Chemistry NCERT Exemplar Solutions Class 12th Chapter Ten

K.E = φ - φ?
φ? = 3 eV = 3 * 1.6 * 10? ¹? J = 4.8 * 10? ¹? J
φ = hc/λ = (6.63 * 10? ³? * 3 * 10? ) / (248 * 10? ) J = 8 * 10? ¹? J
K.E = 8 * 10? ¹? - 4.8 * 10? ¹? = 3.2 * 10? ¹? J
Now using, λ = h / √ (2 K.E m)
λ = (6.63 * 10? ³? ) / √ (2 * 3.2 * 10? ¹? * 9.1 * 10? ³¹) m
λ = (6.63 * 10? ³? ) / (7.63 * 10? ²? ) m = 0.87 * 10? m = 8.7 Å
So, the nearest integer is 9.

T? = 300 K; K? = 1 * 10? ³ s? ¹
T? = 200 K; K? =?
E_a = 11.488 kJ/mol
Using Arrhenius equation:
log (K? /K? ) = (E_a / 2.303R) * [ (T? - T? ) / (T? )]
log (K? / 10? ³) = (11.488 * 10³ / (2.303 * 8.314) * [ (-100) / (6 * 10? )]
log (K? / 10? ³) = -1
K? / 10? ³ = 10? ¹
K? = 10? s? ¹ or K? = 10 * 10? s? ¹

The complex CoCl? ·4NH? is represented as [CoCl? (NH? )? ]Cl.
Here, 4 NH? are neutral monodentate ligands. So, 2 equivalents of ethylene diamine replace 4NH? since ethylene diamine is a didentate ligand.

ΔG° = -9.478 kJ/mol
Using ΔG° = -2.303 RT log K_p
-9.478 * 10³ = -2.303 * 8.314 * 495 log K_p
1 = log K_p ⇒ K_p = 10
Here, for the given reaction A (g)? B (g), K_p = K_c
Initial A = 22 mmol
At equilibrium A = 22 - x mmol; B = x mmol
K_c = [B] / [A] = (x/V) / (22-x)/V) = x / (22-x) = 10
x = 10 (22-x) ⇒ x = 220 - 10x ⇒ 11x = 220 ⇒ x = 20
So, mmol of B at equilibrium are 20.

m = 10 molal
K_b = 0.5 K kg mol? ¹
Using: ΔT_b = I K_b m
and α = (i - 1) / (n - 1)
n for AB? is 3; α = 0.1
0.1 = (i - 1) / (3 - 1) ⇒ I = 1.2
ΔT_b = 1.2 * 0.5 * 10 = 6 °C
So, boiling point of solution = 100 + 6 = 106 °C

A 6.5 molal solution means 6.5 moles of KOH is in 1 kg (1000 g) of solvent (H? O).
Moles of solute, n_B = 6.5
Mass of solute, W_B = 6.5 * 56 = 364 g
Mass of solvent, W_A = 1000 g
Mass of solution = 1364 g
Volume of solution = 1364 / 1.89 mL
Now, molarity = [6.5 / (1364 / 1.89)] * 1000 M = 9 M

Moles of carbon in organic compound = Moles of carbon in CO?
n_c = 420 / 44 moles
Mass of carbon in organic compound = (420 / 44) * 12 = 114.54 g
Moles of hydrogen in compound = 2 * moles of H? O
n_H = 2 * (210 / 18) moles
Mass of hydrogen = 2 * (210 / 18) g = 23.33 g
% of H = (23.33 / 750) * 100 = 3.11%
The nearest integer is 3.

Solubility product of A? X = 4S? ³
Where S? is the solubility of salt A? X.
Solubility product of MX = S? ²
Where S? is the solubility of MX.
Given 4S? ³ = 4 * 10? ¹² ⇒ S? = 10? M
Given S? ² = 4 * 10? ¹² ⇒ S? = 2 * 10? M
So, S? / S? = 10? / (2 * 10? ) = 50

The balanced equation is:
2MnO? + 5C? O? ²? + 16H? → 2Mn²? + 10CO? + 8H? O
So, the value of c (coefficient for H? ) = 16.

Edge length in bcc, a? = 27 Å
Let, Edge length in fcc be a? Å
Now, the same element crystallises in bcc as well as fcc.
For bcc: 4r = √3 a? ⇒ r = (√3 / 4) a?
For fcc: 4r = √2 a? ⇒ r = a? / (2√2)
So, (√3 / 4) a? = a? / (2√2)
(√3 / 4) * 27 = a? / (2√2)
a? = 33.13 Å
The nearest integer is 33.