CBSE Class 11 Chemistry — NCERT Solutions

Given:

• Mass of NaOH = 4 g
• Volume of solution = 250 mL = 0.250 L
• Molar mass of NaOH = 23 + 16 + 1 = 40 g mol⁻¹

Formula:
$$M = \frac{\text{No. of moles of solute}}{\text{Volume of solution in litres}}$$

Step 1: Calculate moles of NaOH.
$$n_{\text{NaOH}} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{4\text{ g}}{40\text{ g mol}^{-1}} = 0.1\text{ mol}$$

Step 2: Calculate molarity.
$$M = \frac{0.1\text{ mol}}{0.250\text{ L}} = 0.4\text{ mol L}^{-1}$$

$$\boxed{M = 0.4\text{ M}}$$

Note: Molarity depends on temperature because volume changes with temperature.

Given:

• Molarity (M) = 3 mol L⁻¹
• Density of solution = 1.25 g mL⁻¹
• Molar mass of NaCl = 23 + 35.5 = 58.5 g mol⁻¹

Step 1: Find mass of NaCl in 1 L of solution.
$$\text{Mass of NaCl} = 3\text{ mol} \times 58.5\text{ g mol}^{-1} = 175.5\text{ g}$$

Step 2: Find mass of 1 L solution.
$$\text{Mass of solution} = 1000\text{ mL} \times 1.25\text{ g mL}^{-1} = 1250\text{ g}$$

Step 3: Find mass of water (solvent).
$$\text{Mass of water} = 1250 - 175.5 = 1074.5\text{ g} = 1.0745\text{ kg}$$

Step 4: Calculate molality.
$$m = \frac{\text{No. of moles of solute}}{\text{Mass of solvent in kg}} = \frac{3\text{ mol}}{1.0745\text{ kg}}$$

$$\boxed{m = 2.79\text{ m}}$$

Note: Molality does not change with temperature since mass is unaffected by temperature.

Given: Various compounds; find oxidation number (O.N.) of the underlined element.

Rules used: Sum of O.N. of all atoms = 0 (neutral molecule) or = charge (ion). O.N. of O = −2 (usually), H = +1 (with non-metals), Na = +1, K = +1, Ca = +2, Al = +3.

(a) NaH₂PO₄ — find O.N. of P

Let O.N. of P = $x$.
$$+1 + 2(+1) + x + 4(-2) = 0$$
$$1 + 2 + x - 8 = 0 \implies x = +5$$

O.N. of P = +5

(b) NaHSO₄ — find O.N. of S

Let O.N. of S = $x$.
$$+1 + 1 + x + 4(-2) = 0$$
$$2 + x - 8 = 0 \implies x = +6$$

O.N. of S = +6

(c) H₂P₂O₇ — find O.N. of P

Let O.N. of P = $x$.
$$2(+1) + 2x + 7(-2) = 0$$
$$2 + 2x - 14 = 0 \implies 2x = 12 \implies x = +5$$

O.N. of P = +5

(d) K₂MnO₄ — find O.N. of Mn

Let O.N. of Mn = $x$.
$$2(+1) + x + 4(-2) = 0$$
$$2 + x - 8 = 0 \implies x = +6$$

O.N. of Mn = +6

(e) CaO₂ — find O.N. of O

This is calcium peroxide. Let O.N. of O = $x$.
$$+2 + 2x = 0 \implies x = -1$$

O.N. of O = −1 (peroxide linkage O–O)

(f) NaBH₄ — find O.N. of B

In NaBH₄, H is bonded to B (more electronegative than H here? Actually B–H: H is −1 when bonded to metals/metalloids in hydrides). Here H = −1.

Let O.N. of B = $x$.
$$+1 + x + 4(-1) = 0$$
$$1 + x - 4 = 0 \implies x = +3$$

O.N. of B = +3

(g) H₂S₂O₇ — find O.N. of S

Let O.N. of S = $x$.
$$2(+1) + 2x + 7(-2) = 0$$
$$2 + 2x - 14 = 0 \implies 2x = 12 \implies x = +6$$

O.N. of S = +6

(h) KAl(SO₄)₂·12H₂O — find O.N. of S

Let O.N. of S = $x$. Consider the formula unit (ignore water of crystallisation for this calculation, or include it — O in water = −2, H = +1).

For the ionic compound: K = +1, Al = +3, each SO₄²⁻ has S with O.N. $x$:
$$x + 4(-2) = -2 \implies x = +6$$

O.N. of S = +6

Concept: Some compounds have elements in non-integral (fractional) or mixed oxidation states, which is rationalised by the actual structure.

(a) KI₃ — O.N. of I

Let O.N. of I = $x$.
$$+1 + 3x = 0 \implies x = -\frac{1}{3}$$

This is a fractional value. Rationalisation: KI₃ is actually $\text{K}^+[\text{I}_3]^-$. In the $\text{I}_3^-$ ion, one I carries −1 and the other two carry 0 (I₂ molecule coordinates to I⁻). So the average is $-1/3$, but structurally one I is −1 and two are 0.

Average O.N. of I = −1/3

(b) H₂S₄O₆ — O.N. of S

Let O.N. of S = $x$.
$$2(+1) + 4x + 6(-2) = 0$$
$$2 + 4x - 12 = 0 \implies 4x = 10 \implies x = +2.5$$

Rationalisation: Tetrathionate ion has the structure $^{-}O_3S-S-S-SO_3^{-}$. The two terminal S atoms have O.N. = +5 and the two middle S atoms (S–S bond) have O.N. = 0. Average = $(5+5+0+0)/4 = +2.5$.

Average O.N. of S = +2.5

(c) Fe₃O₄ — O.N. of Fe

Let O.N. of Fe = $x$.
$$3x + 4(-2) = 0 \implies 3x = 8 \implies x = +\frac{8}{3}$$

Rationalisation: Fe₃O₄ is a mixed oxide = FeO·Fe₂O₃. It contains one Fe²⁺ and two Fe³⁺ ions. Average O.N. = $(2+3+3)/3 = 8/3$.

Average O.N. of Fe = +8/3

(d) CH₃CH₂OH — O.N. of C

For C₁ (–CH₃): Let O.N. = $x_1$. Each H = +1, bonded to C.
$$x_1 + 3(+1) = 0 \text{ (for CH}_3\text{ group, considering C–C bond as zero contribution)}$$
Using the formula approach for each carbon:

• C of CH₃ group: $x_1 + 3(+1) + (-\text{C}) = 0$. Using electronegativity: C–H bonds give H = +1; C–C bond: both same, so 0 contribution.

$$x_1 + 3(+1) + 0 = 0 \implies x_1 = -3$$

• C of CH₂OH group: $x_2 + 2(+1) + 0 + (-2+1) = 0$

$$x_2 + 2 - 1 = 0 \implies x_2 = -1$$

O.N. of C in CH₃ = −3; O.N. of C in CH₂OH = −1

(e) CH₃COOH — O.N. of C

• C of CH₃ group: $x_1 + 3(+1) + 0 = 0 \implies x_1 = -3$
• C of COOH group: $x_2 + 0 + 2(-2) + 1 = 0 \implies x_2 = +3$

O.N. of C in CH₃ = −3; O.N. of C in COOH = +3

(i) Given: Mass of one electron = 9.10939 × 10⁻²⁸ g

Number of electrons that weigh 1 g:
$$N = \frac{1\text{ g}}{9.10939 \times 10^{-28}\text{ g}} = 1.098 \times 10^{27} \text{ electrons}$$

(ii) Mass of one mole of electrons:
$$m = 6.022 \times 10^{23} \times 9.10939 \times 10^{-31}\text{ kg} = 5.486 \times 10^{-7}\text{ kg} \approx 5.49 \times 10^{-4}\text{ g}$$

Charge of one mole of electrons:
$$Q = 6.022 \times 10^{23} \times 1.6022 \times 10^{-19}\text{ C} = 9.65 \times 10^{4}\text{ C} = 96500\text{ C (1 Faraday)}$$

(i) Methane (CH₄): electrons per molecule = 6 (C) + 4×1 (H) = 10 electrons

Total electrons in 1 mole of CH₄:
$$N = 10 \times 6.022 \times 10^{23} = 6.022 \times 10^{24}\text{ electrons}$$

(ii) ¹⁴C has 8 neutrons per atom. Molar mass of ¹⁴C = 14 g/mol.

Moles of ¹⁴C in 7 mg:
$$n = \frac{7 \times 10^{-3}\text{ g}}{14\text{ g/mol}} = 5 \times 10^{-4}\text{ mol}$$

(a) Total number of neutrons:
$$= 5 \times 10^{-4} \times 6.022 \times 10^{23} \times 8 = 2.409 \times 10^{21}\text{ neutrons}$$

(b) Total mass of neutrons:
$$= 2.409 \times 10^{21} \times 1.675 \times 10^{-27}\text{ kg} = 4.035 \times 10^{-6}\text{ kg}$$

(iii) NH₃ has molar mass = 17 g/mol. Protons per molecule = 7 (N) + 3×1 (H) = 10 protons.

Moles of NH₃ in 34 mg:
$$n = \frac{34 \times 10^{-3}\text{ g}}{17\text{ g/mol}} = 2 \times 10^{-3}\text{ mol}$$

(a) Total number of protons:
$$= 2 \times 10^{-3} \times 6.022 \times 10^{23} \times 10 = 1.2044 \times 10^{22}\text{ protons}$$

(b) Total mass of protons:
$$= 1.2044 \times 10^{22} \times 1.6726 \times 10^{-27}\text{ kg} = 2.014 \times 10^{-5}\text{ kg}$$

The answer will NOT change with temperature and pressure because the number of moles (and hence number of protons) depends only on the mass of the sample, not on T and P.

Given: Five organic compounds. We identify the hybridisation of each carbon using the rule: sp³ (4 single bonds), sp² (one double bond or part of aromatic ring), sp (triple bond or two double bonds on same carbon).

(i) CH₂=C=O (Ketene)

• C₁ (=CH₂): forms a double bond with C₂ → $sp^2$ hybridised
• C₂ (=C=): forms two double bonds (one with C₁, one with O) → $sp$ hybridised

(ii) CH₃CH=CH₂ (Propene)

• C₁ (CH₃): four single bonds → $sp^3$ hybridised
• C₂ (CH=): part of C=C double bond → $sp^2$ hybridised
• C₃ (=CH₂): part of C=C double bond → $sp^2$ hybridised

(iii) (CH₂)₂CO (Cyclopropanone)
The ring has three carbons and a carbonyl group:

• C₁ (C=O, carbonyl carbon): forms a double bond with O → $sp^2$ hybridised
• C₂ and C₃ (the two –CH₂– ring carbons): each forms four single bonds → $sp^3$ hybridised

(iv) CH₂=CHCN (Acrylonitrile)

• C₁ (=CH₂): part of C=C → $sp^2$ hybridised
• C₂ (CH=): part of C=C → $sp^2$ hybridised
• C₃ (–C≡N): triple bond with N → $sp$ hybridised

(v) C₆H₆ (Benzene)
All six carbon atoms are part of the aromatic ring with alternating double bonds → all six carbons are $sp^2$ hybridised.

The basic theme of organisation in the periodic table is to classify the elements in such a way that elements with similar properties are grouped together. In the Modern Periodic Table, elements are arranged in order of increasing atomic numbers (Z) in horizontal rows called periods and vertical columns called groups. Elements in the same group have similar valence shell electronic configurations and hence exhibit similar physical and chemical properties. This arrangement reflects the periodicity of properties with increasing atomic number.

Given: Chlorination of methane (free radical reaction).

Concept: Chlorination of methane proceeds via a free radical chain mechanism involving initiation, propagation, and termination steps.

Explanation:

During the chlorination of methane, methyl free radicals ($\cdot CH_3$) are produced in the propagation step:
$$CH_4 + Cl\cdot \rightarrow \cdot CH_3 + HCl$$

In the termination step, two methyl free radicals can combine to form ethane:
$$\cdot CH_3 + \cdot CH_3 \rightarrow CH_3 - CH_3 \text{ (Ethane)}$$

This coupling of two methyl radicals during the termination step accounts for the formation of ethane as a by-product during the chlorination of methane.

Given/Concept: A chemical bond is the force of attraction that holds two atoms together in a molecule or compound.

Formation of a Chemical Bond:

Atoms combine to form chemical bonds in order to attain a state of minimum energy and maximum stability (usually by achieving the nearest noble gas configuration).

There are three main types of chemical bonds:

1. Ionic (Electrovalent) Bond: Formed by the complete transfer of one or more electrons from an electropositive atom to an electronegative atom. The resulting oppositely charged ions attract each other electrostatically. Example: NaCl — Na loses one electron to Cl, forming Na⁺ and Cl⁻.

2. Covalent Bond: Formed by the mutual sharing of electron pairs between two atoms, both of which are short of the noble gas configuration. Example: H₂ — each H atom shares its one electron with the other.

3. Coordinate (Dative) Bond: A special type of covalent bond where both electrons of the shared pair are donated by one atom (the donor) to another (the acceptor). Example: NH₄⁺ — the lone pair of NH₃ is donated to H⁺.

Driving Force: The formation of a bond lowers the potential energy of the system. At the equilibrium bond distance, the energy is at a minimum — the system is most stable.

Conclusion: Chemical bonds form because the bonded state is energetically more stable than the separated atoms.

Correct option: (ii) whose value is independent of path.

A state function is a property whose value depends only on the current state of the system (i.e., initial and final states) and not on the path taken to reach that state. Examples include internal energy (U), enthalpy (H), entropy (S), and Gibbs energy (G).

Given: A liquid–vapour equilibrium in a sealed container at fixed temperature; volume is suddenly increased.

(a) Initial effect on vapour pressure:
When the volume is suddenly increased, the same number of vapour molecules now occupy a larger volume. Therefore, the concentration (and hence the partial pressure) of the vapour decreases initially.

(b) Initial change in rates:

• Rate of evaporation: Evaporation depends on the nature of the liquid and temperature, not on the volume of the container. Hence, the rate of evaporation remains unchanged initially.
• Rate of condensation: Condensation depends on the concentration (number density) of vapour molecules. Since the vapour pressure has decreased, the rate of condensation decreases initially.

Because rate of evaporation > rate of condensation, more liquid evaporates to restore equilibrium.

(c) Final state after equilibrium is restored:
More liquid evaporates until the rate of evaporation once again equals the rate of condensation. At the new equilibrium, the vapour pressure equals the original vapour pressure (since temperature is unchanged and vapour pressure depends only on temperature). Thus, the final vapour pressure is the same as the original vapour pressure.