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4151.

Using Gibb's free energy change Delta G^@=57.34kJ mol^-1 for the reaction X_2Y_(s)=2X^(+)+Y^(2-) (aq) Calculate the solubility product of X_2Y in water at 300 K

Answer»

`10^-10`
`10^-12`
`10^-14`
can not be CALCULATED from the given data

Solution :`57.34kJ mol^-1=-2.303 TIMES 8.3 JK^-1mol^-1 times 300Klog K_(sp)`
`logK_(sp)=(-57.37 times 10^3 J mol^-1)/(2.303 times 8.3 JK^-1 mol^-1 times 300K)`
`log_10K_(sp)=-10`
`THEREFORE K_(sp)=10^-10`
`DeltaG^@=-2.303 RT LOG K_(eq)`
`X_2Y(s) leftrightarrow 2X^(+) (aq)+Y^(2-) (aq)`
`K_(eq)=([X^+]^2[Y^(2-)])/([X_2Y])`
`K_(eq)=[X^+]^2[Y^(2-)]( thereforeX_2Y(s)=1)`
`K_(eq)=K_(sp)`
Discussion
4152.

Which gas can be most readily liquefied :

Answer»

`NH_3`
`Cl_2`
`SO_2`
`CO_2`

ANSWER :C
Discussion
4153.

What are silicates and mention the types of silicates?

Answer»

Solution :Silicates: The MINERAL which contains silicon and oxygen in tetrahedral `[SiO_(4)]^(4-)` UNITS linked together in different PATTERNS are called silicates.
Types of silicates:
(i) Ortho silicates (ii) PYRO silicates
(iii) CYCLIC silicates (iv) Ino silicates
(v) Phyllo silicates (vi) Tecto silicates
Discussion
4154.

When formic acid reacts with PCl_3is forms :

Answer»

FORMYL CHLORIDE
ACETYL chloride
Methyl chloride
Propionyl chloride

Answer :A
Discussion
4155.

Which is the most stable carbocation

Answer»

Iso-propyl
Triphenylmethyl cation
Ethyl cation
`pi`- propyl cation

Solution :Stability of carbonium ions. `Ph_(3)overset(+)(C) gt Ph_(2) overset(+)(C)H gt Phoverset(+)(C)H_(2)` tertiary alkyl `gt` secondary alkyl `gt` primary alkyl `gt` METHYL. More the number of phenyl groups, the greater the DISPERSAL of positive charge and therefore, more the stability of carbonium ion is observed.

Dispersal of positive charge increases with the INCREASE in the number of benzene ring.
Discussion
4156.

Whatweight of CO is required to form Re_(2) (CO)_(10) from 2 . 50 g of Re_(2) O_(7) accordingto theunbalanced reaction : Re_(2) O_(7) + CO to Re_(2) (CO)_(10) + (CO_(2)(Re = 186 . 2 , C = 1 and O = 16)

Answer»

Solution :Suppose the RELATIVE moles of each reactantand product are as follows (just for convenience)
`{:(Re_(2)O_(7)"" + CO "" to ""Re_(2) (CO)_(10)"+ "CO_(2)),(" a molesb molesc molesd moles "):}`
Applying POAC for Re atoms,
`{:(" moles of Re in " Re_(2) O_(7) =" moles of Re in " RE_(2) (CO)_(10)),(" "2 xx "moles of " Re_(2) O_(7) = 2 xx "moles of " Re_(2) (CO)_(10)),("Ra = 2c"),("ora = c" ):}`. . . (i)
ApplyingPOAC for C atoms,
moles of C atoms in CO = molesof C in `Re_(2) (CO)_(10) `
+ moles of C in ` CO_(2)`
`xx` molesof CO ` = 10 xx ` moles of `Re_(2) (CO)_(10) + 1 xx` moles of `CO_(2)`
or b = 10 c + d. . . (II)
moles of O in `Re_(2) O_(7) ` + molesof O in CO
= molesof O in `Re_(2) (CO)_(10) + ` moles of O in `CO_(2)`
`7 xx ` moles of `Re_(2)O_(7) + 1 xx `molesof CO
`= 10 xx` molesof `Re_(2) (CO)_(10) + 2 xx ` moles of `CO_(2)`
or `7a + b = 10 c + 2 d""` . . . (iii)
Fromthe eqns . (i),(ii)and (iii) , we get17 a = b
i.e., `17 xx ` moles of `Re_(2) O_(7)`= moles of CO
`17 xx (2 . 50)/( 484,.4) = ("wt . of CO in G")/(28) [{:(mol. wt. of Re_(2) O_(7) = 484.4),("mol. wt. of CO = 28"):}] `
Wt . of CO = 2 . 46g
Discussion
4157.

Which of the following is used as an antioxidant in food

Answer»

BTX
BHT
BHA
All of these

SOLUTION :BTX is USED in PETROL
Discussion
4158.

Which of the following does not give brick red ppt. with fehling solution ?

Answer»

Formalin
Aceteldehyde
D-Flucose
Acetone

Answer :D
Discussion
4159.

Whichof the following compound gives a ketone with Grignard reagent ?

Answer»

Formaldehyde
ETHYL ALCOHOL
Methylcyanide
Methyl IODIDE

ANSWER :C
Discussion
4160.

What is meant by catalyst poison?

Answer»

Solution :(i) The substances when ADDED to a catalysed REACTION decreases or completely destroys the activity of a catalyst are often KNOWN as CATALYTIC POISONS.
(ii) In the reaction `2SO_2 + O_2 to 2SO_3` with Pt catalyst, the catalyst poison is `AS_2O_3` .
Discussion
4161.

What is diazonium salts?

Answer»

SOLUTION :Ar `N_(2)^(+)X^(-)` is CALLED DIAZONIUM salt, where Ar is an aromatic ring and X may be an ion like `Cl^(-), Br^(-) HSO,_(4)^(-)`. As compound contains two nitrogen atoms `(overset(+)N-=N)` hence the name diazonium salts (di-two, azo nitrogen and onium from ammonium).
Discussion
4162.

Which of the following complex is not hydrate isomer of [Cr(H_(2)O)_(6)]Cl_(3) ?

Answer»

`[Cr(H_(2)O)_(3)Cl_(3)].3H_(2)O`
`[Cr(H_(2)O)_(5)CL]Cl_(2).3H_(2)O`
`[Cr(H_(2)O)_(4)Cl_(2)]Cl.2H_(2)O`
None of these

Answer :A
Discussion
4163.

Which one of the following is the lightest

Answer»

0.2 mole of hydrogen gas
`6.023xx10^(22)` molecules of nitrogen
0.1 g of silver
0.1 mole of oxygen gas

Solution :(a) Moles = given wt./MOLECULAR wt.
Weight of `H_(2)=` moles `xx` molecular wt.
`=0.2xx2=0.4 g`
(b) `6.023xx10^(23)` REPRESENTS 1 mole
Thus `6.023xx10^(22)` will REPRESENT 0.1 mole
Wight of `N_(2)=0.1xx28=2.8=2.8 g`
(C) Weight of silver `=0.1 g`
(d) Weight of oxygen `=32xx0.1=3.2 g`
(e) Weight of water `=1 g`
Thus, 0.1 g silver is the lightest.
Discussion
4164.

Which of the following thermodynamic varibles is extensive ?

Answer»

PRESSURE
Density
Temperature
Mass

Answer :D
Discussion
4165.

Which of the following has the maximum number of unpaired d electrons ?

Answer»

`Ni^(3+)`
`Cu^(+)`
`ZN^(2+)`
`Fe^(2+)`

Solution : The ELECTRONIC CONFIGURATION of the IONS are given below
`Ni^(3+)to1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)4s^(0)3d^(7)`
Discussion
4166.

Valence bond theory for bonding in transition metal complexes was developed by Pauling. From the valence bond point of view, formation of a complex involves reaction between Lewis bases (ligands) and a Lewis acid (metal atom or metal ion) with the formation of coordination covalent (or dative) bonds between them. The model utilizes hybridization of metal s, p and d valence orbitals to account for the observed structures and magnetic properies of complexes. Valence bond theory is able to deal satisfactorily with many stereo chemical and magnetic properies but is has nothing to say about electronic spectra or the reason for the kinetic inertness of chromium (III)and low spin cobalt (III) octahedral complexes. To understand this and more other features of transition metal we must turn to other theories like crystal field theory etc. Pure crystal field theory assumes that the only interaction between the metal ion and the ligands is an electrostatic or ionic one with the ligands being regarded as negative point charges. This theory is quite successful in interpreting many important properties of complexes. The hybridization of [NiCl_(2)(PPh_(3))_(2)]" and "[NiCl_(2).(Pme_(3))_(2)] are respectively (consider PPh_(3) a bulkier ligand than Pme_(3)) :

Answer»

`SP^(3)" and "DSP^(2)`
`sp^(3)" and "sp^(3)`
`dsp^(2)" and "dsp^(2)`
`dsp^(2)" and "sp^(3)`

ANSWER :A
Discussion
4167.

Write the different oxidation states of iron. Why +2 oxidation state of manganes is more stable ? (Zof Mn =25).

Answer»

Solution :The different oxidiation states of iron are +2 and +3.The atomic number of Mn is
`25=1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)4s^(2)3d^(5)`
In + 2 oxidiation STATE,configuration will be
`1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)4s^(2)3d^(5)`
In +3 oxidation state, configuration will be
`1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)4s^(2)3d^(4)`
In `3d^(5)` configuration, Mn has half filled d-orbital and half filled and fully filed orbitals are more stable.

Also, shielding effect is SMALL and electrons are strongly held by nucleus. The exchange energy of these electrons is very LARGE. Thus +2 oxidation state is more stable.
Discussion
4168.

Two reactions, (i) A to Products, follow first order kinetics. The rate of reaction (i) is doubled when temperature is raised from 300 K to 310 K. The half-life for this reaction at 310 K is 30 minutes. At the same temperature, B decomposes twice as fast as A. If the energy of activation for the reaction for the reaction (ii) is half taht of reaction (i), calculate of reaction (ii) at 300 K.

Answer»

Solution :Calculation of ACTIVATION energy of REACTION (i)
`T_(1)=300" K", T_(2)=310" K",k_(1)=k,k_(2)=2K`
`:.""LOG""(k_(2))/(k_(1))=(E_(a))/(2.303" R")((T_(2)-T_(1))/(T_(1)T_(2))), i.e., log2=(E_(a))/(2.303xx8.314)xx(10)/(300xx310)" or "E_(a)=53.60" kJ mol"^(-1)`
Calculation of rate constant of reaction (i) at 310 K
`k=(0.693)/(t_(1//2))=(0.693)/(30" min")=2.31xx10^(-2)min^(-1)`
Rate constant of reaction (ii) at 310 `K=2xx2.31xx10^(-2)min^(-1)=4.62xx10^(-2)min^(-1)`
Energy of activation of reaction (ii) `=(53.60" kJ mol"^(-1))/(2)=26.80" kJ mol"^(-1)`
Aim. To calculate k for reaction (ii) at 300 K
`log""(k_(300" K"))/(k_(300" K"))=(E_(a))/(2.303"R")((T_(2)-T_(1))/(T_(1)T_(2)))`
`:.""log""(4.62xx10^(-2))/(k_(300" K"))=(26.80)/(2.303xx8.314xx10^(-3))xx(10)/(300xx310)=0.0151`
or `""(4.62xx10^(-2))/(k_(300" K"))="Antilog"0.0151=1.035`
or `""k_(300" K")=(4.62xx10^(-2))/(1.035)=4.46xx10^(-2)min^(-1)`
Discussion
4169.

Which of the following is cross linked polymer ?

Answer»

ORLON
MELAMINE
TEFLON
NYLON

SOLUTION :Melamine
Discussion
4170.

The vapour pressure of CS_(2) at 50^(@)C is 845 torr and a solution of 2.0 g sulphur in 100g of CS_(2) has a vapour pressure of 845/9 torr. Calculate the formula of sullphur molecule.

Answer»

Solution :Calculation of molar mas of sulphur.
Accoring to RAOULT's law, `(P_(A)^(@)-P_(S))/P_(S)=(W_(B)xxM_(A))/(M_(B)xxW_(A))`
`P_(A)^(@)854 TORR, P_(S)848.9 torr, W_(B)=2.0g,`
`W_(A)=100 g, M_(A)(CS_(2))=76 g mol^(-1)`
`((854torr-848.9torr))/((848.9torr))=((2.0g)xx(76.g mol^(-1)))/(M_(B)xx(100g))`
`M_(B)=((2.0g)xx(76.g mol^(-1))xx(848.9 torr))/((5.1 torr)xx(100g))=253.0 mol^(-1)`
Calculation of molecular formula of sulphur.
LET the formula of sulphur=`S_(N)`
`NXX(32g mol^(-1))=253.0 g mol^(-1)`
`n=((253g mol^(-1)))/((32g mol^(-1)))=7.9~~8`
Thus, the molecular formula of suphur=`S_(8)`.
Discussion
4171.

Which is the most abundant element in the earth's crust?

Answer»

SOLUTION :ALUMINIUM
Discussion
4172.

Which of the following does not react with oxygen directly? Zn, Ti, Pt, Fe

Answer»

Solution :Platinum is a noble METAL. The sum of its FIRST four ionization enthalpies is very large. Therefore, it does not react with oxygen directly. In contrast Zn, Ti and Fe are active metals and hence directly ract with oxygen to form their RESPECTIVE oxides.
`(2 Zn +O_(2) overset(Delta)rarr 2 ZnO, Ti + O_(2)overset(Delta)rarr TiO_(2), 2 Fe + 3O_(2)overset(Delta)rarr 2 Fe_(2)O_(3))`
Discussion
4173.

Whichof the followingis insolubeisdil HCl ?

Answer»

Aniline
ethyl AMINE
triphenylamine
DIMETHYLAMINE

ANSWER :C
Discussion
4174.

Whch one the following ore is not an ore of Al

Answer»

Anglesite
Mica
Beryl
Orthoclase

Solution :Mica, `KAlSi_(3)O_(10)(OH),_(2),"beryl", Be_(3)Al_(2)So_(6)O_(18)` and orthoclase, `KAlSi_(3)O_(8)` all are of aluminium. Anglesite is `PbSO_(4)`
Discussion
4175.

What is the nature of an antacid ?

Answer»

Solution :SUBSTANCES which REDUCE the role of a HCI by preventing the interaction of histamine with the RECEPTOR present in the stomach wall are called antacids. The most commonly used antacids are cimetidine and RANITIDINE
Discussion
4176.

Which of the following option the first compound has less entropy than second:

Answer»

`(i)` AQUEOUS solution of `1 "M of" MgCI_(2) (II)` aqueous solution of `1 "M of" NaCI`
`(1) Br_(2)` liquid at `25^(@) (ii) Br_(2)` liquid at `20^(@)C`
`(i) HgO` solid `(ii) HgS` solid
`(i) Br_(2)`liquid `(ii) I_(2)` solid

Solution :N//A
Discussion
4177.

The unit of rate constant of second order reaction is…..

Answer»

litre .`"MOL^(-1)","SECOND"^(-1)`
`"mole"^(-1) "litre" "second"^(-1)`
mole/litre .`"second"^(-1)`
`("Mole/litre")^(-1)"second^(-1)"`

ANSWER :B
Discussion
4178.

The spin magnetic moment of cobalt in the compound , Hg[Co(SCN)_(4)] is :

Answer»

`sqrt(3)`
`sqrt(8)`
`sqrt(12)`
`sqrt(24)`

Answer :C
Discussion
4179.

When 6 xx 10^22electrons are used in the electrolysis of a metallic salt, 1.9 gm of the metal is deposited at the cathode. The atomic weight of that metal is 57. So oxidation state of the metal in the salt is

Answer»

`+2`
`+3`
`+1`
`+4`

Answer :B
Discussion
4180.

Using IUPAC norms, write the systematic names of the following: [Pt(NH_(3))_(2)Cl(NH_(2)CH_(3))]Cl

Answer»

SOLUTION :Diamminechlorido(METHYL AMINE)PLATINUM(II) chloride
Discussion
4181.

Which aromatic acid among the following is weaker than simple benzoic acid ?

Answer»




ANSWER :B
Discussion
4182.

What is adsorption ? How does adsorption of a gas on a solid surface vary with (a) temperature (b) pressure. Iilustrate with the help of approprite graphs.

Answer»
Discussion
4183.

Which of the following is not the property of interstitial compound

Answer»

They have HIGH M.P, HIGHER than those of PURE metals
They are very hard, some borides approach diamond in hardness.
They retain metallic conductivity
Their chemical reactivity is very high

ANSWER :D
Discussion
4184.

Zinc is obtained from ZnO by

Answer»

CARBON reduction
Reduction USING silver
Electrochemical process
Acid LEACHING

ANSWER :A
Discussion
4185.

The temperature dependence of the rate of achemical reaction can be accurately explained by Arrhenius equation. With the help of Arrhenius equation, calculate the rate constant for the first order reaction C_2H_5I(g)toC_2H_(4(g))+HI_((G))at700K.Energy of activation (Ea)for the reaction is 209kjmol_(-1)and rate constant at 600K is 1.60xx10^(-5)s^(-1)[Universal gas constant R=8.314JK^(-1)mol^(-1)]

Answer»

Solution :log `k_2= log(1.60xx10^(-5))+(209xx10^3Jmol^(-3)XX100)/(2.303xx8.314xx600xx700)=(0.204-5)+2.599=-2.197`
`therfore k_2`=antilogo-2.197=antilogoof3.803=`6.3535xx10^(-3)s^(-1)`
Discussion
4186.

Which of the following interaction lies in the range of 8-42kH/mol?

Answer»

`H_(2)..H_(2)O`
`HCl..HCl`
`F^(-)…HF`
`HCN…NH_(3)`

SOLUTION :(a)`H_(2)…H_(2)O` Dipole-induced dipole interaction (van der waal.s force) ENERGY < 8kJ/mol
(b) `H-Cl…H-Cl`Dipole-dipole interaction (van der waal.s force) energy < 8kJ/mol
(c) `F^(-)….HF` Ion-dipole H-bond (van der waal.s force) energy < 42kJ/mol
(d) `HCN....NH_(3)` H-bond energy ranges from 8-42kJ/mol.
Discussion
4187.

Which of the following ishighly corrosivesalt ?

Answer»

` FeCl_2`
` +1.1V`
` +0.46 V`
` +0.76 V`

SOLUTION :HgC has corrosive action. If is highly poisonous . It SUBLIMES on heating. It si, therfore, known as corrosive sublimate.
Discussion
4188.

The solubility of iodine in water is greately increased by:

Answer»

ADDING an acid
Boiling the solution
Cooling the solution
Adding POTASSIUM iodide

Answer :D
Discussion
4189.

Which is paramagnetic from the following? Fe^(2+), Zn^(0), Hg^(2+), Ti^(+4)

Answer»

only `FE^(2+)`
`ZN^(0) and TI^(+4)`
`Fe_(2)^(2+) and HG^(2+)`
`Zn^(0) and Hg^(2+)`

Answer :A
Discussion
4190.

The temporary effect in which there is complete tranfer of a shared pair of pi-electrons to one of the atoms joined by a multiple bond on the demand of an attacking reagent is called

Answer»

INDUCTIVE effect
Positive RESONANCE effect
Negative resonance effect
Electromeric effect

Answer :D
Discussion
4191.

What is tonsil ? Write its functions.

Answer»

SOLUTION :Tetra ETHYL LEAD (TEL). It is WIDELY used as an ANTIKNOCK compound.
Discussion
4192.

Which of the following is characteristic of a carbocation?(i) It can rearrange to a more stable carbocation (ii)It can eliminate a hydride ion to form an alkene. (iii) It can add to an alkene to form a new carbocation. (iv) It can abstract a hydride ion from an alkane.

Answer»

(A) (i)
(B) (iii)
(C) (iv)
(D) (II)

Answer :(A,B,D)
Discussion
4193.

When 20g of naphtoic acid (C_(11)H_(8)O_(2)) is dissolved in 50g of benzene (K_(f)=1.72 k kg "mol"^(-1)), a freezing point depression of 2K is observed. The van't Haff factor (i) :-

Answer»

`0.5`
1
2
3

Answer :A
Discussion
4194.

Which allotrope of phosphorus is more reactive and why?

Answer»

SOLUTION :White PHOSPHORUS is more REACTIVE. This is because it is less stable because of angular strain in `P_4` molecule where the ANGLES are only `60^(@)`.
Discussion
4195.

Valence bond theory for bonding in transition metal complexes was developed by Pauling. From the valence bond point of view, formation of a complex involves reaction between Lewis bases (ligands) and a Lewis acid (metal atom or metal ion) with the formation of coordination covalent (or dative) bonds between them. The model utilizes hybridization of metal s, p and d valence orbitals to account for the observed structures and magnetic properies of complexes. Valence bond theory is able to deal satisfactorily with many stereo chemical and magnetic properies but is has nothing to say about electronic spectra or the reason for the kinetic inertness of chromium (III)and low spin cobalt (III) octahedral complexes. To understand this and more other features of transition metal we must turn to other theories like crystal field theory etc. Pure crystal field theory assumes that the only interaction between the metal ion and the ligands is an electrostatic or ionic one with the ligands being regarded as negative point charges. This theory is quite successful in interpreting many important properties of complexes. Select the correct statement about the crystal field theory.

Answer»

Metal-ligand bond in COORDINATION compounds arises PURELY from electrostatic interaction between the metal ion and the ligand.
Metal-ligand bond in coordination compound is purely a covalent bond.
The pattern of SPLITTING of the d-orbitals depends UPON the NATURE of the cystal field.
Both (a) and (c ).

Answer :D
Discussion
4196.

What will be the pH of a solution formed by mixing 40 ml of 0.10 M HCl with 10 ml of 0.45 M NaOH?

Answer»

12
10
8
6

Answer :A
Discussion
4197.

Valence bond theory for bonding in transition metal complexes was developed by Pauling. From the valence bond point of view, formation of a complex involves reaction between Lewis bases (ligands) and a Lewis acid (metal atom or metal ion) with the formation of coordination covalent (or dative) bonds between them. The model utilizes hybridization of metal s, p and d valence orbitals to account for the observed structures and magnetic properies of complexes. Valence bond theory is able to deal satisfactorily with many stereo chemical and magnetic properies but is has nothing to say about electronic spectra or the reason for the kinetic inertness of chromium (III)and low spin cobalt (III) octahedral complexes. To understand this and more other features of transition metal we must turn to other theories like crystal field theory etc. Pure crystal field theory assumes that the only interaction between the metal ion and the ligands is an electrostatic or ionic one with the ligands being regarded as negative point charges. This theory is quite successful in interpreting many important properties of complexes. Which of the following is correct for the complex [Ti(H_(2)O)_(6)]^(3+) ?

Answer»

Hybridization of CENTRAL metal ion of the complex is `sp^(3)d^(2)`.
The complex is paramagnetic containing one UNPAIRED electron.
Hybridization of central metal ion of the complex is `d^(2)sp^(3)`.
Both (B) and (c ).

ANSWER :D
Discussion
4198.

Which of the following drug is analgesic and non-narcotic ?

Answer»

PENICILLIN
ASPIRIN and PARACETAMOL
MORFIN
Varonal

SOLUTION :Aspirin and paracetamol
Discussion
4199.

Which of the following will be obtained in very small amount during this reaction? CH_(3)-CH=CH-CH_(2)-underset(O)underset(||)C-O-H underset((ii)Br_(2)//"C"Cl_(4),Delta)overset((i)AgOH)rarr"Products"

Answer»




`{:(CH_(3)-CH=CH-CH_(2)-C-O-CH_(2)),(""||""|),(""O""CH=CH-CH_(3)):}`

ANSWER :A::B::C::D
Discussion
4200.

What product is expected when tertiary butyl bromide is reacted with sodium ethoxide? Why that product is not formed and how that product can be obtained?

Answer»

Solution :`CH_3-underset(CH_3)underset(|)overset(CH_3)overset(|)(C )-Br+NaOC_2H_5underset(-NaBr)(rarr)CH_3-underset(CH_3)underset(|)overset(CH_3)overset(|)(C )-OC_2H_5`
The EXPECTED product is shown above. But the MAJOR product is 2-methyl-1-propene. This is because sodium ethoxide is a STRONG nucleophile as WELL as a strong buse. Thus elimination REACTION predominates over substitution. The product in the reaction can be obtained as
`CH_3-underset(CH_3)underset(|)overset(CH_3)overset(|)(C )-Br+NaOC_2H_5underset(-NaBr)(rarr)CH_3-underset(CH_3)underset(|)overset(CH_3)overset(|)(C )-OC_2H_5`
Discussion