Question:1 At 300K and 1 atm 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O_2 by volume for complete combustion. After combustion the gases ocuupy 330 mL. Assuming thta the water formed is in liquid form and the volumes were measured at the same temperature and presssure, the formula of the hydrocarbon is:

(1) C_3H_6

(2) C_3H_8

(3) C_4H_8

(4) C_4H_10

Solution:

Volume of N_2 in air = 375x0.8 = 300 ml

Volume of N_2 in air = 375x0.2 = 75 ml

After combustion total volume

330 = V_(N_2)+ V_(CO_2)

330 =300 +15x => x =2

Volume of O_2 used:

15(x+y/4)= 75

=> x+y/4 = 5

=> y = 12

So hydrocarbon is = C_2H_12

BONUS

Question:2 Two closed bulbs of equal volume (V) containing an ideal gas initially at pressure p_i and temperature T_1 are connected through a narrow tube of negligible volume as shown in the figure below. The temperature of one of the bulbs is then raised to T_2. The final pressure p_f is:

(1) p_i((T_1T_2)/(T_1+T_2))

(2) 2p_i(T_1/(T_1+T_2))

(3) 2p_i(T_2/(T_1+T_2))

(4) 2p_i((T_1T_2)/(T_1+T_2))

Solution:

Initial moles and final moles are equal:

(n_T)_i = (n_T)_f

(P_iV)/(RT_1)+ (P_iV)/(RT_1)=(P_fV)/(RT_1)+ (P_fV)/(RT_2)

=> 2 P_i/T_1 = P_f/T_1+ P_f/T_2

=> P_f = (2P_iT_2)/(T_1 + T_2)

Question:3 A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference V esu. If e and m are charge and mass of an electron, respectively, then the value of h/lambda (where lambda is wavelength associated with electron wave)is given by:

(1) meV

(2)2meV

(3) sqrt(meV)

(4) sqrt(2meV)

Solution: As electron of charge 'e' is passed through 'V' volt,kinetic energy of electron becomes =eV

As wavelength of e^- wave (lambda) = h/sqrt(2mK.E)

lambda = h/sqrt(2m eV)

h/lambda = sqrt(2m eV)

Question:4 The species in which the N atom is in a state of sp hybridization is:

(1) NO_2^+

(2) NO_2^-

(3) NO_3^-

(4) NO_2

Solution:

NO_2-> sp^2 Hybridization

NO_2^+ -> sp Hybridization

NO_2^(-) -> sp^2 Hybridization

NO_3^(-) -> sp^2 Hybridization

Question:5 The heats of combustion of carbons and carbon monooxide are -393.5 and -283.5 kJmol^-1, respectively. The heat of formation (in kJ) of carbon monoxide per mole is:

(1) 110.5

(2) 676.5

(3) -676.5

(4) -110.5

Solution:

C_(s) + 1/2 O_2 (g)-> CO_(g); DeltaH_r = DeltaH_f (CO)

=> DeltaH_f =DeltaH_C (C)- DeltaH_C(CO)

=> -393.5+283.5

=> -110 kJ

Question:6 18 g glucose (C_6H_12O_6) is added to 178.2g water. The vapor pressure of water (in torr) for this aqueous solution is:

(1) 7.6

(2) 76.0

(3) 752.4

(4) 759.0

Solution:

Assuming temperature to be 100^0C

Relative lowering of vapour pressure

Equation  (P^0 - P_s)/P^0 = X_"solute" = n/(n+N)

Modified forms of equation is  (P^0 - P_s)/P_s = n/N 

P^0 = 760 torr

P_s = ?

(760-P_s)/P_s = (18//180)/(178.2//18)

=> P_s = 752.4 torr

Question:7 The equilibrium constant at 298 K for a reaction A+B<=> C+D is 100. If the initial concentration of all the four species were 1M each, then equillibrium concentration in D(in "mol" L^-1) will be:

(1) 0.182

(2) 0.818

(3) 1.818

(4) 1.182

Solution:

A+B <=> C+D K=100

Q = (1xx1)/(1xx1)= 1

Q < K so reaction moves in forward reaction

(1+x)^1/(1-x)^2= 100 => (1+x)/(1-x)= 10

=> 1+x = 10-10x => x = 9/11

[D] = 1+x = 1+9/11 = 1.818M

Question:8 Galvanization is applying a coating of :

(1) Pb

(2) Cr

(3) Cu

(4) Zn

Solution: Galvanization is the process of applying a protective zinc coating of steel or iron, to prevent rusting.

Question:9 Decompostion of H_2O_2 follows a first order reaction. In fifty minutes the concentration of H_2O_2 decreases from 0.5 to .125 M in one such decomposition. When the concentration of H_2O_2 reaches 0.05 M, the rate of formation of O_2 will be

(1) 6.93xx10^-2 "mol" min^(-1)

(2) 6.93xx10^-4 "mol" min^(-1)

(3) 2.66 L min^-1 at STP

(4) 1.34xx10^-2 "mol" min^(-1)

Solution:

H_2O_2 (aq) -> H_2O (aq) + 1/2 O_2 (g)

k = 1/t ln (a_0/a_t)

k= 1/50 ln (0.5/0.125)

k= 1/50 ln 4 1/min

("Rate of disappearance of" H_2O_2)/1 = ("Rate ofappearance of" O_2)/(1/2)

("Rate")_(O_2)= 1/2 ("Rate")_(H_2O_2)

=> 1/2 k [H_2O_2]

=> 1/2 xx 1/50 xx ln4 xx 0.05

=> 6.93 xx10^-4 M/min

Question:10 For a linear plot of log(x/m) versus logp in a Freundlich adsorption isotherm, which of the following statements is correct? (k and n are constants)

(1) Both k and 1/n appear in the slope term

(2) 1/n appears as the intercept

(3) Only 1/n appears as the slope

(4) log(1/n) appears as the intercept.

Solution:

According to Freundlich isotherm

x/m = k.p^(1/n)

=> log (x/m) = logk + 1/n log p

So intercept is logk and slope is 1/n

Question:11 Which of the following atoms ahs the highest first ionization energy?

(1) Rb

(2) Na

(3) K

(4) Sc

Solution:

Due to poor shielding of d-electron in Sc, Z_"eff"of Sc becomes more so that so that ionization energy of Sc is more then Na, K and Rb.

Question:12 Which one of the following ores is best concentrated by froth floatation method?

(1) Magnetite

(2) Siderite

(3) Galena

(4) Malachite

Solution:

Froth floatation method is mainly applicable for sulphide ores.

(1) Magnetite ore: Fe_3O_4

(2) Siderite Ore: FeCO_3

(3) Galena ore : PbS (Sulphide Ore)

(4) Malachite ore: Cu(OH)_2, CuCO_3

Question:13 Which of the following statments about water is FALSE?

(1) Water is oxidized to oxygen during photosynthesis.

(2) Water can act both as an acid and as a base

(3) There is extensive intramolecular hydrogen bonding in the condensed phase

(4) Ice formed by heavy water sinks in normal water

Solution: (1) 6CO_2 + 6H_2O -> C_6H_12O_6 + 6O_2 in presence hv and chlorophyll

(2) Water can show amphiprotic nature and hence water can act both as an acid and as a base

(3) There is extensive intermolecular hydrogen bonding in the condensed phase instead of intramolecular H- bonding

(4) Ice formed by heavy watersinks in normal water due to higher density of D_2O then normal water.

Question:14 The main oxides formed on combustion of Li, Na and K in excess of air, respectively:

(1) Li_2O, Na_2O and KO_2

(2) LiO_2, Na_2O_2 and K_2O

(3) Li_2O_2, Na_2O_2 and KO_2

(4) Li_2O, Na_2O_2 and KO_2

Solution:

The stability of the oxide of alkali metals depends upon the comprability of size of cation and anion.

Therefore the main oxide of alkali metals formed on excess of air are as follows:

(a) Li: Li_2O

(b) Na: Na_2O_2

(c) K: KO_2

(d) Rb: RbO_2

(e) Cs: CsO_2

Question:15 The reaction of zinc wiht dilute and concentrated nitric acid, respectively produces:

(1) N_2O and NO_2

(2) NO_2 and NO

(3) NO and N_2O

(4) NO_2 and N_2O

Solution:

Zn + 10HNO_3("dil") -> 4 Zn(NO_3)_2 + N_2O + 5H_2O

Zn + 4HNO_3("conc") -> Zn(NO_3)_2 + 2NO_2 + 2H_2O

Question:16 The pair in which phosphorus atoms have a formal oxidation state od +3 is:

(1) Orthophosphorous and pyrophosphorous acids

(2) Pyrophosphorous and hypophosphoric acids

(3) Orthophosphorous and hypophosphoric acids

(4) Pyrophosphorous and pyrophosphoric acids

Soltuion:

(a) Pyrophosphorous acid: formula H_4P_2O_5; formal oxidation state of phosphorous: +3

(b) Pyrophosphoric acid: formula H_4P_2O_7; formal oxidation state of phosphorous: +5

(c) Orthophosphorous acid : formula H_3PO_3; formal oxidation state of phosphorous: +3

(d) Hypophosphoric acid: formula H_4P_2O_6; formal oxidation state of phosphorous: +4

Both pyrophosphorous and orthophophorous acids have +3 formal oxidation state.

Question:17 Which of the following compounds is metallic and ferromagnetic?

(a) TiO_2

(b) CrO_2

(c) VO_2

(d) MnO_2

Solution:

CrO_2 is metallic as well as ferromagnetic

Question:18 The pair having the same magnetic moment is: [Atomic No: Cr = 24, Mn = 25, Fe= 26, Co =27]

(1) [Cr(H_2O)_6]^(2+) and [CoCl_4]^(2-)

(2) [Cr(H_2O)_6]^(2+) and [Fe(H_2O)_6]^(2+)

(3) [Mn(H_2O)_6]^(2+) and [Cr(H_2O)_6]^(2+)

(4)  [CoCl_4]^(2-) and [Fe(H_2O)_6]^(2+)

Solution:

(a)[Cr(H_2O)_6]^(2+); Cation: Cr^(2+)

Electro configuration:[As]4s^03d^4

No of unpaired electrons: 4

Magnetic moment : sqrt24

(b)[Fe(H_2O)_6]^(2+); Cation: Fe^(2+)

Electro configuration:[As]4s^03d^6

No of unpaired electrons: 4

Magnetic moment : sqrt24

(c)[Mn(H_2O)_6]^(2+); Cation: Mn^(2+)

Electro configuration:[As]4s^03d^5

No of unpaired electrons: 5

Magnetic moment : sqrt35

(a)[CoCl_4]^(2+); Cation: Co^(2+)

Electro configuration:[As]4s^03d^7

No of unpaired electrons: 3

Magnetic moment : sqrt15

Question:19 Which one of the following complexes shows optical isomerism?

(1) [Co(NH_3)_3 Cl_3]

(2) cis[Co(en)_2 Cl_2]Cl

(3) trans[Co(en)_2 Cl_2]Cl

(4) [Co(NH_3)_3 Cl_2]Cl

(en = ethylenediamine)

Solution:

(a) [Co(NH_3)_3 Cl_2]Cl have two geometrical isomers which are opticallyinactive due to presence of plane of symmetry

(b) [Co(NH_3)_3 Cl_3] also have two optically inactive geometrical isomers due to presence of plane of symmetry

(c) Complex cis [Co(en)_2 Cl_2]Cl is optically active due to formation of non-superimposable mirror image.

(d) Complex trans[Co(en)_2 Cl_2]Cl is optically inactive.

Question:20 The concentration of fluoride,lead, nitrate adn iron in a water sample from an underground lake was found to be 1000 ppb, 40 ppb , 100ppm and 0.2 ppm, respectively. This water is unsuitable for drinking due to high concentration of:

(1) Fluoride

(3) Nitrate

(4) Iron

Solution:

Iron: maximum prescribed conc. in drinking water is 0.2 ppm

Fluoride : maximum prescribed conc. in drinking water is 1.5 ppm

Lead: maximum prescribed conc. in drinking water is 50 ppb

Nitrate : maximum prescribed conc. in drinking water is 50 ppm

Hence the concentration of nitrate in a given water exceeds from the upper limit as given above

Question:21 The distillation technique most suited for separating glycerol from spent- lye in the soap industry is:

(1) Simple distillation

(2) Fractional distillation

(3) Steam distillation

(4) Distillation under reduced pressure

Solution:

Distillation under reduced pressure. Glycerol is separated from spent lye in the soap industry by distillation under reduced pressure, as for simple distillation very high temperature is required whihc might decompose the component.

Question:22 The product of the reaction given below is:

Solution:

Question:23 The absolute configuration of

is:

(1) (2R,3S)

(2) (2S,3R)

(3) (2S,3S)

(4) (2R,3R)

Solution:

Question:24 2- chloro-2-methylpentane on reaction with sodium methoxide in methanol yields:

(1) All of these

(2) (a) and (c)

(3) (c) only

(4) (a) and (b)

Solution:

Possible mechanism which takes place is E^2 and SN^1 mechanism. hence possible products are

Question:25 The reaction of propene with HOCl(Cl_2+ H_2O) proceeds through the intermediate:

(1) CH_3-CH^+ - CH_2-OH

(2) CH_3- CH^+ - CH_2-Cl

(3) CH_3- CH(OH)-CH_2^+

(4) CH_3-CHCl-CH_2^+

Solution:

Question:26 In the Hoftmann bromamide degradation reaction, the number of moles of NaOH and Br_2 used per mole of amine produced are:

(1) One mole of NaOH and one mole of Br_2.

(2) Four moles of NaOH and two moles of Br_2.

(3) Two moles of NaOH and two moles of Br_2.

(4) Four moles of NaOH and one mole of Br_2.

Solution: 4 moles of NaOh and one mole of Br_2 is required during production of on mole of one mole during Hoffmann's bromanide degradation reaction

R-C(O)-NH_2 + Br_2+4NaOH -> R-NH_2+ K_2CO_3+ 2NaBr+2H_2O

Question:27 Which of the following statements about low density polythene is FALSE?

(1) Its synthesis high pressure

(2) It is a poor conductor of electricity

(3) Its synthesis requires dioxygen or a peroxide initiator as a catalyst

(4) It is used in the manufacture of buckets, dust-bins etc.

Solution: Low density polythene: it is obtained by the polymerisation of ethene under hight pressure of 1000-200 atm at a tempearture of 350K to 570K in th epressure of traces of dioxygen or a peroxide initiator. Low density polythene is chemically inert and poor conductor of electricity. It is used for manufacture squeeze bottels, toys and flexible pipes

Question:28 Thiol group is present in:

(1) Cytosine

(2) Cystine

(3) Cysteine

(4) Methionine

Solution: Among 20 naturally occuring amino acids "Cysteine" has '-SH' or thiol functional group

General formula of amino acid : R-CH(NH_2)-COOH

Value of R -CH_2-SH  in cysteine.

Question:29 Which of the following is an anionic detergent?

(1) Sodium stearate

(2) Sodium lauryl sulphate

(3) Cetyltrimethyl ammonium bromide

(4) Glyceryl oleate

Solution:

Sodium lauryl sulphate is example of anionic detergent. These are sodium salts of sulphonated long chain alcohols or hydrocarbons.

Cetyltrimethyl ammonium bromide is an example of cationic detergent.

Question:30 The hottest region of Bunsen flame shown in the figure below is:

(1) region 1

(2) region 2

(3) region 3

(4) region 4

Solution: Region 2 is the hottest region of the Bunsen Burner Flame