Chemistry Test 2 with solutions

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Illustration 1: One mole of a mixture of CO and CO₂ requires exactly 20 gms of NaOH to convert all the CO₂ into Na₂CO₃. How many more gms of NaOH would it require for conversion into Na₂CO₃ if the mixture (one mole) is completely oxidised
to CO₂.

(A) 60 gm (B) 80 gm

(C) 40 gm (D) 20 gm

Solution: Moles of NaOH =

Moles of CO₂ = [Q ‘n’ factor for CO₂ = 2]

Moles of CO =

Moles of CO₂ produced = from CO

Moles of NaOH extra =

Mass of NaOH extra = 60

Illustration 2: One litre of 0.1 M CuSO₄ solution is electrolysed till the whole of copper is deposited at cathode. During the electrolysis a gas is released at anode. The volume of the gas is

(A) 112ml (B) 254 ml

(C) 1120 ml (D) 2240 ml

Solution: When copper is deposited at cathode, oxygen gas is released at anode

Equivalents of copper = 0.1 ´ 2 = 0.2

Equivalents of oxygen = 0.2

Volume of oxygen at NTP = 0.2 ´ 5600 = 1120 ml

Illustration 3: In a reaction

FeS₂ + MnO₄ + H⁺ ¾® Fe³⁺ + SO₂ + Mn²⁺ + H₂O

The equivalent mass of FeS₂ would be equal to

(A) Molar mass (B)

(C) (D)

Solution: Fe²⁺ ¾® Fe³⁺ + e^–

S₂^–2 ¾® 2S⁺⁴ + 10e^–

–––––––––––––––––

FeS₂ ¾® 2S⁺⁴ + Fe³⁺ + 11e^–

\ Equivalent mass of FeS₂ =

Illustration 4: Equal volumes of 0.2M HCl and 0.4M KOH are mixed. The concentration of the principal ions in the resulting solution are

(A) [K⁺] = 0.4M, [Cl^–] = 0.2M, [H⁺] = 0.2M

(B) [K⁺] = 0.2M, [Cl^–] = 0.1M, [OH^–] = 0.1M

(C) [K⁺] = 0.1M, [Cl^–] = 0.1M, [OH^–] = 0.1M

(D) [K⁺] = 0.2M, [Cl^–] = 0.1M, [OH^–] = 0.2M

Solution: Resulting solution in alkaline since M(KOH) > M(HCl), hence the molarity of KOH after reaction

= = 0.1M

KOH + HCl KCl + H₂O

t = 0 0.4 0.2

t = t 0.2 0 0.2

0.1 0.1 due to dilution

\ [K⁺] = 0.1 + 0.1 = 0.2M

[Cl^–] = 0.1M

[OH^–] = 0.1M

Illustration 5: To prepare a solution that is 0.5M KCl starting with 100ml of 0.4M HCl

(A) Add 0.75gm KCl (B) Add 20 ml of water

(C) Add 0.10 mol of KCl (D) Evaporate 10 ml water

Solution: a) 100ml of 0.4M KCl = mol = 0.04 mol (initial)

= 0.04 ´ 74.5 (initially) = 2.98gm

= 3.73g (after) = 0.05 mol in 100 ml

= 0.5M

\ (A) is true

b) Addition of 20 ml water will decrease molarity that will be less than 0.4 M

\ (B) is false

c) 04 (initial as in (a)) + 0.10 (added)

= 0.14 mol KCl in 100 ml

= 1.4 M

\ (C) is false

d) 04 mol KCl in 90 ml solution (after 10 ml water evaporated) = 0.044 M

\ (D) is also false

Illustration 6: For the reaction

+ ¾® CaHPO₄ + 2H₂O

Which are true statements

(A) Equivalent weight of H₃PO₄ is 49

(B) Resulting mixture is neutralised by 1 mol of KOH

(C) CaHPO₄ is an acid salt

(D) 1 mol of H₃PO₄ is completely neutralised by 1.5 mol of Ca(OH)₂.

Solution: a) By given reaction

H₃PO₄ º 2OH^–

\ Equivalent weight = = 49

\ (A) is true

b) Only one acidic H in CaHPO₄ hence neutralised by 1 mol of KOH

\ (B) is true

c) One acidic H in CaHPO₄ makes it acidic salt.

\ (C) is true

d) H₃PO₄ º 3H⁺ º 3OH^– º5 Ca(OH)₂

\ (D) is true

Illustration 7: The brown ring complex compound is formulated as [Fe(H₂O)₅NO]SO₄. The oxidation number of iron is

(A) 1 (B) 2

(C) 3 (D) 0

Solution: [Fe(H₂O)₅NO]SO₄ [Fe(H₂O)₅NO]²⁺ + SO₄^2–

In this complex NO transfers one of its electron to Fe (which is initially +2)

This gives +1 charge on NO and +1 charge on Fe. Fe has thus three unpaired electrons as confirmed by its magnetic moment which is B.M.

\ (A)

Illustration 8: An organic compound contains 4% sulphur minimum molecular weight is

(A) 200 (B) 400

(C) 800 (D) 16000

Solution: 4 gm sulphurs is in 100g compound hence 32gm sulphur is in
= 800 g compound.

\ (C)

Illustration 9: If 20 ml of 0.2 M K₃[Fe(CN)₆] is reduced by some equivalents of N₂H₄, then calculate numbers of moles of N₂H₄ required for the above reaction the answer is

(A) 10^–5 (B) 10^–3

(C) 10^–6 (D) 10^–2

Solution: N₂H₄ + K₃[Fe(CN)₆ + KOH ¾® K₄[Fe(CN)₆] + N₂ + H₂O

\ Number of equivalents of

K₃[Fe(CN)₆] =

Number of moles of

N₂H₄= = 10^–3

\ (B)

Illustration 10: 0.7gm of Na₂CO₃.XH₂O were dissolved in water and the volume was made to 100 ml, 20 ml of this solution required 19.8 ml of N/10 HCl for complete neutralisation. The value of x is

(A) 7 (B) 3

(C) 2 (D) 5

Solution: Meq. of Na₂CO₃.xH₂O in 20 ml = 19.8 ´

\ Meq. Of Na₂CO₃.xH₂O in 100 ml = 19.8 ´ ´ 5

\ = 19.8 ´ ´ 5

or =

M = 141.41

\ 23 ´2 + 12 + 3 ´ 16 + 18x = 141.41

x = 2

\ (C)

Illustration 11: The Vander Wall’s constant b is equal to

(A) The molecular volume of 1 mol of the gas

(B) Two times the molecular volume of 1 mole of the gas

(C) Three times the molecular volume of 1 mole of the gas

(D) Four times the molecular volume of 1 mole of the gas

Solution: ‘b’ is the volume occupied by 1mole of a gas molecules. Excluded volume for 2 gas molecules

= p (2r)³ = pr³

\ Excluded volume / molecule =

\ b = ´ N_A

\ b = 4 ´ volume of 1 mole of molecules

Illustration 12: What will be percentage of ‘free volume’ available in 1 mole of H₂O_(l) at 760 mm and 373 K. Density of H₂O_(g) at 373 K is 0.958 gm/ml.

(A) 0.0613% (B) 99.9386%

(C) 0.0543% (D) 99.9457%

Solution: Volume occupied by 1 mole of H₂O_(g) at 373 K =

= = 30.62L

Volume of 1 mole of H₂O_(l)

= = 18.789 ml

% of volume occupied by liquid water

= 0.0613%

% of free volume

= 100 – 0.0613 = 99.9386%

Illustration 13: One mole of each monoatomic, diatomic and triatomic gases are mixed, C_p/C_V for the mixture is

(A) 1.40 (B) 1.428

(C) 1.67 (D) None of these

Solution: C_v =

C_p =

= 1.428

Illustration 14: The vapour pressure of water at 20°C is 17.5 torr. What will be the no. of moles of water present in one litre of air at 20°C and 40% relative humidity.

(A) 4.2 ´ 10^–4 mole (B) 4.2 ´ 10^–6 mole

(C) 3.82 ´ 10^–4 mole (D) 3.82 ´ 10^–6 mole

Solution: Relative humidity (RH) =

\ Partial pressure of H₂O = RH ´ Vapour pressure of H₂O

= = 7 torr

= = 0.0092 atm

Now PV = nRT

= 0.000382 mole

= 3.82 ´ 10^–4 mole

Illustration 15: The compressibility factor of a gas is less than unity at STP. Therefore

(A) V_m > 22.4 litre (B) V_m < 22.4 litre

(C) V_m = 22.4 litre (D) V_m = 44.8 litre

Solution: Z =

Ar Z < 1 (given)

\ or PV < nRT

P = 1 atm

V = V_m

R = 0.082

T = 273

\ V_m < 0.0821 ´ 273

= V_m < 22.4 litre

Illustration 16: The composition of the equilibrium mixture (Cl₂ 2Cl), attained at 1200°C, is determined by measuring the rate of effusion through a narrow aperture. At 1.80 mm Hg pressure, the mixture effuses 1.16 times as fast as Krypton under the same conditions. What will be the fraction of chlorine molecules dissociated into atoms (Atomic mass of Kr = 84)

(A) 13% (B) 13.7%

(C) 26% (D) 26.4%

Solution:

or M = 62.425

for Cl₂ 2Cl

initially 1 0

at equilibrium (1 – a) 2a

i (Van’t Hoff factor) = 1 – a + 2a = 1 + a

or 1 + a =

on solving a = 0.137

or a = 13.7%

Illustration 17: A spherical balloon of 21 cm diameter and 4.851 L volume is to be filled with hydrogen at NTP from a cylinder containing the gas at 20 atm and 27°C the cylinder can hold 2.82 litre of water at N.T.P. what will be the number of balloons that can be filled up?

(A) 5 (B) 10

(C) 15 (D) 20

Solution: Volume of one balloon = 4.821 litre

Volume of n balloons = 4.821 n litre

Total volume of hydrogen in the cylinder at NTP

V = = 51.324 litre

Actual volume of H₂ to be transferred to balloons

= 51.324 – 2.82 = 48.504 litre

(\ 2.82 L is retained in the cylinder)

\ NO. of balloon, n = » 10

Illustration 18: The numerical value of for a gas at critical condition is … times of at normal condition

(A) 4 (B) 8/3

(C) 3/8 (D) 1/4

Solution: At critical temperature

\ P_CV_C= RT_C

\ for a gas at critical temperature is times of that gas at NTP

Illustration 19: What will be the molecular diameter of Helium if Vander Waal’s constant, b = 24 ml mole^–1?

(A) 2.71Å (B) 2.71mm

(C) 5.42Å (D) 542Å

Solution: b = 4V_m

(V_m is the volume occupied by one mole of gas)

or 24 = 4 ´ N₀ ´ pr³

or r = = 1.355 ´ 10^–8 cm

\ d = 2r = 2 ´ 1.355 ´ 10^–8 cm

= 2.71Å

Illustration 20: The density of phosphorus vapour at 310°C and 775 torr is 2.64g dm^–3. What is the molecular formula of phosphorus?

(A) P (B) P₂

(C) P₄ (D) P₈

Solution:

or M =

Put d = 2.64gm dm^–3

R= 0.0821 dm³ atm K^–1 mol^–1

P = atm

and T = 310 + 273 = 583K

We get M = 123.9g mol^–1

No. of P atoms in a molecule = = 3.99» 4

Thus the molecular formula of phosphorus = P₄

Illustration 21: The nucleus of an atom is located at x = y = z = 0. If the probability of finding an s-orbital electron in a tiny volume assumed x = a, y = z = 0 is 1 ´ 10^–5, what is the probability of finding of the electron in the same sized volume around x = z = 0, y = a?

(A) 1 ´ 10^–5 (B) 1 ´ 10^–5a

(C) 4 ´ 10^–5a² (D) 1 ´ 10^–5 ´ a^–1

Solution: Since s-orbital is spherically symmetrical and has equi-distance from the nucleus. Hence the probability is identical

\ (A)

Illustration 22: What is the probability at the second site if the electron were in a p_z orbital for the data given in question number (1)

(A) 1 ´ 10^–5 (B) 2 ´ 10^–5

(C) 4 ´ 10^–5 (D) 0

Solution: P_Z has a node at z = 0, hence probability of finding of e^– in this volume is zero

\ (D)

Illustration 23: Following ions will be coloured if Aufbaun rule is not followed

(A) Cu⁺² (B) Fe⁺²

(C) Se⁺³ (D) (A), (B) true

Solution: Ion with unpaired electrons in d or f-orbitals will be coloured

\ (D)

Illustration 24: If wavelength is equal to distance travelled by the electron in one second, then

(A) (B)

(C) (D)

Solution: Since, the distance travelled in one second by velocity = vc_m = l

Q l =

\ l = Þ l =

\ (D)

Illustration 25: Number of photons of light of wavelength 4000Å required to provide 1.00J of energy is

(A) 2.01 ´ 10¹⁸ (B) 12.01 ´ 10³¹

(C) 1.31 ´ 10¹⁷ (D) None is correct

Solution: Q E = nhn where n = no. of photons

\ E =

\ n = = 2.01 ´ 10¹⁸

\ (A)

Illustration 26: When a certain metal was irradiated with a light of frequency 3.2 ´ 10¹⁶ Hz the photoelectrons emitted had twice the kinetic energy as did photo electrons emitted when the same metal was irradiated with light frequency 2.0 ´ 10¹⁶Hz. Hence threshold frequency is

(A) 1.6 ´ 10¹⁶ Hz (B) 0.8 ´ 10¹⁶ Hz

(C) 8 ´ 10¹⁵ Hz (D) 8 ´ 10¹⁶Hz

Solution: Q hn = hn₀ + KE

\ KE = h(n – n₀)

Again 2(KE)₁ = (KE)₂

2h(n₁– n₀) = (n₂ – n₀) ´ h Þ 2(n – n₀) = (n₂ – n₀)

\ Þ 2n₁ – n₂= n₀

Since n₁ = 2 ´ 10¹⁶ Hz

n₂ = 3.2 ´ 10¹⁶ Hz

\ n₀ = 8 ´ 10¹⁵ Hz

\ (C)

Illustration 27: The radial distribution curve of 2s sub-level consists of ‘a’ nodes, a is

(A) 1 (B) 3

(C) 2 (D) 0

Solution: Since the radial nodes = n – l – 1

Hence a = 2 – 0 – 1 = 1

\ (A)

Illustration 28: Uncertainty in position and momentum are equal. Uncertainty in velocity is

(A) (B)

(C) (D) None

Solution: Given Dp = Dx

Since Dp ´ Dx »

(Dp)² »

Dp »

\ Dv =

\ (C)

Illustration 29: If the radius of first Bohr orbit of H-atom is x, then de-Broglie wavelength of electron in 3^rd orbit is nearly:

(A) 2px (B) 6px

(C) 9x (D)

Solution: Q mvr₃ =

mv = Þ mv =

\ l = = 6px

\ (B)

Illustration 30: If the shortest wavelength of H-atom in Lyman series is x, then the longest wavelength in Balmer Series of He⁺ is

(A) (B)

(C) (D)

Solution: For the shortest wave length of Lymann series in H-atom

Þ = R_H

Let the longest wavelength of Balmer series is l

\ l_max =

\ (A)

Illustration 31: For the reaction 2NO + Br₂ ¾® 2NOBr, the following mechanism has been given

fast

NO + Br₂ NOBr₂

NOBr₂ + NO 2NOBr

Hence rate law is;

(A) k[NO]²[Br₂] (B) k[NO][Br₂]

(C) k[NOBr₂] [NO] (D) k[NO][Br₂]²

Solution: For the reaction

NOBr₂ + NO 2NOBr

\ rate = k₁[NOBr₂] [NO] …(1)

But from the reaction

NO + Br₂ NOBr₂

K_equilibrium = Þ [NOBr₂] = K_equilibrium [NO] [Br₂] …(2)

Putting equation (2) in equation (1)

rate = k₁ k_equilibrium [NO]²[Br₂]

= k[NO]²[Br₂], where k = k₁ k_equilibrium

\(A)

Illustration 32: Following is the graph between (a – x)^–1 and time for second order reaction,
Q = tan^–1 (0.5), OA = 2L mol^–1.

Hence rate at the start of the reaction is:

(A) 1.25 L mol^–1 min^–1 (B) 0.5 mol lit^–1 min^–1

(C) 0.125 L mol^–1 min^–1 (D) 1.25 L mol^–1 min^–1

Solution: For second order reaction,

Kt =

\ (a – x)^–1 = Kt +

\ k = tanq (\ tanq = 0.5 from question)

\ K = 0.5 mol lit^–1 min^–1

and = 2 L mol^–1

\ (C)

Illustration 33: Half-life period in Q.N. = (2), above is

(A) 1.386 min (B) 4 min

(C) 16 min (D) 2 min

Solution: t_1/2 (for 2^nd order reaction) =

\ (B)

Illustration 34: The rate constant for the reaction

2N₂O₅ ¾® 4NO_2(g) + O_2(g) is 3 ´ 10^–5 sec^–1. If the rate is 2.4 ´ 10^–5 mol lit^–1 sec^–1, the concentration of N₂O₅ in mol L^–1 is;

(A) 1.4 (B) 1.2

(C) 0.04 (D) 0.8

Solution: The decomposition of N₂O₅ is a first order reaction

\ rate = k[N₂O₅]

2.4 ´ 10^–5 = 3 ´ 10^–5[N₂O₅]

\ [N₂O₅] = = 0.8 mol L^–1

\ (D)

Illustration 35: Following is the graph between logT₅₀ and log a (where a = initial concentration) for a given reaction at 27°C. Hence order is;

(A) 0 (B) 1

(C) 2 (D) 3

Solution: T₅₀ µ a^(1–n) where n = order of reaction

Þ T₅₀ = k a^{(1 – n)}

or, logT₅₀= logk + (n – 1)log a

from question, q = tan^–1(m)

Þ n – 1 = 1

Þ n = 2

\ (C)

Illustration 36: For the reaction

is equal to

(A) k₁(a – y) – k₂(a – y) (B) k₂(a – y) – k₁ (a– y)

(C) k₁(a – y) + k₂(a – y) (D) –k₁(a – y) – k₂(a – y)

Solution: = k₁(a – y) + k₂(a – y)

\ = – k₁(a – y) – k₂(a – y)

\ (D)

Illustration 37: There are two radionuclei A & B A is an a-emitter and B is a b-emitter, their disintegration constant are in the ratio of 1:2. What should be the number of atoms of two at time t = 0. So that probability of getting of a and b-particles are same at time t = 0

(A) 2 : 1 (B) 1:2

(C) 1:4 (D) 4:1

Solution: Since the rate of disintegrations are same

l_AN_A = l_B×N_B Þ =

\ (A)

Illustration 38: At radioactive equilibrium, the ratio between the atoms of two radioactive elements A and B was found to be 3.1 ´ 10⁹ : 1 respectively. If T₅₀ of the element A is 2 ´ 10¹⁰ years, then T₅₀ of the element B is

(A) 6.2 ´ 10⁹ years (B) 6.45 years

(C) 3 ´ 10⁸ years (D) None

Solution: At radioactive equilibrium

t_50(B) = = = years

\ (B)

Illustration 39: The no. of b-particles emitted during the charge

_ax^C ¾® _dy^b is

(A) (B) + c

(C) – a (D) – c

Solution: _ax^c ¾® _dy^b + m × ₂He⁴ + n × _–1b⁰

Hence a = d + 2m – n

\n = d + 2m – a = – a

\ (C)

Illustration 40: One mole of A present in a closed vessel undergoes decay as

_zA^m ¾® _z–4B^{m – 8} + 2 × ₂He⁴

The volume of He collected at NTP after 20 days (t_1/2(A) = 10 days) is

(A) 11.2 L (B) 22.4 L

(C) 33.6 L (D) 67.2L

Solution: Q = 2ⁿ where n = no. of half lives

= = 2² Q n = = 2

\ N = mole

\ decayed moles = moles

But moles of He formed = moles

= ´ 22.4 L at NTP

= 33.6 L at NTP

\ (C)

Illustration 41: Which one of the following does not of dissolve in conc. H₂SO₄?

(A) CH₃ – C = C – CH₃ (B) CH₃ – CH₂ – C º CH

(C) CH º CH (D) CH₂ = CH₂

Solution: If CH º CH were to dissolve in H₂SO₄ a bisulphite salt of vinyl carbocation H₂C = C⁺H would be formed. The more s-character in the positively charged ‘C’ less stable is the carbocation and less likely to be formed.

\(C)

Illustration 42: Which one of the following compounds will give in the presence of peroxide a product different from that obtained in the absence of peroxide?

(A) 1-butane (B) 1-butene, HBr

(C) 2-butene, HCl (D) 2-butene, HBr

Solution: Peroixde effect is observed when unsymmetrical alkene is treated with HBr only (and not with HCl and HI).

\(B)

Illustration 43: Which of the following alkene on acid catalysed hydration form 2-methyl propan-2-ol.

(A) (CH₃)₂CH = CH₂(B) CH₃ – CH = CH₂

(C) CH₃ – CH = CH – CH₃ (D) CH₃ – CH₂ – CH = CH₂

Solution: Addition of H₂O occurs according to Markownikoff’s rule.

\ (A)

Illustration 44: Which of the following compounds yields only one product on monobromination?

(A) Neopentane (B) Toluene

(C) Phenol (D) Aniline

Solution: CH₃ – CH₃ has twelve equivalent 1°H. Hence H forms only one product on monobromination.

\(A)

Illustration 45: Aqueous solution of the following compounds are electrolysed. Acetylene gas is obtained from.

(A) Sodium fumarate (B) Sopdium maleate

(C) Sodium succinate (D) Both (A) and (B)

Solution:

\ (D)

Illustration 46: Dehydration of butan-2-ol with conc. H₂SO₄ gives preferred product.

(A) but-1-ene (B) but-2-ene

(C) propene (D) ethane

Solution: CH₃ – – – CH₃ CH₃ – CH = CH – CH₃ (80%)

+ CH₃ – CH₂ – CH = CH₂ (20%)

This is in accordance with saytzeff rule.

\ (B)

Illustration 47: CH₃ – C º C – CH₃ ‘X’. What is X

(A) CH₃CH₂CH = CH₂ (B) CH₃CH₂C º CH

(C) CH₃ – CH = CH – CH₃ (D) CH₂ = C = CH – CH₃

Solution: Isomerisation occurs, when 2-butyne is treated with NaNH₂, it converts into terminal alkyne (1-butyne).

\ (B)

Illustration 48: Identify the compound ‘Y’ in the following sequence of reactioin

HC º CH

*(A)*		*(B)*
*(C)*		*(D)*	*CH₃COOH*

Solution:

\ (A)

Illustration 49: Dehydration of 1-butanol gives 2-butene as a major product, by which of the following intermediate the compound 2-butene obtained

*(A)*			*(B)*
*(C)*			*(D)*
*Solution:*

\ (C)

Illustration 50: The principal organic compound formed in the reaction

CH₂ = CH(CH₂)COOH + HBr …………. is

(A)	CH₃ –– (CH₃)₈COOH	(B)	CH₂ = CH(CH₂)₈COBr
(C)	–CH₂(CH₂)₈COOH	(D)	CH₂ = CH(CH₃)₇ – – COOH

Solution: Follows the peroxide effect

\ CH₂ = CH(CH₂)₈COOH – CH₂(CH₂)₈COOH

\ (C)

Illustration 51: Kc for reaction [Ag(CN)₂]^–Ag⁺+2CN^–, the equilibrium constant at 25°C is 4.0´10^-19 then the silver ion concentration in a solution which was originally 0.1 molar in KCN and 0.03 molar in AgNO₃ is

(A) 7.5´10¹⁸ (B) 7.5´10^-18

(C) 7.5´10¹⁹ (D) 7.5´10^-19

Solution 2KCN + AgNO₃ ¾¾® Ag(CN)₂^– + KNO₃+ K⁺

0.1 0.03 0 0 0

(0.1-0.06) 0 0.03 0.03 0.03

[Ag(CN)₂]^– = 0.03

Now [Ag(CN)₂]^– Ag⁺ + 2CN^–

0.03 0 0.04 (left from KCN)

(0.03-a) ‘a’ (0.04+a)

a is very – very small

\0.03 -a » 0.03

and 0.04 + a » 0.04

\Kc= 4´10^-19=

\ a = 7.5´10^-18 Ans

Illustration 52 Given A ¾® B + C, DH= -10 Kcals, the energy of activation of backward reaction is 15 Kcals mol^-1 if the energy of activation of forward reaction in the presence of a catalyst is 3 K Cal mol^-1, the catalyst will increase the rate of reaction at 300 K by the number of times equal to

(A) e^3.33 (B) e^4.21

(C) e^-2.7 (D) e^2.303

Solution DH = Ea (for forward reaction) –Ea (of backward reaction)

-10 = Ea for FR-15

\ Ea for FR = 5 kcals mol^-1

the catalyst decreases the Ea by 2 k Cals mol^-1

Illustration 53: A vessel contains H₂(g) at 2 atm pressure, when H₂S(g) at a pressure of 4 atm is introduced into the vessel. Where reaction

8H₂S(g) 8H₂+S₈(s)

Occurs at a temperature of 1000 K. It is found that

= , then

(A) max^m wt of solid formed is 32 gm

(B) max^m wt of solid formed is 0.32 gm

(C) K_p= K_cRT

(D) K_c=256

Solution: Under identical conditions of volume and temperature Pµn

so initially

8H₂S(g) 8H₂(g) + S₈(s)

Initially 2 1

at equilibrium (2-x) (1+x)

But at equilibrium

\ Kp =

Since Dn = 8-8 =0

\ Kp = Kc = 256

Illustration 54: At a certain temperature the following equilibrium is established

CO(g) + NO₂(g) CO₂(g) + NO(g)

One mole of each of the four gases is mixed in one litre container and the reaction is allowed to reach equilibrium state. When excess of baryta water is added to the equilibrium mixture, the wt of white precipitate obtained is 236.4 gm. The equilibrium constant Kc of the following reaction is

(A) 1.2 (B) 2.25

(C) 2.1 (D) 3.6

Solution CO(g) + NO₂(g) CO₂(g) + NO(g)

t = 0 1 1 1 1

t=t (1-x) (1-x) (1+x) (1+x) at equilibrium

CO₂+Ba(OH)₂ ¾® BaCO₃+H₂O

Moles of BaCO₃ =

\ moles of CO₂ at equilibrium =1.2

(1+x) = 1.2

x = 0.2

\ Kc=

Illustration 55 Rate of disappearance of the reactant A at two different temperatures is given by A B

heat of reaction in the given temperature range, when equilibrium is set up is

(A)

(B)

(C)

(D) None

Solution: At 300°K, Kc =

At 400°K, Kc =

DH =

\ (C) is correct

Illustration 56: Liquid NH₃ ionizes to a slight extent at -60°C. Its ionic product

The number of NH₂^– ions present per ml of pure liquid NH₃are

(A) 300 ions (B) 400 ions

(C) 600 ions (D) 500 ions

Solution: 2NH₃ NH₄⁺+ NH₂^–

K = [NH₄⁺] [NH₂^–] = x² = 10^-30

\ x = 10—^{15—24591580810328902198alkjdfakslkdlsfjadsklfjadsklfjadsklfjasdlkfjdaslkfjas dkladsklfj}^-15 M = [NH₂^–]

[NH₂^–] = 10^-15´ ´6.023´10²³

= 600 ions/ml

Illustration 57: When sulphur in the from of S₈ is heated at 900 K, the initial pressure of one atm falls by 29% at equilibrium. This is because of conversion of some S₈ to S₂. Find the value of equilibrium constant for this reaction

(A) 1.16 atm³ (B) 0.71 atm³

(C) 2.55 atm³ (D) 5.1 atm³

Solution: S₈(g) 4S₂(g)

at t =0 1 0

at equilibrium 1-0.29 4´0.29

= 0.71 atm =1.16 atm

K_p =

Illustration 58: At equilibrium, constants for the reaction at 1000° K are

CoO(s) + H₂(g) Co(s) + H₂O(g) K₁ = 65

CoO(s) + CO(g) Co(s) + CO₂(g) K₂= 500

The equilibrium constant for the reaction at 1000° K

CO(g) + H₂O(g) CO₂(g) + H₂(g)

(A) 0.13 (B) 7.69

(C) 0.0179 (D) 69.3

Solution: K₁ = [H₂O]/[H₂] = 65

K₂ = [CO₂] / [CO(g) ] = 500

Now, for the required reaction

To obtain this, divide K₂ by K₁, we get

Illustration 59: 92 gms of ethyl alcohol were treated with 120 gms acetic acid and equilibrium was established, when 117.34 gms of ester and 24 gm of water were formed. The equilibrium constant is

(A) 4 (B) 8

(C) 10 (D) 15

Solution: C₂H₅OH + CH₃COOH CH₃COOC₂H₅ + H₂O

t =0 92gm 120 gm 0 0

92/46=2 126/60=2 117.34 gm 24 gm

at equi. (2-1.33) (2-1.33)

Kc=

Illustration 60: For the reaction

2NOCl(g) 2NO(g) + Cl₂(g)

the values of DH° and DS° at 298 K are 77.2 kJ mol^-1 and 122 JK^-1 mol^-1 respectively. The standard equilibrium constant at the same temperature is

(A) 0.695´10^-8 (B) 6.95´10^-8

(C) 69.5´10^-8 (D) 695´10^-8

Solution: Using the relation DG° = DH^° – TDS^°

DG° = 77200-298´122

= 40844 J mol^-1

let the equilibrium constant be K^°_c

we know that

DG° = -2.303 RT logK^°_c

log K^°_c = -7158

K^°_c = 6.95´10^-8

Illustration 61: An acidic indicator HIn (K_in = 10^–6) ionises as HIn H⁺ + In^–. The acid colour predominates over the basic colour when HIn is at least 10 times more concentrated than In^– ion. On the other hand basic colour predominates over the acid colour when the In^– ion is at least 5 times more concentrated than HIn. Hence pH range of the indicator is

(A) 5.0 – 6.7 (B) 7.0 – 8.7

(C) 5.3 – 7.0 (D) 7.0 – 8.1

Solution: pH = pK_in + , where pK_in = 6

For acid colour to predominates

\ pH £ 6 – 1 i.e 5.0

for base colour to predominate ³ 5

\ pH ³ 6 + log5 Þ pH ³ 6.7

\ (A)

Illustration 62: The correct statement amongst the following is

(A) A strong electrolyte remains completely dissociated at all dilutions

(B) Upon dilution the degree of dissociation of a weak electrolyte and number of ions per unit volume of its solution both increase.

(C) A strong electrolyte is completely ionised at all dilutions but not completely dissociated.

(D) pH of solution of a weak acid decreases with dilution.

Solution: “Complete dissociation” implies that interionic attraction has completely ceased to exist. This condition in the case of a solution of strong electrolyte is achieved only at infinite dilution when concentration of solution tends to zero.

\ (C)

Illustration 63: 4M solution of a weak monobasic acid (x% ionized and pH = 3.0) is diluted to 1 M by adding water (distilled). Percentage ionisation and pH of solution after dilution will be respectively.

(A) 2x and 2.7 (B) 0.25x and 3.3

(C) 0.5x and 2.7 (D) 2x and 3.3

Solution: Solution of weak acid being concentrated, we can use the approximate

a = i.e. a µ

Thus, decreasing the conc. to one fourth, a will be doubled. Doubling of a means doubling of percentage ionisation. Hence % ionisation will be 2x.

[H⁺] = i.e. [H⁺] µ

If the conc. is decreased to one fourth of its original value H⁺ ion conc. will be halved. Thus, after dilution [H⁺] = 0.5 ´ 10^–3M

\ pH = 3.3

\ (D)

Illustration 64: pH of a buffer solution changes from 6.20 to 6.17 when 0.003 mole of acid is added to 500 mL of the buffer. The buffer capacity of the system is, therefore

(A) 0.1 (B) 0.3

(C) 0.2 (D) 0.4

Solution: Buffer capacity =

= = 0.2

\ (C)

Illustration 65: K_sp of CaSO₄ is 2.4 ´ 10^–5 at 25°C. In a solution containing Ca²⁺ ions the precipitation of CaSO₄ begins to occur when SO₄^2– ion concentration in the solution is made just to exceed the value of 4.8 ´ 10^–3M. Hence concentration of Ca²⁺ ion in the solution is

(A) 200 ppm (B) 40 ppm

(C) 400 ppm (D) 100 ppm

Solution: [Ca²⁺] [SO₄^2–] = 2.4 ´ 10^–5

[Ca²⁺] = = 5 ´ 10^–3M

1 L solution contains 5 ´ 10^–3 mole of Ca²⁺ ions

1000 L solution will contain 5 mole i.e. 200 g Ca²⁺ ions

Taking density of aqueous solution to be unity

1000 kg i.e. 10⁶ g solution contains 200g Ca²⁺ions

Conc. of Ca²⁺ ion = 200 ppm

\ (A)

Stoichiometry

How many grams of phosphoric acid would be needed to neutralise 100gm of magnesium hydroxide?

(Molecular weight of H₃PO₄ = 98 and Mg(OH)₂ = 58.3 gm)

(A) 66.7 (B) 252gm

10 L of hard water required 0.56 gm of lime (CaO) for removing hardness. Hence temporary hardness in ppm of CaCO₃is

(A) 100 (B) 200

20 ml of xM HCl neutralises completely 10 ml of 0.1 M NaHCO₃ solution and further 5 ml of 0.2 M Na₂CO₃ solution in the presence of methyl orange at end point, the value of x is

(A) 0.167M (B) 0.133M

When 10 ml of ethyl alcohol (d = 0.7893 gm/ml) is mixed with 20 ml of water (d = 0.9971 g/ml) at 25°C, the final solution has a density 0.9571 gm/ml. The percentage change in total volume on mixing is

(A) 3.1% (B) 2.4%

0.635 gm of a cupric salt was dissolved in water and excess of KI was added in the solution and the liberated iodine required 25ml of N/5 Na₂S₂O₃ solution. The weight percentage of copper in the salt is

(A) 25% (B) 40%

The chloride of a metal (M) contains 65.5% of chlorine 100 ml of the chloride of the metal at STP weight 0.72gm. The molecular formula of the metal chloride is

(A) MCl₃ (B) MCl

22.4 litres of H₂S and 22.4 litre of SO₂ both at STP are mixed together. The amount of sulphur precipitated as a result of chemical calculation is

(A) 16gm (B) 23gm

In the mixture of NaHCO₃ and Na₂CO₃ volume of a given HCl required is x ml with phenolphthalein indicator and further y ml required with methyl orange indicator. Hence volume of HCl for complete reaction of NaHCO₃ is

(A) 2x (B) y

A sample of oleum is labelled 109%. The % of free SO₃ in the sample is

(A) 40% (B) 80%

7.65 ´ 10^–3 moles of NaNO₃ require 3.06 ´ 10^–2 moles of a reducing agent to get reduced to NH₃. What is the n-factor of the reducing agent?

(A) 2 (B) 3

How many grams of sodium bicarbonate are require to neutralise 10ml of 0.902 M vinegar?

(A) 8.4 gm (B) 1.5gm

1 gm mole of oxalic acid is treated with conc. H₂SO₄. The resultant gaseous mixture is passed through a solution of KOH. The mass of KOH consumed will be

(A) 28gm (B) 56gm

NH₃ + OCl^– ¾® N₂H₄ + Cl^–

On balancing the above equation in basic solution using integral coefficients, which of the following whole number will be the coefficient of N₂H₄?

(A) 1 (B) 2

A certain compound has the molecular formula X₄O₆. If 10 gm of X₄O₆ has 5.72g X, atomic mass of X is

(A) 32 amu (B) 37 amu

In the following reaction

28NO₃^– + 3As₂S₃ + 4H₂O ¾® 6AsO₄^3– + 28NO + 9SO₄^2– + 8H⁺

equivalent weight of As₂S₃ (with molecular weight M) is

(A) (B)

Gaseous State

At constant pressure what would be the percentage decrease in the density of an ideal gas for a 10% increase in the temperature.

(A) 10% (B) 9.1%

A small hole is pricked in a container containing a mixture of CH₄ and O₂ and placed in vacuum. What should be the molar ratio of methane to oxygen in the container when the two gases are effusing out in the ratio of 2:1

(A) (B)

The Van der Wall’s equation is (V – nb) = nRT for n moles where ‘a’ and ‘b’ are Van der Wall’s constant which of the following statements are true about ‘a’ and ‘b’ when the temperature of the gas is too low

(A) Both remains same (B) ‘a’ remains same, b varies

The tube in the figure is shielded at both end and heated upto double the original temperature both side of Hg column gases are packed with increasing temperature the Hg column

(A) Shift towards ‘B’ (B) Shifts towards A

The molar specific heat at constant volume of mixture of gases A and B is 4.33 cal. A is monoatomic and B is diatomic then the ratio of moles of A and B is

(A) 1:1 (B) 2:1

Rate of diffusion of two gases are r_A and r_B, molecular weights are M_A and M_B then partial pressure P_Ais (if number of moles are n_A and n_B).

(A) (B)

There are two gases A and B having degree of freedom f_A and f_B, the difference in energy supplied for increment in temperature of one mole gas by 50°C is (in cal)

(A) 50 (f_A – f_B) (B) 100 (f_A– f_B)

When an ideal gas undergoes unrestricted expansion no cooling takes place because the molecules

(A) Exert no attractive forces on each other (B) Do work equal to loss of KE

If two gases of molecular weights M_A and M_B at temperatures T_A, T_B and T_AM_B= T_BM_A, then which property has the same magnitude for both the gases.

(A) Density (B) Pressure

At low pressure Vander Wall’s equation for 3 moles of a real gas will have its simplified form

(A) (B)

for two gases A and B with molecular weights M_A and M_B, its observed that a certain temperature T, the mean velocity of A is equal to the root mean square velocity of B. Thus the mean velocity of A can be made equal to the mean velocity of B if

(A) A is at temperature T, and B at T¢×T > T¢

(B) A is lowered to a temperature T₂ = T

(D) Both A and B are placed at lower temperature

The quantity represents the

(A) Number of molecules in the gas (B) Mass of the gas

The rms velocity of hydrogen is times the rms velocity of nitrogen. If the temperature of the gas

(A) T(H₂) = T(N₂) (B) T(H₂) > T(N₂)

1 lt of N₂ and l of O₂ at the same temperature and pressure were mixed together what is the relation between the masses of two gases in the mixture.

(A) (B)

Consider a mixture of SO₂ and O₂ kept at room temperature compared to the oxygen molecule, the SO₂ molecule will hit the wall with

(A) Smaller average speed (B) Greater average speed

Atomic Structure

Wave function vs distance from nucleus graph of an orbital is given below:

The number of nodal sphere of this orbital is

(A) 1 (B) 2

For an electron in a hydrogen atom the wave function, y is proportional to , where a₀ is the Bohr’s radius. What is the ratio of probability of finding the electron at the nucleus to the probability of finding it at a₀.

(A) e (B) e²

What transition in He⁺ion shall have the same wave number as the first line in Balmer series of H-atom

(A) 3 ® 2 (B) 6 ® 4

The potential energy of the electron in an orbit of H-atom would be

(A) – mv² (B)

An electron is moving with a kinetic energy of 4.55 ´ 10^–25 Joules. What will be the de-Broglie wave length for this electron?

(A) 5.28 ´ 10^–7 m (B) 7.28 ´ 10^–7m

If each orbital can hold a maximum of 3-electrons. The number of elements in 4^th period of periodic table is:

(A) 48 (B) 57

The no. of orbital in a sub-shell is equal to

(A) n² (B) 2l

Which of the following curves may represent the speed of the electron in a hydrogen atom as a function of principal quantum no. n.

(A) a (B) b

The ratio of the energy of the electron in ground state of hydrogen to that of the electron in the first excited state of Be⁺³is

(A) 1 : 4 (B) 1 : 8

The electronic transition from n = 2 to n = 1 will produce the shortest wavelength in

(A) H-atom (B) D-atom

The wave number of first line of Balmer series of hydrogen is 152,000 cm^–1. The wavelength of first Balmer line of Li⁺²ion is

(A) 15,200 cm^–1 (B) 60,800 cm^–1

The dissociation energy of H₂ is 430.53 kJ mol^–1. If H₂ is dissociated by illumination with radiation of wavelength 253.7 nm. The fraction of the radiant energy which will converted into kinetic energy is given by

(A) 8.86% (B) 2.33%

The orbital angular momentum of an electron in 2s-orbital is

(A) (B) zero

Photoelectric emission is observed from a surface for frequencies v, and v₂ of the incident radiation (where v₁ > v₂). If the maximum kinetic energy of the photoelectrons in two cases are in the ratio of 1:k, then the threshold frequency v₀ is given by

(A) (B)

The difference between nth and (n + 1)th Bohr’s radius of H-atom is equal to its (n – 1)th Bohr’s radius. The value of n is

(A) 1 (B) 2

Chemical Kinetics

A radioactive isotopes x with half-life of 1.37 ´ 10⁹ years decays to y which is stable. A sample of rock from the moon was found to contain both the elements x and y in the ratio 1:7. What is the age of the rock.

(A) 1.96 ´ 10⁸ years (B) 3.85 ´ 10⁹ years

A radioactive mixture containing a short lived species A and short lived species B. Both emitting a-particles at a given instant, emits at rate 10,000 a-particles per minute. 10 minutes later, it emits at the rate of 7000 particles per minute. If half lives of the species are 10 min and 100 hours respectively, then the ratio of activities of A : B in the initial mixture was

(A) 3:7 (B) 4:6

₉₂U²³⁸ (IIIB) undergoes following emissions:

₉₂U²³⁸

Which is/are correct statements

(A) A will be of IB group (B) A will be of III B group

The electron in a hydrogen atom makes transition from M shell to L, the ratio of magnitude of initial to final acceleration of electron is:

(A) 9:4 (B) 81:16

In order to determine the volume of blood in an animal without killing it, a 1.00 ml sample of an aqueous solution containing tritium is injected into the animal blood stream. The sample injected has an activity of 1.8 ´ 10⁶ cps (counts per second). After sufficient time for the sample to be completely mixed with the animal blood due to normal blood circulation, 2.00 ml of blood are withdrawn from animal and the activity of the blood sample withdrawn is found to be 1.2 ´ 10⁴ cps. Calculate the volume of the animal blood.

(A) 300ml (B) 200ml

A radioactive isotope is being produced at a constant rate x. Half-life of the radioactive substance is y. After sometimes no. of radioactive nuclei becomes constant, the value of this constant is

(A) (B) xy

Number of nuclei of a radioactive substance at time t = 0 are 1000 and 900 at time = 2 sec. Number of nuclei at t = 4s will be

(A) 800 (B) 810

There are two radioactive substances A and B. Decay constant of B is twice that of A. Initially both have equal no. of nuclei. After n-half lives of A, the rate of disintegration of both becomes equal, the value of n is

(A) 1 (B) 2

A radioactive nuclide is produced at a constant rate of a-per second. It decay constant is l. If N₀ be the no. of nuclei at time t = 0, then maximum no. of nuclei possible are

(A) (B) N₀ +

For the reaction

R – X + OH^–‑ ¾® R – OH + X^–, the rate expression is given as rate

= 4.7 ´ 10^–5 [R – X] [OH^–] + 0.24 ´ 10^–5 [R – X]

What % of R – X react by S_N2 mechanism when [OH^–] = 0.001 M

(A) 1.9 (B) 4.7

The mechanism of the reaction

2NO + O₂ ¾® 2NO₂ is

k₁

NO + NO N₂O₂ (fast)

k_–1

N₂O₂ + O₂ 2NO₂ (slow)

the rate constant for the reaction is

(A) k₂ (B) k₂×k₁×k_–1

The inversion of cane sugar proceeds with half-life of 600 minutes at pH = 5 for any concentration of sugar. However if pH = 6, the half-life changes to 60 minute. The rate law expression for sugar inversion can be written as

(A) r = k×[sugar]² [H⁺]⁰ (B) r = k× [sugar]¹ [H⁺]⁰

Rate of chemical reaction is nA ¾® Product, is doubled when the concentration of A is increased four times. If the half-life time of the reaction at any temperature is 16 minutes, then the time required for 75% of the reaction to complete is

(A) 24.0 minutes (B) 27.3 minutes

The rate of a chemical reaction generally increases rapidly even for small temperature increase because of rapid increase in the

(A) Collision frequency

(B) Fraction of molecules with energies in excess of the activation energy

(D) Average kinetic energy of the molecule

Rate constant, k = 1.8 ´ 10⁴mol^–1 Ls^–1 and E_a = 2 ´ 10² kJ mol^–1, when T ® ¥, then the value of A is

(A) 1.8 ´ 10⁴ kJ mol^–1 (B) 1.8 ´ 10⁴ mol^–1L sec^–1

Chemical Equilibrium

For the reaction A + B C + D, equilibrium concentrations of [C] = [D] = 0.5M if we start with 1 mole each of A and B. Percentage of A converted into C if we start with 2 mole of A and 1 mole of B is

(A) 25% (B) 40%

CH₃ – CO – CH_3(g) CH₃ – CH_3(g) + CO_(g)

Initial pressure of CH₃COCH₃ is 100mm when equilibrium is set up, mole fraction of CO_(g) is hence K_p is

(A) 100mm (B) 50mm

For the reaction

CaCO_3(s) CaO_(s) + CO_2(g)

Equilibrium constant K_p is 1.642 atm at 1000K, if 40 gm of CaCO₃ was put into a 10 L flask, percentage of calcium carbonate would remain unreacted at the equilibrium.

(A) 66% (B) 34%

If equilibrium constant of

CH₃COOH + H₂O CH₃COO^– + H₃O⁺

Is 1.8 ´ 10^–5, equilibrium constant for

CH₃COOH + OH^– ¾® CH₃COO^– + H₂O is

(A) 1.8 ´ 10^–9 (B) 1.8 ´ 10⁹

PCl₅ is 40% dissociated when pressure is 2 atmosphere it will be 80% dissociated when pressure is approximately

(A) 0.2 atm (B) 0.5 atm

66. For the reaction: 2HI_(g) H_2(g) + I_2(g), the degree of dissociated (a) of HI_(g) is related to equilibrium constant K_p by the expression

(A) (B)

For which of the following reactions, the degree of dissociation cannot be calculated from the vapour density data.
i) 2HI_(g) H_2(g) + I_2(g)
ii) 2NH_3(g) N_2(g) + 3H_2(g)

iii) 2NO_(g) N_2(g) + O_2(g)

iv) PCl_5(g) PCl_3(g) + Cl_2(g)

(A) (i) and (iii) (B) (ii) and (iv)

Van’t Hoff equation giving the effect of temperature on chemical equilibrium is represented as

(A) (B)

Pure ammonia is placed in a vessel at temperature where its degree of dissociation (a) is appreciable at equilibrium.

(A) K_p does not change with pressure (B) a does not change with pressure

Steam starts reacting with iron at high temperature to give hydrogen gas and Fe₃O₄. The correct expression for equilibrium constant is

(A) (B)

X nY, X decomposes to give Y (in one litre vessel) if degree of dissociation is a then K_C and its unit.

(A) ,mol^n–1 lit^n–1(B) ,molⁿ litⁿ

Solubility of a solute in a solvent (say H₂O) is dependent on temperature as given by

S = Ae^–^DH/RT where DH is heat of reaction

Solute + H₂O Solution, DH = ±X

For a given solution variation of log S with temperature is shown graphically. Hence solute is

(A) CuSO₄.5H₂O (B) NaCl

For the reaction A_(g) B_(g)+ C_(g)

(A) K_p = a³P (B) K_p = a²(K_p + P + 1)

The values of K_C for the following reactions are given as below

A B, K_C =1, B C, K_C = 3 and C D, K_C = 5

Evaluate the value of K_C for A D

(A) 15 (B) 5

For the reaction (1) and (2)

A B + C

D 2E

Given: : : 9 : 1

If the degree of dissociation of A and D be same then the total pressure at equilibrium (1) and (2) are in the ratio.

(A) 3:1 (B) 36:1

Ionic Equilibrium

If the degree of ionization of water be 1.8 ´ 10^-9 at 298K. Its ionization
constant will be

(A) 1.8 ´ 10^-16 (B) 1 ´10^-14

When a solution of benzoic acid was titrated with NaOH the pH of the solution when half the acid neutralized was 4.2. Dissociation constant of the acid is

(A) 6.31 ´ 10^-5 (B) 3.2 ´ 10^–5

10^-2 mole of NaOH was added to 10 litre of water. The pH will change by

(A) 4 (B) 3

79. For an aqueous solution to be neutral it must have

(A) pH = 7 (B) [H⁺]=[OH^–]

If an aqueous solution at 25°C has twice as many OH^– as pure water its pOH will be

(A) 6.699 (B) 7.307

What would be the pH of an ammonia solution if that of an acetic acid solution of equal strength is 3.2? Assume dissociation constant for NH₃ & acetic acid are equal.

(A) 3.2 (B) 6.4

The pH of an aqueous solution of 0.1M solution of a weak monoprotic acid which is 1% ionised is

(A) 1 (B) 2

pH of a 10^–10 M NaOH is nearest to

(A) 10 (B) 7

The weight CH₃COONa needed to add in 1 litre CH₃COOH to obtain a buffer solution of pH value 4 is (Given K_a = 1.8 ´ 10^–5)

(A) 1.2 g (B) 1.47g

A dilute HCl solution saturated with H₂S has pH value 3 then the conc. of S^{– –} is (Given K₁ = 1 ´ 10^–7, K₂ = 1.3 ´ 10^–13)

(A) 2 ´ 10^–13 (B) 2.4 ´ 10^–13

Which is the strongest acid

(A) H₃AsO₄ (B) H₂AsO₄^–

In an aqueous solution of triprotic acid H₃A which is true

(A) [H⁺] = 3[A^3–] (B) [H⁺] > 3[A^3–]

If K₁ & K₂ be first and second ionisation constant of H₃PO₄and K₁>> K₂ which is incorrect.

(A) [H⁺] = [H₂ PO₄^–] (B) [H⁺] =

Which one of the following anion does not hydrolyse

(A) H^– (B) CN^–

If the ionic product of water varies with temperature as follows and the density of water be nearly constant for this range of temperature the process
H⁺ + OH^‑H₂O is

Temp. °C	0	10	25	40	50
K_w	0.114´10^–14	0.292´10^–14	1.008 ´10^–14	2.91´10^–14	5.474 ´10^–14

(A) Exothermic (B) Endothermic

Stoichiometry

C 2. B
C 4. A
C 6. A
C 8. D
A 10. A
C 12. D
A 14. A
D

Gaseous State

B 17. A
D 19. C
C 21. A
A 23. A
D 25. A
B 27. A
C 29. C
A

Atomic Structure

A 32. D
B 34. A
B 36. B
C 38. C
A 40. D
D 42. A
B 44. B
B

Chemical Kinetics

C 47. C
B 49. B
A 51. A
B 53. A
A 55. A
D 57. C
B 59. B
B

Chemical Equilibrium

6 D 62. B
B 64. B
A 66. D
A 68. C
A 70. C
D 72. D
C 74. A
B

Ionic Equilibrium

A 77. A
A 79. B
A 81. D
C 83. B
B 85. D
A 87. B
D 89. A
A

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79. For an aqueous solution to be neutral it must have

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