Solution of Chemistry paper (JEE 2016: Mains)

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 = 1.818!!M

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

(2) Lead

(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