# Chemistry Test 2 with solutions

** **

** **

*Illustration 1: One mole of a mixture of CO and CO _{2} requires exactly 20 gms of NaOH to convert all the CO_{2} into Na_{2}CO_{3}. How many more gms of NaOH would it require for conversion into Na_{2}CO_{3} if the mixture (one mole) is completely oxidised*

to CO_{2}.

* (A) 60 gm (B) 80 gm*

* (C) 40 gm (D) 20 gm*

** Solution:** Moles of NaOH =

Moles of CO_{2} = [Q ‘n’ factor for CO_{2} = 2]

Moles of CO =

Moles of CO_{2} produced = from CO

Moles of NaOH extra =

Mass of NaOH extra = **60**

*Illustration **2: One litre of 0.1 M CuSO _{4} 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*

* FeS _{2} + MnO_{4} + H^{+} *

*¾*

*®*

*Fe*^{3+}+ SO_{2}+ Mn^{2+}+ H_{2}O* The equivalent mass of FeS _{2} would be equal to*

* (A) Molar mass (B) *

* (C) (D) *

** Solution:** Fe

^{2+}¾® Fe

^{3+}+ e

^{–}

S_{2}^{–2} ¾® 2S^{+4} + 10e^{–}

–––––––––––––––––

FeS_{2} ¾® 2S^{+4} + Fe^{3+} + 11e^{–}

\ Equivalent mass of FeS_{2} =

*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 + H_{2}O

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**®** CaHPO _{4} + 2H_{2}O*

* Which are true statements *

* (A) Equivalent weight of H _{3}PO_{4} is 49*

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

* (C) CaHPO _{4} is an acid salt*

* (D) 1 mol of H _{3}PO_{4} is completely neutralised by 1.5 mol of Ca(OH)_{2}.*

** Solution:** a) By given reaction

H_{3}PO_{4} º 2OH^{–}

\ Equivalent weight = = 49

** ****\ (A) is true**

- b) Only one acidic H in CaHPO
_{4}hence neutralised by 1 mol of KOH

** ****\ (B) is true**

- c) One acidic H in CaHPO
_{4}makes it acidic salt.

** ****\ (C) is true**

- d) H
_{3}PO_{4}º 3H^{+}º 3OH^{–}º5 Ca(OH)_{2}

** ****\ (D) is true**

*Illustration 7: The brown ring complex compound is formulated as [Fe(H _{2}O)_{5}NO]SO_{4}. The oxidation number of iron is*

* (A) 1 (B) 2*

* (C) 3 (D) 0*

** Solution:** [Fe(H

_{2}O)

_{5}NO]SO

_{4}[Fe(H

_{2}O)

_{5}NO]

^{2+}+ SO

_{4}

^{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 K _{3}[Fe(CN)_{6}] is reduced by some equivalents of N_{2}H_{4}, then calculate numbers of moles of N_{2}H_{4} required for the above reaction the answer is*

* (A) 10 ^{–5} (B) 10^{–3}*

* (C) 10 ^{–6} (D) 10^{–2}*

** Solution:** N

_{2}H

_{4}+ K

_{3}[Fe(CN)

_{6}+ KOH ¾® K

_{4}[Fe(CN)

_{6}] + N

_{2}+ H

_{2}O

\ Number of equivalents of

K_{3}[Fe(CN)_{6}] =

Number of moles of

N_{2}H_{4}= = 10^{–3}

** ****\ (B)**

*Illustration **10: 0.7gm of Na _{2}CO_{3}.XH_{2}O 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 Na

_{2}CO

_{3}.xH

_{2}O in 20 ml = 19.8 ´

\ Meq. Of Na_{2}CO_{3}.xH_{2}O 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} = pr^{3}

\ 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 H _{2}O_{(l)} at 760 mm and 373 K. Density of H_{2}O_{(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 H

_{2}O

_{(g)}at 373 K =

= = 30.62L

Volume of 1 mole of H_{2}O_{(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 H_{2}O = RH ´ Vapour pressure of H_{2}O

= = 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 (Cl _{2} 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

or M = 62.425

for Cl_{2} 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 H_{2} 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_{C}V_{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 ´ N_{0} ´ pr^{3}

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) P _{2}*

* (C) P _{4} (D) P_{8}*

*Solution:*

or M =

Put d = 2.64gm dm^{–3}

R= 0.0821 dm^{3} 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 = P_{4}

*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*^{–5}a* (C) 4 **´** 10 ^{–5}a^{2} (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 **´** 10 ^{18} (B) 12.01 *

*´*

*10*^{31}* (C) 1.31 **´** 10 ^{17} (D) None is correct*

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

\ E =

\ n = = 2.01 ´ 10^{18}

**\ (A)**

*Illustration 26: When a certain metal was irradiated with a light of frequency 3.2 **´** 10 ^{16} 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 *

*´*

*10*^{16}Hz. Hence threshold frequency is* (A) 1.6 **´** 10 ^{16} Hz (B) 0.8 *

*´*

*10*^{16}Hz* (C) 8 **´** 10 ^{15} Hz (D) 8 *

*´*

*10*^{16}Hz** Solution: ** Q hn = hn

_{0}+ KE

\ KE = h(n – n_{0})

Again 2(KE)_{1} = (KE)_{2}

2h(n_{1}^{ }– n_{0}) = (n_{2} – n_{0}) ´ h Þ 2(n – n_{0}) = (n_{2} – n_{0})

\ Þ 2n_{1} – n_{2}= n_{0}

Since n_{1} = 2 ´ 10^{16} Hz

n_{2} = 3.2 ´ 10^{16} Hz

\ n_{0} = 8 ´ 10^{15} 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) 2**p**x (B) 6**p**x*

* (C) 9x (D) *

** Solution:** Q mvr

_{3}=

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 + Br _{2} *

*¾*

*®*

*2NOBr, the following mechanism has been given** **fast** *

* NO + Br _{2} NOBr_{2}*

* NOBr _{2} + NO 2NOBr*

* Hence rate law is;*

* (A) k[NO] ^{2}[Br_{2}] (B) k[NO][Br_{2}]*

* (C) k[NOBr _{2}] [NO] (D) k[NO][Br_{2}]^{2}*

** Solution:** For the reaction

NOBr_{2} + NO 2NOBr

\ rate = k_{1}[NOBr_{2}] [NO] …(1)

But from the reaction

NO + Br_{2} NOBr_{2}

K_{equilibrium} = Þ [NOBr_{2}] = K_{equilibrium} [NO] [Br_{2}] …(2)

Putting equation (2) in equation (1)

rate = k_{1} k_{equilibrium} [NO]^{2}[Br_{2}]

= k[NO]^{2}[Br_{2}], where k = k_{1} 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*

* 2N _{2}O_{5} *

*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 N_{2}O_{5}in mol L^{–1}is;* (A) 1.4 (B) 1.2*

* (C) 0.04 (D) 0.8*

** Solution:** The decomposition of N

_{2}O

_{5}is a first order reaction

\ rate = k[N_{2}O_{5}]

2.4 ´ 10^{–5} = 3 ´ 10^{–5}[N_{2}O_{5}]

\ [N_{2}O_{5}] = = 0.8 mol L^{–1}

** ****\ (D)**

*Illustration 35: Following is the graph between logT _{50} 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:** T

_{50}µ a

^{(1–n)}where n = order of reaction

Þ T_{50} = k a^{(1 – n)}

^{ }or, logT_{50}= 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) k _{1}(a – y) – k_{2}(a – y) (B) k_{2}(a – y) – k_{1} (a– y)*

* (C) k _{1}(a – y) + k_{2}(a – y) (D) –k_{1}(a – y) – k_{2}(a – y)*

** Solution:** = k

_{1}(a – y) + k

_{2}(a – y)

\ = – k_{1}(a – y) – k_{2}(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_{A}N_{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 **´** 10 ^{9} : 1 respectively. If T_{50} of the element A is 2 *

*´*

*10*^{10}years, then T_{50}of the element B is* (A) 6.2 **´** 10 ^{9} years (B) 6.45 years*

* (C) 3 **´** 10 ^{8 } years (D) None*

** Solution:** At radioactive equilibrium

t_{50(B) } = = = years

** ****\ (B)**

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

* _{a}x^{C} *

*¾*

*®*

_{d}y^{b}is* (A) (B) + c*

* (C) – a (D) – c*

*Solution:*_{a}x^{c} ¾® _{d}y^{b} + m × _{2}He^{4} + n × _{–1}b^{0}

Q

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*

* _{z}A^{m} *

*¾*

*®*

_{z–4}B^{m – 8}+ 2

*×*

_{2}He^{4}* 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 = 2

^{n}where n = no. of half lives

= = 2^{2} 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. H _{2}SO_{4}?*

* (A) CH _{3} – C = C – CH_{3} (B) CH_{3} – CH_{2} – C *

*º*

*CH** (C) CH **º** CH (D) CH _{2} = CH_{2}*

** Solution:** If CH º CH were to dissolve in H

_{2}SO

_{4}a bisulphite salt of vinyl carbocation H

_{2}C = 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) (CH _{3})_{2}CH = CH_{2 }(B) CH_{3} – CH = CH_{2}*

* (C) CH _{3} – CH = CH – CH_{3} (D) CH_{3} – CH_{2} – CH = CH_{2}*

** Solution:** Addition of H

_{2}O 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:** CH

_{3}– CH

_{3}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. H _{2}SO_{4} gives preferred product.*

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

* (C) propene (D) ethane*

** Solution:** CH

_{3}– – – CH

_{3}CH

_{3}– CH = CH – CH

_{3}(80%)

+ CH_{3} – CH_{2} – CH = CH_{2} (20%)

This is in accordance with saytzeff rule.

**\ (B)**

*Illustration 4**7: CH _{3}*

*– C*

*º*

*C – CH*_{3}‘X’. What is X* (A) CH _{3}CH_{2}CH = CH_{2} (B) CH_{3}CH_{2}C *

*º*

*CH** (C) CH _{3} – CH = CH – CH_{3} (D) CH_{2} = C = CH – CH_{3}*

** Solution:** Isomerisation occurs, when 2-butyne is treated with NaNH

_{2}, 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) |
CH_{3}COOH |

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*

* CH _{2} = CH(CH_{2})COOH + HBr …………. is*

(A) |
CH_{3} –– (CH_{3})_{8}COOH |
(B) |
CH_{2} = CH(CH_{2})_{8}COBr |

(C) |
–CH_{2}(CH_{2})_{8}COOH |
(D) |
CH_{2} = CH(CH_{3})_{7} – – COOH |

** Solution:** Follows the peroxide effect

* *\ CH_{2} = CH(CH_{2})_{8}COOH – CH_{2}(CH_{2})_{8}COOH

** ****\ (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 AgNO_{3}is* (A) 7.5**´**10 ^{18} (B) 7.5*

*´*

*10*^{-18}* (C) 7.5**´**10 ^{19} (D) 7.5*

*´*

*10*^{-19}** Solution ** 2KCN + AgNO

_{3}¾¾® Ag(CN)

_{2}

^{–}+ KNO

_{3}+

_{ }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, **D**H= -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 H _{2}(g) at 2 atm pressure, when H_{2}S(g) at a pressure of 4 atm is introduced into the vessel. Where reaction *

* 8H _{2}S(g) 8H_{2}+S_{8}(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_{c}RT*

* (D) K _{c}=256*

** Solution: ** Under identical conditions of volume and temperature Pµn

so initially

8H_{2}S(g) 8H_{2}(g) + S_{8}(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) + NO _{2}(g) CO_{2}(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) + NO

_{2}(g) CO

_{2}(g) + NO(g)

t = 0 1 1 1 1

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

CO_{2}+Ba(OH)_{2} ¾® BaCO_{3}+H_{2}O

Moles of BaCO_{3} =

\ moles of CO_{2} 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 NH _{3} ionizes to a slight extent at -60*

*°*

*C. Its ionic product** *

* The number of NH _{2}^{–} ions present per ml of pure liquid NH_{3 }are *

* (A) 300 ions (B) 400 ions *

* (C) 600 ions (D) 500 ions *

** Solution: ** 2NH

_{3}NH

_{4}

^{+}+ NH

_{2}

^{–}

K = [NH_{4}^{+}] [NH_{2}^{–}] = x^{2} = 10^{-30}

\ x = 10—^{15—24591580810328902198alkjdfakslkdlsfjadsklfjadsklfjadsklfjasdlkfjdaslkfjas dkladsklfj }^{-15} M = [NH_{2}^{–}]

[NH_{2}^{–}] = 10^{-15}´ ´6.023´10^{23}

= 600 ions/ml

*Illustration 57: When sulphur in the from of S _{8} is heated at 900 K, the initial pressure of one atm falls by 29% at equilibrium. This is because of conversion of some S_{8} to S_{2}. Find the value of equilibrium constant for this reaction *

* (A) 1.16 atm ^{3} (B) 0.71 atm^{3}*

* (C) 2.55 atm ^{3} (D) 5.1 atm^{3}*

** Solution: ** S

_{8}(g) 4S

_{2}(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) + H _{2}(g) Co(s) + H_{2}O(g) K_{1} = 65*

* CoO(s) + CO(g) Co(s) + CO _{2}(g) K_{2}= 500*

* The equilibrium constant for the reaction at 1000**°** K*

* CO(g) + H _{2}O(g) CO_{2}(g) + H_{2}(g)*

* (A) 0.13 (B) 7.69*

* (C) 0.0179 (D) 69.3*

** Solution: ** K

_{1}= [H

_{2}O]/[H

_{2}] = 65

K_{2} = [CO_{2}] / [CO(g) ] = 500

Now, for the required reaction

To obtain this, divide K_{2} by K_{1}, 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:** C

_{2}H

_{5}OH + CH

_{3}COOH CH

_{3}COOC

_{2}H

_{5}+ H

_{2}O

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) + Cl _{2}(g) *

* the values of **D**H**°** and **D**S**°** 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^{–3}M

\ 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 CaSO_{4} is 2.4 *

*´*

*10*^{–5}at 25°C. In a solution containing Ca^{2+}ions the precipitation of CaSO_{4}begins to occur when SO_{4}^{2–}ion concentration in the solution is made just to exceed the value of 4.8

*´*

*10*^{–3}M. Hence concentration of Ca^{2+}ion in the solution is* (A) 200 ppm (B) 40 ppm*

* (C) 400 ppm (D) 100 ppm*

** Solution:** [Ca

^{2+}] [SO

_{4}

^{2–}] = 2.4 ´ 10

^{–5}

[Ca^{2+}] = = 5 ´ 10^{–3}M

1 L solution contains 5 ´ 10^{–3} mole of Ca^{2+} ions

1000 L solution will contain 5 mole i.e. 200 g Ca^{2+} ions

Taking density of aqueous solution to be unity

1000 kg i.e. 10^{6} g solution contains 200g Ca^{2+ }ions

Conc. of Ca^{2+} ion = 200 ppm

** ****\ (A)**

**Stoichiometry**

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

(Molecular weight of H_{3}PO_{4} = 98 and Mg(OH)_{2} = 58.3 gm)

(A) 66.7 (B) 252gm

(C) 112gm (D) 168 gm

- 10 L of hard water required 0.56 gm of lime (CaO) for removing hardness. Hence temporary hardness in ppm of CaCO
_{3}is

(A) 100 (B) 200

(C) 10 (D) 20

- 20 ml of xM HCl neutralises completely 10 ml of 0.1 M NaHCO
_{3}solution and further 5 ml of 0.2 M Na_{2}CO_{3}solution in the presence of methyl orange at end point, the value of x is

(A) 0.167M (B) 0.133M

(C) 0.15M (D) 0.2M

- 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%

(C) 1% (D) None of these

- 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 Na
_{2}S_{2}O_{3}solution. The weight percentage of copper in the salt is

(A) 25% (B) 40%

(C) 50% (D) 63.5%

- 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) MCl_{3} (B) MCl

(C) MCl_{2} (D) MCl_{4}

- 22.4 litres of H
_{2}S and 22.4 litre of SO_{2}both at STP are mixed together. The amount of sulphur precipitated as a result of chemical calculation is

(A) 16gm (B) 23gm

(C) 48gm (D) 96gm

- In the mixture of NaHCO
_{3}and Na_{2}CO_{3}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 NaHCO_{3}is

(A) 2x (B) y

(C) (D) (y – x)

- A sample of oleum is labelled 109%. The % of free SO
_{3}in the sample is

(A) 40% (B) 80%

(C) 60% (D) 9%

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

(A) 2 (B) 3

(C) 4 (D) 5

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

(A) 8.4 gm (B) 1.5gm

(C) 0.758 gm (D) 1.07gm

- 1 gm mole of oxalic acid is treated with conc. H
_{2}SO_{4}. The resultant gaseous mixture is passed through a solution of KOH. The mass of KOH consumed will be

(A) 28gm (B) 56gm

(C) 84gm (D) 112 gm

- NH
_{3}+ OCl^{–}¾® N_{2}H_{4}+ Cl^{–}

On balancing the above equation in basic solution using integral coefficients, which of the following whole number will be the coefficient of N_{2}H_{4}?

(A) 1 (B) 2

(C) 3 (D) 4

- A certain compound has the molecular formula X
_{4}O_{6}. If 10 gm of X_{4}O_{6}has 5.72g X, atomic mass of X is

(A) 32 amu (B) 37 amu

(C) 42 amu (D) 98 amu

- In the following reaction

28NO_{3}^{–} + 3As_{2}S_{3} + 4H_{2}O ¾® 6AsO_{4}^{3–} + 28NO + 9SO_{4}^{2–} + 8H^{+}

equivalent weight of As_{2}S_{3} (with molecular weight M) is

(A) (B)

(C) (D)

**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%

(C) 11% (D) 12.09%

- A small hole is pricked in a container containing a mixture of CH
_{4}and O_{2}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)

(C) 2:1 (D) 1:2

- 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

(C) ‘a’ varies, ‘b’ remains same (D) both 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

(C) Remain same (D) Start to vibrate

- 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

(C) 1:2 (D) 3:1

- Rate of diffusion of two gases are r
_{A}and r_{B}, molecular weights are M_{A}and M_{B}then partial pressure P_{A}is (if number of moles are n_{A}and n_{B}).

(A) (B)

(C) (D) None of these

- 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})

(C) 200 (f_{A} + f_{B}) (D) 50

- 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

(C) Colide without loss of energy (D) Are about the inversion temperature

- If two gases of molecular weights M
_{A}and M_{B}at temperatures T_{A}, T_{B}and T_{A}M_{B}= T_{B}M_{A}, then which property has the same magnitude for both the gases.

(A) Density (B) Pressure

(C) KE per mole (D) V_{rms}

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

(A) (B)

(C) (D)

- 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 T_{2} = T

(C) Both A and B are raised to a higher temperature

(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

(C) Number of moles of the gas (D) Translational energy of the gas

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

(A) T(H_{2}) = T(N_{2}) (B) T(H_{2}) > T(N_{2})

(C) T(H_{2}) < T(N_{2}) (D) T(H_{2}) = T(N_{2})

- 1 lt of N
_{2}and l of O_{2}at the same temperature and pressure were mixed together what is the relation between the masses of two gases in the mixture.

(A) (B)

(C) (D)

- Consider a mixture of SO
_{2}and O_{2}kept at room temperature compared to the oxygen molecule, the SO_{2}molecule will hit the wall with

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

(C) Greater kinetic energy (D) Greater mass

**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

(C) 3 (D) 4

- For an electron in a hydrogen atom the wave function, y is proportional to , where a
_{0}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 a_{0}.

(A) e (B) e^{2}

(C) (D) Zero

- 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

(C) 5 ® 2 (D) 7 ® 5

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

(A) – mv^{2} (B)

(C) (D)

- 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^{–7}m

(C) 2 ´ 10^{–10}m (D) 3 ´ 10^{–5} m

- 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

(C) 27 (D) 36

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

(A) n^{2} (B) 2l

(C) 2l +1 (D) m

- 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

(C) c (D) d

- 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
^{+3}is

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

(C) 1 : 16 (D) 16 : 1

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

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

(C) He^{+} ion (D) Be^{+3} ion

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

(A) 15,200 cm^{–1} (B) 60,800 cm^{–1}

(C) 76,000 cm^{–1} (D) 1,36,8000 cm^{–1}

- The dissociation energy of H
_{2}is 430.53 kJ mol^{–1}. If H_{2}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%

(C) 1.3% (D) 100%

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

(A) (B) zero

(C) (D)

- Photoelectric emission is observed from a surface for frequencies v, and v
_{2}of the incident radiation (where v_{1}> v_{2}). If the maximum kinetic energy of the photoelectrons in two cases are in the ratio of 1:k, then the threshold frequency v_{0}is given by

(A) (B)

(C) (D)

- 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

(C) 3 (D) 4

**Chemical Kinetics**

- A radioactive isotopes x with half-life of 1.37 ´ 10
^{9}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 ´ 10^{8} years (B) 3.85 ´ 10^{9} years

(C) 4.11 ´ 10^{9} years (D) 3.06 ´ 10^{9} 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

(C) 6:4 (D) None

_{92}U^{238}(IIIB) undergoes following emissions:

_{92}U^{238}

Which is/are correct statements

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

(C) B will be of IA group (D) C will be of IIIA 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

(C) 4:9 (D) 16:81

- 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 ´ 10
^{6}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 ´ 10^{4}cps. Calculate the volume of the animal blood.

(A) 300ml (B) 200ml

(C) 250 ml (D) 400 ml

- 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

(C) (ln2)xy (D)

- 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

(C) 790 (D) 700

- 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

(C) 4 (D) None

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

(A) (B) N_{0} +

(C) N_{0} (D) + N_{0}

- 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_{N}2 mechanism when [OH^{–}] = 0.001 M

(A) 1.9 (B) 4.7

(C) 2.8 (D) 4.9

- The mechanism of the reaction

2NO + O_{2} ¾® 2NO_{2} is

k_{1}

NO + NO N_{2}O_{2} (fast)

k_{–1}

N_{2}O_{2} + O_{2} 2NO_{2} (slow)

the rate constant for the reaction is

(A) k_{2} (B) k_{2}×k_{1}×k_{–1}

(C) k_{2}×k_{1} (D)

- 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}

(C) r = k× [sugar]^{1} [H^{+}]^{1} (D) r = k× [sugar]^{0} [H^{+}]^{1}

- 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

(C) 48 minutes (D) 49.4 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

(C) Activation energy

(D) Average kinetic energy of the molecule

- Rate constant, k = 1.8 ´ 10
^{4}mol^{–1}Ls^{–1}and E_{a}= 2 ´ 10^{2}kJ mol^{–1}, when T ® ¥, then the value of A is

(A) 1.8 ´ 10^{4} kJ mol^{–1} (B) 1.8 ´ 10^{4} mol^{–1}L sec^{–1}

(C) 1.8 ´ 10^{4}mol L^{–1} sec^{–1} (D) 2.4 ´ 10^{3} kJ mol^{–1} 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%

(C) 66.66% (D) 33.33%

- CH
_{3}– CO – CH_{3(g)}CH_{3}– CH_{3(g)}+ CO_{(g)}

Initial pressure of CH_{3}COCH_{3} is 100mm when equilibrium is set up, mole fraction of CO_{(g)} is hence K_{p} is

(A) 100mm (B) 50mm

(C) 25mm (D) 150mm

- 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 CaCO_{3} was put into a 10 L flask, percentage of calcium carbonate would remain unreacted at the equilibrium.

(A) 66% (B) 34%

(C) 50% (D) 25%

- If equilibrium constant of

CH_{3}COOH + H_{2}O CH_{3}COO^{–} + H_{3}O^{+}

Is 1.8 ´ 10^{–5}, equilibrium constant for

CH_{3}COOH + OH^{–} ¾® CH_{3}COO^{–} + H_{2}O is

(A) 1.8 ´ 10^{–9} (B) 1.8 ´ 10^{9}

(C) 5.55 ´ 10^{–10} (D) 5.55 ´ 10^{10}

- PCl
_{5}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

(C) 0.3 atm (D) 0.6 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)

(C) (D)

- 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)}_{3(g)}+ Cl_{2(g)}

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

(C) (i) and (ii) (D) (iii) and (iv)

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

(A) (B)

(C) (D)

- 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

(C) (NH_{3}) does not change with pressure (D) [H_{2}] < [N_{2}]

- Steam starts reacting with iron at high temperature to give hydrogen gas and Fe
_{3}O_{4}. The correct expression for equilibrium constant is

(A) (B)

(C) (D)

- 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) ,mol^{n} lit^{n}

(C) ,K_{C} is unit less (D) ,K_{C} is unit less

- Solubility of a solute in a solvent (say H
_{2}O) is dependent on temperature as given by

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

Solute + H_{2}O Solution, DH = ±X

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

(A) CuSO_{4}.5H_{2}O (B) NaCl

(C) Sucrose (D) CaO

- For the reaction A
_{(g)}B_{(g) }+ C_{(g)}

(A) K_{p} = a^{3}P (B) K_{p} = a^{2}(K_{p} + P + 1)

(C) K_{p} = a^{2} (K_{P} + P) (D) K_{p} = a^{2}

- 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

(C) 3 (D) 1

- 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

(C) 1:1 (D) 0.5:1

**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}

(C) 1 ´ 10^{-16} (D) 1.67 ´ 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}

(C) 8.7 ´ 10^{–8} (D) 6.42 ´ 10^{–4}

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

(A) 4 (B) 3

(C) 11 (D) 7

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

(A) pH = 7 (B) [H^{+}]=[OH^{–}]

(C) [H^{+}] = (D) [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

(C) 7 (D) 6.98

- 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 NH
_{3}& acetic acid are equal.

(A) 3.2 (B) 6.4

(C) 9.6 (D) 10.8

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

(A) 1 (B) 2

(C) 3 (D) 11

- pH of a 10
^{–10}M NaOH is nearest to

(A) 10 (B) 7

(C) 4 D) 10

- The weight CH
_{3}COONa needed to add in 1 litre CH_{3}COOH to obtain a buffer solution of pH value 4 is (Given K_{a}= 1.8 ´ 10^{–5})

(A) 1.2 g (B) 1.47g

(C) 1.8 g (D) 4 g

- A dilute HCl solution saturated with H
_{2}S has pH value 3 then the conc. of S^{– –}is (Given K_{1}= 1 ´ 10^{–7}, K_{2}= 1.3 ´ 10^{–13})

(A) 2 ´ 10^{–13} (B) 2.4 ´ 10^{–13}

(C) 3 ´ 10^{–15} (D) 1.3 ´ 10^{–15} M

- Which is the strongest acid

(A) H_{3}AsO_{4} (B) H_{2}AsO_{4}^{–}

(C) HAsO_{4}^{—} (D) AsO_{4}^{—}

- In an aqueous solution of triprotic acid H
_{3}A which is true

(A) [H^{+}] = 3[A^{3–}] (B) [H^{+}] > 3[A^{3–}]

(C) [H^{+}] < 3[A^{3–}] (D) [H^{+}] = [OH^{–}]

- If K
_{1}& K_{2}be first and second ionisation constant of H_{3}PO_{4 }and K_{1}>> K_{2}which is incorrect.

(A) [H^{+}] = [H_{2} PO_{4}^{–}] (B) [H^{+}] =

(C) K_{2} = [HPO_{4}^{– –}] (D) [H^{+}]= 3[A_{3}^{–}]

- Which one of the following anion does not hydrolyse

(A) H^{–} (B) CN^{–}

(C) NO_{2}^{– –} (D) S^{– –}

- 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^{‑ }H_{2}O 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

(C) Can’t say (D) Ionization

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**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**