kg}{\mathrm{m}}^{-1}{\mathrm{s}}^{-2}$$
5. 5
   $$=1.01332\times 10^5\mathrm{kg}{\mathrm{m}}^{-1}{\mathrm{s}}^{-2}$$
6. 6Since 1 N = 1 kg m s⁻² and 1 Pa = 1 N m⁻² = 1 kg m⁻¹ s⁻², the numerical value is already in pascal:
7. 7
   $$P=1.01332\times 10^5\mathrm{Pa}$$

Step-by-step solution

1. 1The SI unit of mass is the kilogram (kg).
2. 21 kilogram is defined as the mass equal to the mass of the international prototype of kilogram — a platinum-iridium cylinder preserved at the International Bureau of Weights and Measures at Sèvres, France.

Step-by-step solution

1. 1Recall each SI prefix and its power of ten: micro = 10⁻⁶, deca = 10, mega = 10⁶, giga = 10⁹, femto = 10⁻¹⁵.
2. 2Match them one by one: (i) micro ↔ 10⁻⁶, (ii) deca ↔ 10, (iii) mega ↔ 10⁶, (iv) giga ↔ 10⁹, (v) femto ↔ 10⁻¹⁵.

Step-by-step solution

1. 1Significant figures are those meaningful digits that are known with certainty.
2. 2They indicate the uncertainty in an experiment or a calculated value. For example, if 15.6 mL is the result of an experiment, then 15 is certain while 6 is uncertain, and the total number of significant figures is 3.
3. 3Hence, significant figures are defined as the total number of digits in a number, including the last digit that represents the uncertainty of the result.

Step-by-step solution

1. 1(i) 1 ppm means 1 part out of 1 million (10⁶) parts.
2. 2
   $$\text{mass}\%=\frac{15}{10^6}\times 100=1.5\times 10^{-3}\%$$
3. 3(ii) From the result above, 100 g of the sample contains 1.5 × 10⁻³ g of CHCl₃, i.e. 1000 g of the sample contains 1.5 × 10⁻² g of CHCl₃.
4. 4Molality = moles of solute per kilogram of solvent:
5. 5
   $$m=\frac{1.5\times 10^{-2}\mathrm{g}}{M_{\mathrm{CHC}{\mathrm{l}}_3}}$$
6. 6Molar mass of chloroform, CHCl₃:
7. 7
   $$M=12.00+1.00+3(35.5)=119.5\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$$
8. 8
   $$m=\frac{1.5\times 10^{-2}}{119.5}\mathrm{m}=1.255\times 10^{-4}\mathrm{m}$$

Step-by-step solution

1. 1(i) Move the decimal point four places right: 0.0048 = 4.8 × 10⁻³.
2. 2(ii) 234,000 = 2.34 × 10⁵.
3. 3(iii) 8008 = 8.008 × 10³.
4. 4(iv) 500.0 = 5.000 × 10² (the decimal point is moved two places left).
5. 5(v) 6.0012 is already between 1 and 10, so 6.0012 = 6.0012 × 10⁰.

Step-by-step solution

1. 1(i) 0.0025: leading zeros are not significant; only 2 and 5 count → 2 significant figures.
2. 2(ii) 208: all three digits are significant (the embedded zero counts) → 3.
3. 3(iii) 5005: embedded zeros count → 4.
4. 4(iv) 126,000: trailing zeros without a decimal point are not significant → 3.
5. 5(v) 500.0: the decimal point makes the trailing zero significant → 4.
6. 6(vi) 2.0034: embedded zeros count → 5.

Step-by-step solution

1. 1(i) 34.216: the fourth significant digit is 2 (< 5), so keep 34.2.
2. 2(ii) 10.4107: the fourth significant digit is 1, so 10.4.
3. 3(iii) 0.04597: significant digits are 4, 5, 9; the next digit 7 ≥ 5 rounds 9 up to 10, giving 0.0460.
4. 4(iv) 2808: the fourth digit is 8; rounding to three figures gives 2810.

Step-by-step solution

1. 1(a) Fix the mass of dinitrogen at 28 g, then scale the dioxygen masses: the data become 32 g, 64 g, 32 g and 80 g of dioxygen.
2. 2These masses of dioxygen bear a simple whole-number ratio 2 : 4 : 2 : 5. Hence the data obey the law of multiple proportions.
3. 3Statement of the law: if two elements combine to form more than one compound, then the masses of one element that combine with a fixed mass of the other element are in the ratio of small whole numbers.
4. 4(b)(i) Convert 1 km to mm and pm:
5. 5
   $$1\mathrm{km}=1\mathrm{km}\times \frac{1000\mathrm{m}}{1\mathrm{km}}\times \frac{100\mathrm{cm}}{1\mathrm{m}}\times \frac{10\mathrm{mm}}{1\mathrm{cm}}=10^6\mathrm{mm}$$
6. 6
   $$1\mathrm{km}=1\mathrm{km}\times \frac{1000\mathrm{m}}{1\mathrm{km}}\times \frac{1\mathrm{pm}}{10^{-12}\mathrm{m}}=10^{15}\mathrm{pm}$$
7. 7(b)(ii):
8. 8
   $$1\mathrm{mg}=1\mathrm{mg}\times \frac{1\mathrm{g}}{1000\mathrm{mg}}\times \frac{1\mathrm{kg}}{1000\mathrm{g}}=10^{-6}\mathrm{kg}$$
9. 9
   $$1\mathrm{mg}=1\mathrm{mg}\times \frac{1\mathrm{g}}{1000\mathrm{mg}}\times \frac{1\mathrm{ng}}{10^{-9}\mathrm{g}}=10^6\mathrm{ng}$$
10. 10(b)(iii): 1 mL = 1 cm³ = 10⁻³ L.
11. 11
    $$1\mathrm{mL}=1\mathrm{c}{\mathrm{m}}^3=1\mathrm{c}{\mathrm{m}}^3\times \frac{1\mathrm{dm}\times 1\mathrm{dm}\times 1\mathrm{dm}}{10\mathrm{cm}\times 10\mathrm{cm}\times 10\mathrm{cm}}=10^{-3}\mathrm{d}{\mathrm{m}}^3$$

Step-by-step solution

1. 1Time taken = 2.00 ns = 2.00 × 10⁻⁹ s.
2. 2Speed of light = 3.0 × 10⁸ m s⁻¹.
3. 3Distance = speed × time:
4. 4
   $$d=(3.0\times 10^8\mathrm{m}{\mathrm{s}}^{-1})(2.00\times 10^{-9}\mathrm{s})$$
5. 5
   $$=6.00\times 10^{-1}\mathrm{m}=0.600\mathrm{m}$$

Step-by-step solution

1. 1A limiting reagent determines the extent of a reaction: it is the reactant which is consumed first, stopping the reaction and limiting the amount of product formed.
2. 2The balanced reaction shows 1 atom (or 1 mol) of A reacts with 1 molecule (or 1 mol) of B₂.
3. 3(i) 200 molecules of B₂ react with 200 atoms of A, leaving 100 atoms of A unused. Hence B₂ is the limiting reagent.
4. 4(ii) 2 mol of A reacts with only 2 mol of B₂, so 1 mol of B₂ remains unused. Hence A is the limiting reagent.
5. 5(iii) 100 atoms of A combine with all 100 molecules of B₂; the mixture is stoichiometric, so there is no limiting reagent.
6. 6(iv) 2.5 mol of B₂ combines with only 2.5 mol of A, leaving 2.5 mol of A. Hence B₂ is the limiting reagent.
7. 7(v) 2.5 mol of A combines with only 2.5 mol of B₂, leaving 2.5 mol of B₂. Hence A is the limiting reagent.

Step-by-step solution

1. 1Balance the chemical equation:
2. 2
   $${\mathrm{N}}_{2(\mathrm{g})}+3{\mathrm{H}}_{2(\mathrm{g})}⟶2N{\mathrm{H}}_{3(\mathrm{g})}$$
3. 31 mole (28 g) of dinitrogen reacts with 3 moles (6 g) of dihydrogen to give 2 moles (34 g) of ammonia.
4. 4Dihydrogen required for 2.00 × 10³ g of N₂:
5. 5
   $$\frac{6\mathrm{g}}{28\mathrm{g}}\times 2.00\times 10^3\mathrm{g}=428.6\mathrm{g}$$
6. 6Given dihydrogen = 1.00 × 10³ g, which exceeds 428.6 g, so N₂ is the limiting reagent.
7. 7Mass of ammonia produced from 2000 g of N₂:
8. 8
   $$m_{\mathrm{N}{\mathrm{H}}_3}=\frac{34\mathrm{g}}{28\mathrm{g}}\times 2000\mathrm{g}=2428.57\mathrm{g}$$
9. 9(ii) N₂ is the limiting reagent and H₂ is the excess reagent, so H₂ remains unreacted.
10. 10(iii) Mass of dihydrogen left unreacted:
11. 11
    $$1.00\times 10^3\mathrm{g}-428.6\mathrm{g}=571.4\mathrm{g}$$

Step-by-step solution

1. 1Molar mass of Na₂CO₃:
2. 2
   $$M=(2\times 23)+12.00+(3\times 16)=106\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$$
3. 31 mole of Na₂CO₃ means 106 g of Na₂CO₃, so 0.50 mol means:
4. 4
   $$0.50\text{mol}=\frac{106\mathrm{g}}{1\text{mol}}\times 0.50\text{mol}=53\mathrm{g}$$
5. 5A 0.50 M solution contains 0.50 mol of Na₂CO₃ per litre of solution, i.e. 53 g of Na₂CO₃ dissolved and made up to 1 L.

Step-by-step solution

1. 1Write the balanced reaction of dihydrogen with dioxygen (Gay-Lussac's law of gaseous volumes applies):
2. 2
   $$2{\mathrm{H}}_{2(\mathrm{g})}+{\mathrm{O}}_{2(\mathrm{g})}⟶2{\mathrm{H}}_2{\mathrm{O}}_{(\mathrm{g})}$$
3. 3Two volumes of dihydrogen react with one volume of dioxygen to give two volumes of water vapour.
4. 4Hence ten volumes of dihydrogen react with five volumes of dioxygen to produce ten volumes of water vapour.

Step-by-step solution

1. 1(i) Using 1 pm = 10⁻¹² m:
2. 2
   $$28.7\mathrm{pm}=28.7\times 10^{-12}\mathrm{m}=2.87\times 10^{-11}\mathrm{m}$$
3. 3(ii) Using 1 pm = 10⁻¹² m:
4. 4
   $$15.15\mathrm{pm}=15.15\times 10^{-12}\mathrm{m}=1.515\times 10^{-11}\mathrm{m}$$
5. 5(iii) 25365 mg = 2.5365 × 10¹ g (since 1000 mg = 1 g). Then convert grams to kilograms:
6. 6
   $$25365\mathrm{mg}=2.5365\times 10^1\mathrm{g}\times \frac{1\mathrm{kg}}{10^3\mathrm{g}}=2.5365\times 10^{-2}\mathrm{kg}$$

Step-by-step solution

1. 1(i) 1 g of Au:
2. 2
   $$N_{\mathrm{Au}}=\frac{6.022\times 10^{23}}{197}=3.06\times 10^{21}\text{atoms}$$
3. 3(ii) 1 g of Na:
4. 4
   $$N_{\mathrm{Na}}=\frac{6.022\times 10^{23}}{23}=2.62\times 10^{22}\text{atoms}$$
5. 5(iii) 1 g of Li (atomic mass 7):
6. 6
   $$N_{\mathrm{Li}}=\frac{6.022\times 10^{23}}{7}=8.6\times 10^{22}\text{atoms}$$
7. 7(iv) 1 g of Cl₂ (molar mass 35.5 × 2 = 71 g mol⁻¹) as molecules, remembering each Cl₂ molecule has two Cl atoms:
8. 8
   $$N_{\mathrm{Cl}}=2\times \frac{6.022\times 10^{23}}{71}=2\times 8.48\times 10^{21}=1.70\times 10^{22}\text{atoms}$$
9. 9Compare: Li gives 8.6 × 10²² atoms, the largest of the four.

Step-by-step solution

1. 1Mole fraction of C₂H₅OH:
2. 2
   $$x_{{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}}=\frac{n_{{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}}}{n_{{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}}+n_{{\mathrm{H}}_2\mathrm{O}}}=0.040$$
3. 3Number of moles of water in 1 L of water (density of water = 1 g mL⁻¹, so 1 L = 1000 g):
4. 4
   $$n_{{\mathrm{H}}_2\mathrm{O}}=\frac{1000\mathrm{g}}{18\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}}=55.55\text{mol}$$
5. 5Substitute into the mole-fraction equation:
6. 6
   $$\frac{n_{{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}}}{n_{{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}}+55.55}=0.040$$
7. 7
   $$n_{{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}}=0.040n_{{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}}+2.222\Rightarrow 0.96n_{{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}}=2.222$$
8. 8
   $$n_{{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}}=\frac{2.222}{0.96}=2.314\text{mol}$$
9. 9Molarity (based on 1 L of solution):
10. 10
    $$\text{M}=\frac{2.314\text{mol}}{1\mathrm{L}}=2.314\mathrm{M}$$

Step-by-step solution

1. 11 mole of carbon atoms = 6.022 × 10²³ atoms of carbon and also equals 12 g of carbon.
2. 2Mass of one atom:
3. 3
   $$m=\frac{12\mathrm{g}}{6.022\times 10^{23}}=1.993\times 10^{-23}\mathrm{g}$$

Step-by-step solution

1. 1(i) For multiplication and division, the result has as many significant figures as the factor with the fewest. The least precise number is 0.112 (3 significant figures).
2. 2Hence the answer to (i) should have 3 significant figures.
3. 3(ii) For 5 × 5.364: here 5 is an exact, counted number, so the precision is governed by 5.364, which has 4 significant figures.
4. 4Hence the answer to (ii) should have 4 significant figures.
5. 5(iii) For 0.0125 + 0.7864 + 0.0215: in addition, the result is reported to the least number of decimal places — here each term has four decimal places.
6. 6The sum 0.8204 therefore has 4 significant figures.

Step-by-step solution

1. 1Molar mass of argon is the abundance-weighted sum of the isotopic molar masses:
2. 2
   $${\bar{M}}_{\mathrm{Ar}}=\left(35.96755\times \frac{0.337}{100}\right)+\left(37.96272\times \frac{0.063}{100}\right)+\left(39.9624\times \frac{99.600}{100}\right)\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$$
3. 3
   $$=0.1212+0.0239+39.8026=39.948\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$$

Step-by-step solution

1. 1(i) 1 mole of Ar contains 6.022 × 10²³ atoms:
2. 2
   $$N=52\times 6.022\times 10^{23}=3.13\times 10^{25}\text{atoms of Ar}$$
3. 3(ii) 1 atom of He has mass 4 u, so 52 u of He corresponds to:
4. 4
   $$N=\frac{52}{4}=13\text{atoms of He}$$
5. 5(iii) 4 g of He contains 6.022 × 10²³ atoms:
6. 6
   $$N=\frac{6.022\times 10^{23}\times 52}{4}=7.83\times 10^{24}\text{atoms of He}$$

Step-by-step solution

1. 1(i) 1 mole (44 g) of CO₂ contains 12 g of carbon, so the carbon in 3.38 g CO₂ is:
2. 2
   $$m_{\mathrm{C}}=\frac{12\mathrm{g}}{44\mathrm{g}}\times 3.38\mathrm{g}=0.9217\mathrm{g}$$
3. 318 g of water contains 2 g of hydrogen, so the hydrogen in 0.690 g of water is:
4. 4
   $$m_{\mathrm{H}}=\frac{2\mathrm{g}}{18\mathrm{g}}\times 0.690\mathrm{g}=0.0767\mathrm{g}$$
5. 5Total mass of the compound = 0.9217 + 0.0767 = 0.9984 g, giving 92.32% C and 7.68% H.
6. 6Moles of carbon and hydrogen:
7. 7
   $$n_{\mathrm{C}}=\frac{92.32}{12.00}=7.69,n_{\mathrm{H}}=\frac{7.68}{1}=7.68$$
8. 8Ratio C : H = 7.69 : 7.68 ≈ 1 : 1. Hence the empirical formula of the gas is CH.
9. 9(ii) 10.0 L of the gas at STP weighs 11.6 g. Therefore 22.4 L (1 mole at STP) weighs:
10. 10
    $$M=\frac{11.6\mathrm{g}}{10.0\mathrm{L}}\times 22.4\mathrm{L}=25.984\mathrm{g}\approx 26\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$$
11. 11(iii) Empirical formula mass of CH = 12 + 1 = 13 g.
12. 12
    $$n=\frac{\text{molar mass}}{\text{empirical formula mass}}=\frac{26}{13}=2$$
13. 13Molecular formula = (CH)₂ = C₂H₂.

Step-by-step solution

1. 1A 0.75 M HCl solution has 0.75 mol of HCl per litre. Molar mass of HCl = 1 + 35.5 = 36.5 g mol⁻¹, so 1 L contains:
2. 2
   $$0.75\times 36.5=27.375\mathrm{g}\mathrm{of}\mathrm{HCl}$$
3. 3HCl present in 25 mL of solution:
4. 4
   $$m_{\mathrm{HCl}}=\frac{27.375\mathrm{g}}{1000\mathrm{mL}}\times 25\mathrm{mL}=0.6844\mathrm{g}$$
5. 5From the equation, 2 mol of HCl (2 × 36.5 = 71 g) react with 1 mol of CaCO₃ (100 g).
6. 6Mass of CaCO₃ reacting with 0.6844 g of HCl:
7. 7
   $$m_{\mathrm{CaC}{\mathrm{O}}_3}=\frac{100\mathrm{g}}{71\mathrm{g}}\times 0.6844\mathrm{g}=0.9639\mathrm{g}$$

Step-by-step solution

1. 11 mol of MnO₂ has mass 55 + 2(16) = 87 g, and from the equation it reacts with 4 mol of HCl, i.e. 4 × 36.5 = 146 g of HCl.
2. 2Therefore, HCl reacting with 5.0 g of MnO₂:
3. 3
   $$m_{\mathrm{HCl}}=\frac{146\mathrm{g}}{87\mathrm{g}}\times 5.0\mathrm{g}=8.39\mathrm{g}\approx 8.4\mathrm{g}$$