Londhe Chemistry Classes ( LCC)

Date : TEST ID: 3

Time : 00:60:00 CHEMISTRY
Marks :
1.SOLID STATE,1.THE SOLID STATE,2.SOLUTIONS,2.SOLUTIONS AND COLLIGATIVE PROPERTIES,3.CHEMICAL
THERMODYNAMICS AND ENERGETICS,3.IONIC EQUILIBRIA ,4.CHEMICAL THERMODYNAMICS ,5.CHEMICAL
KINETICS,6.CHEMICAL KINETICS

Single Correct Answer Type 10. In which case Raoult's law is not applicable?
1. Which of the following will show a negative a) 1 M NaCl b) 1 M urea
deviation from Raoult's law? c) 1 M glucose d) 1 M sucrose
a) Acetone-benzene b) Acetone-ethanol 11. The solubiiity product of Al(OH)3 is 2.7 ×
c) Benzene-methanol d) Acetone-chloroform 10−11 . Its solubility in gL−1 and pH of this
2. When mango is placed in 1 N HCl solutions, it solution are respectively (Atomic mass of Al =
a) Swells b) Shrinks 27u )
c) Remains unchanged d) Bursts a) 7.8 × 10−2 , 11.5 b) 7.8 × 10−3 , 9.5
3. Desalination can be carried out by ______ c) 7.8 × 10−2 , 9.5 d) 7.8 × 10−3 , 11.5
a) Diffusion 12. The ionisation constant of benzoic acid is
b) Reverse osmosis 6.46 × 10−5 and K sp for silver benzoate is
c) Osmosis 2.5 × 10−13 . How many times is silver
d) Semi-permeable membrane benzoate more soluble in a buffer of pH 3.19 as
4. Which of the following concentration terms compared to its solubility in pure water?
is/are independent of temperature? a) 1.5 b) 2.8
a) Molality b) Molality and mole c) 3.2 d) 2.3
fraction 13. NaOH(aq), HCl(aq) and NaCl(aq)
c) Molarity and mole d) Molality and concentration of each is 10−3 M. Their pH will
fraction normality be respectively
5. Sugar dissolved in water is a ________ type of a) 10,6,2 b) 11,3,7
solution c) 10,2,6 d) 3,4,7
a) Solid in solid b) Solid in gas 14. The pK a of HCN is 9.30. The pH of a solution
c) Solid in liquid d) Gas in solid prepared by mixing 2.5 moles of KCN and 2.5
6. If α is the degree of dissociation of Na2 SO4 , the moles of HCN in water and making up the total
van’t Hoff’s factor (i) used for calculating the volume of 500 mL, is
molecular mass is ______ a) 9.30 b) 7.30
a) 1 + α b) 1 − α c) 1 + 2α d) 1 − 253 c) 10.30 d) 8.30
7. What would happen if a thin slice of sugar beet 15. Then pKa of HCN is 9.30. The pH of a solution
is placed in a concentrated solution of NaCl in prepared by mixing 2.5 moles of KCN and 2.5
water? moles of HCN in water and making up the total
a) Sugar beet will lose water from its cells volume of 500 mL is
b) Sugar beet will lose water from solution a) 9.30 b) 7.30
c) Sugar beet will neither absorb nor lose c) 10.30 d) 8.30
water 16. What will be the pH and % a respectively for
d) Sugar beet will dissolve in solution the salt BA of 0.1M concentration? Given, K a
8. A 500 g toothpaste sample has 0.2 g fluoride. for HA = 10−6 and K b for BOH = 10−6 .
What is the concentration of F − in ppm? a) 7,10% b) 5,10%
a) 200 b) 250 c) 400 d) 1000 c) 5,0.1% d) 7,1%
9. What is the normality of 1 M solution of H3PO4 17. The pK b for fluoride ion at 25∘ C is 10.83, the
? ionisation constant of hydrofluoric acid at this
a) 0.33 N b) 1 N c) 2 N d) 3 N temperature is

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 a) 2.72 × 10−5 b) 3.52 × 10−3 c) Change in enthalpy d) Change in free
c) 6.76 × 10 −4 d) 5.38 × 10−2 energy
18. A buffer solution is prepared in which the conc. 26. C + O2 ⟶ CO2 ; ∆H = −396 kJ mol−1
NH3 is 0.30 M and the concentration of Heat liberated in formation of 0.154 kg of CO2
NH4+ is 0.20 M. If the equilibrium constant, Kb is _________
for NH3 equals 18×10−5 , what is the pH of this a) 1386 kJ b) 13.86 kJ c) 138.6 kJ d) 1386 J
solution? 27. If 900 J/g of heat is exchanged at boiling point
a) 8.73 b) 9.08 of water, then what is increase in entropy?
c) 9.44 d) 11.72 a) 43.4 JK −1 mol−1 b) 87.2 JK −1 mol−1
19. A 100 Ml, 0.1 M solution of ammonium acetate c) 900 JK −1 mol−1 d) Zero
is diluted by adding 100 mL of water. The pH 28. Equal volumes of molar KOH and are
of the resulting solution will be neutralized by dilute H2SO4 solution and x kcal
(pKs of acetic acid is nearly equal to pKb of and y kcals of heat are liberated respectively.
NH4OH) Which of the following is true?
a) 4.9 b) 5.9 a) x = y b) x=1/3 y
c) 7.0 d) 10.0 c) x=1/2 y d) x=1/4 y
20. Three sparingly soluble salt that have same 29. Conversion of oxygen to ozone represented by
solubility products are given below. the equation 3O2 → 2O3 is an endothermic
I. A2 X reaction. Enthalpy change ∆H accompanying
II. AX the reaction __________
III. AX3 a) Is negative
Their solubilities order in a saturated solution b) Is positive
will be c) Is zero
a) ∥> | >∥ I ∥ b) ∥ l ∣>∥> 1 d) Is either negative or zero
c) ∥ il > | >∥ d) ∥> ||| > | 30. The heat energy required to ionise the
21. Calculate the free energy change for following molecules is given as follows
1 ΔH1 ΔH2
2CuO(s) → Cu2 O(s) + O2(g) N2 (𝑔) ⟶ N2⊕ (𝑔), O2 (𝑔) ⟶ O+
2 (𝑔)
2 ΔH3 Δ𝐻4
(Given: ∆H = 145.6 J mol−1 , T = 300 K ∆S = Li2 (𝑔) ⟶ Li⊕ ⊕
2 (𝑔), C2 (𝑔) ⟶ C2 (𝑔)
116 J mol−1 K −1 ) The correct decreasing order of energy in
a) 110.8 kJ mol−1 b) 221.5 kJ mol−1 terms of heat
c) 55.4 kJ mol−1 d) 145.6 kJ mol−1 Δ𝐻 > Δ𝐻3 > Δ𝐻2 Δ𝐻 > Δ𝐻3 > Δ𝐻1
a) 1 b) 2
22. ΔH for combustion of ethane and ethyne are > Δ𝐻4 > Δ𝐻4
−341.1 and −310.0kcal, respectively. What Δ𝐻 > Δ𝐻4 > Δ𝐻1 Δ𝐻 > Δ𝐻1 > Δ𝐻4
c) 3 d) 3
will be the ratio of calorific values of ethane > Δ𝐻2 > Δ𝐻2
and ethyne, respectively? 31. For a chemical reaction at 27℃, the activation
a) 1: 0.95 b) 0.65: 2 energy is 600 R. The ratio of the rate constants
c) 0.95: 1 d) 0.002: 1 at 327℃ to that of at 27℃ will be _________
23. Calculate the temperature at which a) 2 b) 40 c) e d) e2
∆G = -5.2 kJ mol-1, ∆H = 145.6 kJ mol-1 32. D) Order of a Reaction can be
and ∆S = 216 JK-1 mol-1 for a chemical reaction a) Fractional b) Zero
a) 6980C b) 4250C c) 650 K d) 6500C c) Negative d) Both (A) and (B)
24. 1 mole of conc. HCl requires x moles of dil. 33. The disintegration constant of radium with
NaOH for neutralization and 1 mole of conc. half-life 1600 yr is
H2SO4 requires moles of same dil. NaOH. Then a) 2.12 × 10−4 yr −1 b) 4.33 × 10−4 yr −1
which of the following equation is true? c) 3.26 × 10 yr
−3 −1
d) 4.33 × 10−12 yr −1
1 1
a) x = 2y b) x = y c) x = 2 y d) x = 3 y 34. For a chemical reaction ... A... can never be a
25. Mixing of non-reacting gases is generally fraction. Here, A refers to
accompanied by ________ a) half-life b) molecularity
a) Decrease in entropy b) Increase in entropy c) order d) rate constant

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35. During the course of a chemical reaction, the 43. Which type of crystal should be softest and
rate constant _________ have lowest melting point?
a) Increases as the reaction proceeds a) Ionic Crystals b) Molecular Crystals
b) Remains constant c) Metallic Crystals d) Covalent Crystals
c) Decreases as the reaction proceeds 44. A metallic element crystallizes in simple cubic
d) First decreases followed by an increase lattice. Each edge length of the unit cell is 3 Å.
36. The rate constant is given by the equation k = The density of the element is 8 g/cc. Number of
−Ea
pze RT . Which factor should register a decrease unit cells in 108 g of the metal is _________
for the reaction to proceed more rapidly? (Molar mass of the metal = 108 g/mol)
a) T b) z c) Ea d) p a) 1.33 × 1020 b) 2.7 × 1022
37. The half-life of a first order reaction having c) 5 × 1023 d) 2 × 1024
rate constant k = 1.7 × 10−5 s−1 is ______ 45. Potassium fluoride has NaCl type structure.
a) 12.1 hr b) 9.7 hr c) 11.3 hr d) 1.8 hr What is the distance between K + and F − ions if
38. Diazonium salt decomposes as C6 H5 N2+ Cl− → cell edge is ‘a’ cm?
C6 H5 Cl + N2 at 0℃. The evolution of N2 a) 2a cm b) a/2 cm c) 4a cm d) a/4 cm
becomes two times faster when initial 46. Which of the following crystal will exhibit the
concentration of the salt is doubled. Thus it is relation a = b = c and α = β = γ = 90∘ ?
________ a) Diamond b) Cinnabar
a) A first order reaction c) Calcite d) Ice
b) A pseudo order reaction 47. If ' a ' stands for the edge length of the cubic
c) Independent of square of initial systems, simple cubic, body centred cubic and
concentration of reactant face centred cubic, then the ratio of radii of the
d) A zero order reaction spheres in these systems will be respectively,
1 1 1 1
39. The reaction, X → Product follows first order a) a: √3 a: a b) 2 a: √3a: √2 a
2 4 2√2
kinetics. In 40 minutes the concentration of X
c) 1 a: √3 a: √2 a d) 1a: √3a: √2a
changes from 0.1 M to 0.025 M. Then the rate 2 2 2
of reaction when concentration of X is 0.01 M 48. If r is the radius of the atom in close packing
will be ________ and A is the radius of the octahedral voids than
a) 1.73 × 10−4 M min−1 b) 3.47 × 10−5 M min−1 r/R is equal to
c) 3.47 × 10−4 M min−1 d) 1.73 × 10−5 M min−1 a) 9.1 b) 2.41
40. The rate of a reaction A + 3B → C, the rate of c) 3.22 d) 4.76
formation of C is
a) The same as rate of consumption of A 49. Which of the following statement is NOT
b) The same as the rate of consumption of B CORRECT?
c) Twice the rate of consumption of A a) The number of carbon atoms in a unit cell of
d) 3/2 times the rate of consumption of B diamond is 4
41. In a face centered cubic arrangement of A and The number of Bravais lattices in which a
b)
B atoms, if A atoms are at the corner of the unit crystal can be categorised is 14
cell and B atoms at the face centres, and one of c) The fraction of the total volume occupied by
the A atom is missing from one corner in unit the atoms in a primitive cell is 0.48
cell. Then the simplest formula of compound is d) Molecular solids are generally volatile
________ 50. The property of a substance due to presence of
a) A7 B3 b) AB3 c) A7 B24 d) A7/8 B3 unpaired electrons is called
42. How many space lattices are obtainable from a) Para magnetism b) Diamagnetism
the different crystal systems? c) Polymerisation d) Polymorphosis
a) 7 b) 14 c) 32 d) 230

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 Londhe Chemistry Classes ( LCC)
Date : TEST ID: 3
Time : 00:45:00 CHEMISTRY
Marks : 50
1.SOLID STATE,1.THE SOLID STATE,2.SOLUTIONS,2.SOLUTIONS AND COLLIGATIVE PROPERTIES,3.CHEMICAL
THERMODYNAMICS AND ENERGETICS,3.IONIC EQUILIBRIA ,4.CHEMICAL THERMODYNAMICS ,5.CHEMICAL
KINETICS,6.CHEMICAL KINETICS

: ANSWER KEY :
1) d 2) b 3) b 4) b 29) b 30) c 31) c 32) d
5) c 6) c 7) a 8) c 33) b 34) b 35) b 36) c
9) d 10) a 11) a 12) c 37) c 38) a 39) c 40) c
13) b 14) a 15) a 16) a 41) c 42) b 43) d 44) c
17) c 18) c 19) c 20) a 45) b 46) a 47) a 48) b
21) a 22) c 23) b 24) c 49) c 50) a
25) b 26) a 27) a 28) a

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 Londhe Chemistry Classes ( LCC)
Date : TEST ID: 3
Time : 00:45:00 CHEMISTRY
Marks : 50
1.SOLID STATE,1.THE SOLID STATE,2.SOLUTIONS,2.SOLUTIONS AND COLLIGATIVE PROPERTIES,3.CHEMICAL
THERMODYNAMICS AND ENERGETICS,3.IONIC EQUILIBRIA ,4.CHEMICAL THERMODYNAMICS ,5.CHEMICAL
KINETICS,6.CHEMICAL KINETICS

: HINTS AND SOLUTIONS :

Single Correct Answer Type Raoult's law is not applicable, if the total number
1 (d) of particles of solute changes in the solution due
to association or dissociation.

11 (a)
Let s be the solubility of AI(OH)3
Acetone and chloroform will show a negative Al(OH)3 ⇌ Al3+ (aq) +
deviation due to their association after mixing. 3OH − (aq)
Concentration at t = 0 1 0
4 (b) 0
Molality and mole fraction depends only upon Concentration
weight but not on volume. Hence, these are not at equilibrium 1-S S
affected by change in temperature. 3S
K sp = [Al3+ ][OH]−3 = (S)(3S)3 = 27S4
6 (c)
Na2 SO4 → 2Na+ + SO2− K sp 27 × 10−11
4 S4 = = = 1 × 10−12
1 0 0 27 27 × 10
S4 = 1 × 10−3 mol L−1
(1 − α) 2α α
Molar mass of Al(OH)3 is 78 g. Therefore,
∴ i = (1 − α) + α + 2α = 1 + 2α
7 (a) solubility of Al(OH)3 in g
Beet cell solution will be less concentrated than L−1 = 1 × 10−3 × 78gL−1
concentrated NaCl solution. So, solvent molecules = 78 × 10−3 gL−1
(water) will move from dilute solution of beet = 7.8 × 10−2 gL−1
cells to concentrated NaCl solution, due to pH of the solution
osmosis. Cell membrane will act as S = 1 × 10−3 molL−1
semipermeable membrane [OH − ] = 3S = 3 × 1 × 10−3 = 3 × 10−3
8 (c) pOH = 3 − log 3
ppm means part per million pH = 14 − pOH = 11 + log 3
To find parts by mass of fluoride ion in 1 million = 11.4771
parts by mass of the toothpaste: 12 (c)
500 g toothpaste contains 0.2 g fluoride, 1000000 C6 H5 COOAg(s) ⇌ C6 H5 COO− + Ag + ; K 1 = K sp
g toothpaste contains 𝑥 g fluoride, 1
C6 H5 COO− + H + ⇌ C6 H5 COOH; K 2 =
500 0.2 Ka
then 1000000 = 𝑥
C6 H5 COOAg(s) + H ⇌ C6 H5 COOH + Ag + ; K 3
+
6 4
0.2 × 10 20 × 10 K sp
𝑥= = = 4 × 102 = 400 ppm =
500 5 × 102 Ka
9 (d)
[C6 H5 COOH][Ag + ] S ⋅ S S2 K sp
H3PO4 is tribasic acid. K3 = + = + = + =
[H ] [H ] [H ] Ka
10 (a)
where, S is the solubility of C6 H5 COOAg.
In a buffer of pH = 3.19
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 log[H + ] = −3.19 = 4̅. 81 HF + H2 O ⇌ H3 O+ + F −
[H+ ] = antilog 4̅ . 81 = 6.46 × 10−4 [H3 O+][F−]
K= [HF]
….(iii)
S2 Ksp Ksp ×[H+ ]
[H+ ]
= or S2 = K b ⋅ K = [H3 O+ ][OH− ] = K w
Ka Ka

2.5×10−13 ×6.46×10−4
Kw
S=√ or =K
6.46×10−5 Kb
S = √2.5 × 10−13 × 10
Taking log on both sides, we get
S = 1.6 × 10−6 M (in buffer)
In aqueous solution, solubility of C6 H5 COOAg : log K = log K w − log K b
K sp = [C6 H5 COO− ][Ag + ] = S ⋅ S = S2
S = √K sp = √2.5 × 10−13 = 5 × 10−7 M = −pK w + log K b
S(C6 H5 COOAg) in buffer 1.6 × 10−6
= = −14 + 10.83 = −3.17
S(C6 H5 COOAg) in aqueous solution 5.0 × 10−7
= 3.2 or K = 6.76 × 10−4
C6 H5 COOAg is 3.2 times more soluble in buffer
than in pure water. 18 (c)
13 (b) Given, conc. of NH3 (base) = 0.30M
NaOH = [OH− ] = 10−3 Conc. of NH4+ (salt) = 0.20M
[H + ][OH− ] = 10−14 K b = 1.8 × 10−5
[H + ] = 10−11 [ salt ]
∵ pOH = pK b + log
pH = − log[H + ] = − log[10−11 ] = 11 [ base ]
HCl(aq) = [H + ] = 10−3 0.20
⇒ 4.74 + log = 4.74 + (0.301 − 0.477)
pH = − log[10−3 ] = 3 0.30
NaCl(aq) = Neutral; [H + ] = [OH− ] = 10−7 = 4.74 − 0.176 = 4.56
i.e . pH = 7 ⇒ pH = 14 − 4.56 = 9.44
14 (a) 19 (c)
[KCN] 2.5 Ammonium acetate is a salt of weak acid and
pH = pK a + log = 9.3 + log = 9.30 weak base. When solutions are of weak and weak
[HCN] 2.5
15 (a) base, pH of solution is calculated from the
[salt] following relationship.
∵ pH = pK a + log 1 1
[acid] pH = 7 + ( pK a − pK b )
As [salt] = [acid] 2 2
∵ pK a ≈ pK b (given)
∴ pH = pK a = 9.30
∴ pH of solution = 7
16 (a)
1 1 1 20 (a)
∵ pH = pK w + pK a − pK b 1
Ksp 3
2 2 2 S1 = [ 4 ] for A2 Xi
1 1 1
= × 14 + × 6 − × 6 1
2 2 2 K 4
S∥l = √K sp for AX; S∥II = [ sp ] for AX3
pH = 7 27
∴ S∥ > S1 > S∥II
Kw 10−14
α=√ = √ −6 = √10−2 21 (a)
Ka × Kb 10 × 10−6 ∆G = ∆H − T∆S = 145600 − 300 × 116
= 10−1 = 0.1 or 10% = 145600 − 34800
17 (c) = 110.8kJ/mol
F + H2 O ⇌ HF + OH− 22 (c)
[HF][OH− ] Calorific value is heat produced by 1 g of fuel.
Kb = [F−]
…….(i)
−341.1
Calorific value of ethane = = −11.37kcal/g
Kw = [H3 O+ ][OH− ]= 10−14 …..(ii) 30
−310.0
Given, pK b = 10.83 Calorific value for ethyne = = −11.92kcal/
26
Dissociation of HF in water is represented by the g
equation, ∴ Ratio of calorific values of ethane and ethyne

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 = 11.37: 11.92 = 0.95: 1 concentration of reactants, therefore remains
constant
23 (b) 36 (c)
216
-5.2 = 145.6 – T (1000) or T = 698 K = 4250C For the reaction to proceed more rapidly, k should
25 (b) increase which can be brought about by decreases
Mixing of non-reacting gases increases in Ea
randomness and so increases entropy 37 (c)
26 (a) 0.693 0.693
t1/2 = = s = 40764.7 s
Heat liberated in the formation of 1 mole k 1.7 × 10−5
40764.7
(i. e. 44 × 10−3 kg) of CO2 = 396 kJ = hr = 11.32 hr ≅ 11.3 hr
3600
∴ Heat liberated in the formation of 0.154 kg of
38 (a)
0.154 × 396 d d d[N2 ]
CO2 = = 1386 kJ rate = − dt [C6 H5 N2+ Cl− ] = dt [C6 H5 Cl] =
44 × 10−3 dt
27 (a) rate of evolution of N2 becomes two times faster
(900 × 18) than the initial i.e., (rate)2 = 2(rate)1 ; when
∆vap S = = 43.4 JK −1 mol−1
373 concentration of salt is doubled i.e.,
29 (b) [C6 H5 N2+ Cl− ]2 = 2[C6 H5 N2+ Cl− ]1
For endothermic reaction, ∆H = +ve Let the order of reaction be 𝑥
30 (c) ∴ rate = k[C6 H5 N2+ Cl− ]𝑥
Enthalpy of ionisation depends upon the size of (rate)1 = k[C6 H5 N2+ Cl− ]1𝑥
molecules. Larger the size lesser will be ionisation (rate) 2 = k[C6 H5 N2+ Cl− ]2𝑥 = k 2𝑥 [C6 H5 N2+ Cl− ]1𝑥
energy. (rate)1 k[C6 H5 N2+ Cl− ]𝑥
∴ =
(rate)2 k. 2𝑥 [C6 H5 N2+ Cl− ]𝑥
31 (c)
1 1
Activation energy (Ea ) = 600 R; = 𝑥 ∴𝑥=1
2 2
T1 = 27℃ = 300 K
∴ The reaction is first order reaction
T2 = 327℃ = 600 K
39 (c)
k 2 Ea T2 − T1
ln = [ ] t = 40 min, [X]0 = 0.1 M, [X] = 0.025 M
k1 R T1 × T2
2.303 0.1
600R 600 − 300 ∴k= log
= [ ] t 0.025
R 300 × 600 2.303 2.303
300 = × log 4 = × 2 log 2
= 600 [ ] 40 40
300 × 600 2.303 × 2 × 0.3010
=1 = = 0.03467 min−1
40
k2 ∴ Rate = k[𝑥]
∴ = e1 = e
k1 = (0.03467 × 0.01)M min−1
33 (b) = 3.467 × 10−4 M min−1
0.693
λ= ≈ 3.47 × 10−4 M min−1
t12
44 (c)
where, λ = disintegration constant = ? z×M 1 × 108
N0 = =
d × a3 8 × (3 × 10−8 )3
t12 = 1600yr 108 × 10+24
= = 0.5 × 1024 = 5 × 1023
216
0.693
∴ λ = 1600 = 4.33 × 10−4 yr −1 46 (a)
Diamond is a cubic crystal system that has a =
34 (b) b = c intercept and α = β = γ = 90∘ crystal
For a chemical reaction, molecularity can never be angles.
a fraction.
47 (a)
35 (b) For simple cubic, r = 2
a

Rate constant is independent of concentration of

√3a
reactants. It is the rate of reaction at unit For body centred cubic, r = 4

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 a R r rvoid
For face centred cubic, r = 2√2 For OV, r (i. e. r+ or ) = 0.414
− r
Therefore, the ratio of radii of scc, bcc and fcc.
r 1
∴ = 0.414 = 2.41
a √3 a R
= : a: 2√2
2 4
49 (c)
48 (b) The fraction of the total volume occupied by the
atoms in a primitive cell is 0.52

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