Some Basic Concepts of Chemistry — NCERT Solutions

Scientific notation: $N \times 10^n$ where $1 \leq N < 10$

(i) $0.0048 = \boxed{4.8 \times 10^{-3}}$

(ii) $234000 = \boxed{2.34 \times 10^{5}}$

(iii) $8008 = \boxed{8.008 \times 10^{3}}$

(iv) $500.0 = \boxed{5.000 \times 10^{2}}$

(v) $6.0012 = \boxed{6.0012 \times 10^{0}}$ (already in scientific notation)

(i) 0.0025:
Leading zeros are not significant. Only 2 and 5 are significant.
$$\boxed{2\text{ significant figures}}$$

(ii) 208:
All digits including the zero between 2 and 8 are significant.
$$\boxed{3\text{ significant figures}}$$

(iii) 5005:
Zeros between non-zero digits are significant.
$$\boxed{4\text{ significant figures}}$$

(iv) 126,000:
Trailing zeros in a whole number without decimal point are ambiguous; conventionally taken as not significant.
$$\boxed{3\text{ significant figures}}$$

(v) 500.0:
The decimal point makes all trailing zeros significant.
$$\boxed{4\text{ significant figures}}$$

(vi) 2.0034:
All digits are significant (zeros between non-zero digits count).
$$\boxed{5\text{ significant figures}}$$

(i) 34.216 → 3 significant figures:
The 4th digit is 1 (< 5), so drop it.
$$\boxed{34.2}$$

(ii) 10.4107 → 3 significant figures:
The 4th digit is 1 (< 5), so drop.
$$\boxed{10.4}$$

(iii) 0.04597 → 3 significant figures:
Leading zeros are not significant. The significant digits are 4, 5, 9, 7.
Rounding to 3 sig figs: 4th significant digit is 7 (≥ 5), so round up 9 → 10, carry over.
$$0.04597 \rightarrow \boxed{0.0460}$$

(iv) 2808 → 3 significant figures:
The 4th digit is 8 (≥ 5), so round up.
$$2808 \rightarrow \boxed{2810}$$

(a) Law obeyed:

For a fixed mass of dinitrogen (14 g), the masses of dioxygen that combine are 16 g, 32 g (cases i and ii).
Ratio of O₂ = 16 : 32 = 1 : 2 (simple whole number ratio)

For 28 g of dinitrogen, masses of O₂ are 32 g and 80 g (cases iii and iv).
Ratio of O₂ = 32 : 80 = 2 : 5 (simple whole number ratio)

This obeys the Law of Multiple Proportions.

Statement: When two elements combine to form two or more different compounds, the masses of one element that combine with a fixed mass of the other element bear a simple whole number ratio to one another.

(b) Unit conversions:

(i) 1 km = ? mm = ? pm
$$1\text{ km} = 10^3\text{ m} = 10^3 \times 10^3\text{ mm} = \boxed{10^6\text{ mm}}$$
$$1\text{ km} = 10^3\text{ m} = 10^3 \times 10^{12}\text{ pm} = \boxed{10^{15}\text{ pm}}$$

(ii) 1 mg = ? kg = ? ng
$$1\text{ mg} = 10^{-3}\text{ g} = 10^{-3} \times 10^{-3}\text{ kg} = \boxed{10^{-6}\text{ kg}}$$
$$1\text{ mg} = 10^{-3}\text{ g} = 10^{-3} \times 10^{9}\text{ ng} = \boxed{10^{6}\text{ ng}}$$

(iii) 1 mL = ? L = ? dm³
$$1\text{ mL} = \boxed{10^{-3}\text{ L}}$$
$$1\text{ mL} = 1\text{ cm}^3 = (10^{-1}\text{ dm})^3 = 10^{-3}\text{ dm}^3 = \boxed{10^{-3}\text{ dm}^3}$$

Given:

• Speed of light, $c = 3.0 \times 10^8\text{ m s}^{-1}$
• Time, $t = 2.00\text{ ns} = 2.00 \times 10^{-9}\text{ s}$

Formula: Distance = Speed × Time

$$d = c \times t = 3.0 \times 10^8\text{ m s}^{-1} \times 2.00 \times 10^{-9}\text{ s}$$

$$= 6.0 \times 10^{-1}\text{ m}$$

$$= \boxed{0.60\text{ m}}$$

Reaction: A + B₂ → AB₂

1 atom/mol of A reacts with 1 molecule/mol of B₂ in a 1:1 ratio.

(i) 300 atoms of A + 200 molecules of B₂:
A needed for 200 B₂ = 200 atoms. A available = 300 (excess).
Limiting reagent: B₂

(ii) 2 mol A + 3 mol B₂:
B₂ needed for 2 mol A = 2 mol. B₂ available = 3 mol (excess).
Limiting reagent: A

(iii) 100 atoms of A + 100 molecules of B₂:
Ratio is exactly 1:1. Both are consumed completely.
No limiting reagent (both are in stoichiometric amounts)

(iv) 5 mol A + 2.5 mol B₂:
A needed for 2.5 mol B₂ = 2.5 mol. A available = 5 mol (excess).
Limiting reagent: B₂

(v) 2.5 mol A + 5 mol B₂:
B₂ needed for 2.5 mol A = 2.5 mol. B₂ available = 5 mol (excess).
Limiting reagent: A

Balanced equation: $\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)$

Given:

• Mass of N₂ = $2.00 \times 10^3$ g
• Mass of H₂ = $1.00 \times 10^3$ g

Step 1: Calculate moles of reactants.
$$n_{\text{N}_2} = \frac{2000}{28} = 71.43\text{ mol}$$
$$n_{\text{H}_2} = \frac{1000}{2} = 500\text{ mol}$$

Step 2: Identify limiting reagent.
From stoichiometry, 1 mol N₂ requires 3 mol H₂.
71.43 mol N₂ requires $71.43 \times 3 = 214.3$ mol H₂.
Available H₂ = 500 mol > 214.3 mol.

∴ N₂ is the limiting reagent.

(i) Mass of NH₃ produced:
1 mol N₂ gives 2 mol NH₃.
$$n_{\text{NH}_3} = 2 \times 71.43 = 142.86\text{ mol}$$
$$\text{Mass of NH}_3 = 142.86 \times 17 = \boxed{2428.6\text{ g} \approx 2.43 \times 10^3\text{ g}}$$

(ii) Yes, H₂ will remain unreacted.

(iii) Mass of unreacted H₂:
$$\text{H}_2\text{ consumed} = 71.43 \times 3 = 214.3\text{ mol}$$
$$\text{H}_2\text{ remaining} = 500 - 214.3 = 285.7\text{ mol}$$
$$\text{Mass of H}_2\text{ remaining} = 285.7 \times 2 = \boxed{571.4\text{ g}}$$

0.50 mol Na₂CO₃:
This refers to a definite amount (number of moles) of Na₂CO₃.
$$\text{Mass} = 0.50\text{ mol} \times 106\text{ g mol}^{-1} = 53\text{ g of Na}_2\text{CO}_3$$
No mention of volume or solvent — it is simply a quantity of substance.

0.50 M Na₂CO₃:
This refers to a concentration — a solution containing 0.50 moles of Na₂CO₃ dissolved in enough water to make 1 litre of solution.
$$\text{It contains } 0.50\text{ mol Na}_2\text{CO}_3\text{ per litre of solution}$$

Key difference: 0.50 mol is an amount of substance, while 0.50 M is a concentration (moles per litre of solution).

Balanced equation:
$$2\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(g)$$

By Gay-Lussac's Law of Gaseous Volumes, volumes react in the same ratio as molar coefficients.

From the equation: 2 volumes H₂ + 1 volume O₂ → 2 volumes H₂O

Given: 10 volumes H₂ and 5 volumes O₂.

Check stoichiometry: 10 volumes H₂ requires $\frac{10}{2} = 5$ volumes O₂. Exactly 5 volumes O₂ are available — both are completely consumed.

$$\text{Volumes of H}_2\text{O produced} = 10\text{ volumes}$$

$$\boxed{10\text{ volumes of water vapour are produced.}}$$

The basic SI unit of length is metre (m) and of mass is kilogram (kg).

(i) 28.7 pm:
$$1\text{ pm} = 10^{-12}\text{ m}$$
$$28.7\text{ pm} = 28.7 \times 10^{-12}\text{ m} = \boxed{2.87 \times 10^{-11}\text{ m}}$$

(ii) 15.15 pm:
$$15.15\text{ pm} = 15.15 \times 10^{-12}\text{ m} = \boxed{1.515 \times 10^{-11}\text{ m}}$$

(iii) 25365 mg:
$$1\text{ mg} = 10^{-6}\text{ kg}$$
$$25365\text{ mg} = 25365 \times 10^{-6}\text{ kg} = \boxed{2.5365 \times 10^{-2}\text{ kg}}$$

Concept: Number of atoms = $\frac{\text{mass}}{\text{molar mass}} \times N_A$ (for monoatomic) or $\times 2N_A$ (for diatomic)

For 1 g of each substance:

(i) Au: Molar mass = 197 g mol⁻¹
$$n = \frac{1}{197} = 5.08 \times 10^{-3}\text{ mol atoms}$$

(ii) Na: Molar mass = 23 g mol⁻¹
$$n = \frac{1}{23} = 4.35 \times 10^{-2}\text{ mol atoms}$$

(iii) Li: Molar mass = 6.941 g mol⁻¹
$$n = \frac{1}{6.941} = 1.44 \times 10^{-1}\text{ mol atoms}$$

(iv) Cl₂: Molar mass = 71 g mol⁻¹; each molecule has 2 atoms
$$n = \frac{1}{71} \times 2 = 2.82 \times 10^{-2}\text{ mol atoms}$$

Comparison: Li (0.144) > Na (0.0435) > Cl₂ (0.0282) > Au (0.00508)

$$\boxed{\text{(iii) 1 g of Li has the largest number of atoms.}}$$

Given:

• Mole fraction of ethanol, $x_{\text{ethanol}} = 0.040$
• Density of water = 1 g mL⁻¹ = 1 kg L⁻¹
• Molar mass of ethanol (C₂H₅OH) = 46 g mol⁻¹
• Molar mass of water = 18 g mol⁻¹

Step 1: Find mole fraction of water.
$$x_{\text{water}} = 1 - 0.040 = 0.960$$

Step 2: Assume 1 mol total. Then moles of ethanol = 0.040 mol, moles of water = 0.960 mol.

Step 3: Find mass of water (solvent).
$$\text{Mass of water} = 0.960 \times 18 = 17.28\text{ g} = 0.01728\text{ L}$$
(using density = 1 g mL⁻¹)

Step 4: Calculate molarity.
$$M = \frac{\text{moles of ethanol}}{\text{volume of solution in L}}$$

Volume of solution ≈ volume of water (dilute solution) = $\frac{17.28\text{ g}}{1\text{ g mL}^{-1}} = 17.28\text{ mL} = 0.01728\text{ L}$

$$M = \frac{0.040\text{ mol}}{0.01728\text{ L}} = \boxed{2.31\text{ mol L}^{-1}}$$

Concept: 1 mole of ¹²C = 12 g and contains $6.022 \times 10^{23}$ atoms.

$$\text{Mass of one }^{12}\text{C atom} = \frac{12\text{ g}}{6.022 \times 10^{23}}$$

$$= \frac{12}{6.022 \times 10^{23}}\text{ g}$$

$$= \boxed{1.993 \times 10^{-23}\text{ g}}$$

(i) $\dfrac{0.02856 \times 298.15 \times 0.112}{0.5785}$:

For multiplication/division, the answer should have the same number of significant figures as the number with the least significant figures.

• 0.02856 → 4 sig figs
• 298.15 → 5 sig figs
• 0.112 → 3 sig figs
• 0.5785 → 4 sig figs

Least = 3 significant figures.
$$\boxed{3\text{ significant figures}}$$

(ii) $5 \times 5.364$:

• 5 is an exact number (or 1 sig fig)
• 5.364 → 4 sig figs

If 5 is exact (counting number), the answer has 4 significant figures.
$$\boxed{4\text{ significant figures}}$$

(iii) $0.0125 + 0.7864 + 0.0215$:

For addition/subtraction, the answer should have the same number of decimal places as the number with the fewest decimal places.

• 0.0125 → 4 decimal places
• 0.7864 → 4 decimal places
• 0.0215 → 4 decimal places

All have 4 decimal places, so the answer has 4 decimal places.
$$0.0125 + 0.7864 + 0.0215 = 0.8204$$
$$\boxed{4\text{ significant figures}}$$

Formula:
$$\text{Average molar mass} = \sum (\text{fractional abundance} \times \text{isotopic molar mass})$$

$$M = \frac{0.337}{100} \times 35.96755 + \frac{0.063}{100} \times 37.96272 + \frac{99.600}{100} \times 39.9624$$

$$= 0.00337 \times 35.96755 + 0.00063 \times 37.96272 + 0.99600 \times 39.9624$$

$$= 0.12121 + 0.02392 + 39.8025$$

$$= \boxed{39.948\text{ g mol}^{-1}}$$

Avogadro's number: $N_A = 6.022 \times 10^{23}$ mol⁻¹

(i) 52 moles of Ar:
$$\text{Number of atoms} = 52\text{ mol} \times 6.022 \times 10^{23}\text{ mol}^{-1}$$
$$= \boxed{3.131 \times 10^{25}\text{ atoms}}$$

(ii) 52 u of He:
Molar mass of He = 4 u = 4 g mol⁻¹
$$\text{Number of atoms} = \frac{52\text{ u}}{4\text{ u/atom}} = 13\text{ atoms}$$
$$= \boxed{13\text{ atoms}}$$

(iii) 52 g of He:
$$n = \frac{52\text{ g}}{4\text{ g mol}^{-1}} = 13\text{ mol}$$
$$\text{Number of atoms} = 13 \times 6.022 \times 10^{23} = \boxed{7.829 \times 10^{24}\text{ atoms}}$$

Step 1: Find mass of C and H from combustion products.

Moles of CO₂ = $\frac{3.38}{44} = 0.0768$ mol → Moles of C = 0.0768 mol
Mass of C = $0.0768 \times 12 = 0.9216$ g

Moles of H₂O = $\frac{0.690}{18} = 0.0383$ mol → Moles of H = $0.0383 \times 2 = 0.0767$ mol
Mass of H = $0.0767 \times 1 = 0.0767$ g

Step 2: Find mole ratio C : H.
$$C : H = 0.0768 : 0.0767 \approx 1 : 1$$

(i) Empirical formula: $\boxed{\text{CH}}$

Step 3: Find molar mass.

At STP, 1 mole of gas occupies 22.4 L.
$$\text{Moles of gas in 10.0 L} = \frac{10.0}{22.4} = 0.4464\text{ mol}$$
$$\text{Molar mass} = \frac{11.6\text{ g}}{0.4464\text{ mol}} = \boxed{26.0\text{ g mol}^{-1}}$$

Step 4: Find molecular formula.

Empirical formula mass of CH = 12 + 1 = 13 g mol⁻¹
$$n = \frac{26.0}{13} = 2$$

(iii) Molecular formula: $\boxed{\text{C}_2\text{H}_2}$ (ethyne/acetylene)

Given:

• Volume of HCl = 25 mL = 0.025 L
• Molarity of HCl = 0.75 M
• Molar mass of CaCO₃ = 100 g mol⁻¹

Step 1: Find moles of HCl.
$$n_{\text{HCl}} = 0.75\text{ mol L}^{-1} \times 0.025\text{ L} = 0.01875\text{ mol}$$

Step 2: From stoichiometry, 2 mol HCl reacts with 1 mol CaCO₃.
$$n_{\text{CaCO}_3} = \frac{0.01875}{2} = 9.375 \times 10^{-3}\text{ mol}$$

Step 3: Find mass of CaCO₃.
$$\text{Mass} = 9.375 \times 10^{-3}\text{ mol} \times 100\text{ g mol}^{-1}$$
$$= \boxed{0.9375\text{ g}}$$

Given:

• Mass of MnO₂ = 5.0 g
• Molar mass of MnO₂ = 55 + 32 = 87 g mol⁻¹
• Molar mass of HCl = 1 + 35.5 = 36.5 g mol⁻¹

Step 1: Find moles of MnO₂.
$$n_{\text{MnO}_2} = \frac{5.0}{87} = 0.05747\text{ mol}$$

Step 2: From stoichiometry, 4 mol HCl reacts with 1 mol MnO₂.
$$n_{\text{HCl}} = 4 \times 0.05747 = 0.2299\text{ mol}$$

Step 3: Find mass of HCl.
$$\text{Mass of HCl} = 0.2299\text{ mol} \times 36.5\text{ g mol}^{-1}$$
$$= \boxed{8.39\text{ g}}$$