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1701.

Titanium is purified by ……..

Answer»

MOND process
Van-Arkel Method
Zone refining
Electrolytic refining

Answer :B
Discussion
1702.

When electric current is passed through acidified water, 1 12 mL of hydrogen gas at STP collected at the cathode in 965 seconds. The current passed in amperes is

Answer»

`1.0`
`0.5`
`0.1`
`2.0`

SOLUTION :112 mL of `H_(2)` at `STP =(2XX 112G)/(22400)`
(Since 22400 mL at STP =M.wt)
Amount deposiled `=(Eq. wt xxi XXT)/(96500)`
`therefore (2xx 112)/(22400) =(1xx965xxi)/(96500)`
`i=1` AMP
Discussion
1703.

Which of the following on hydrolysis forms acetic acid?

Answer»

`CH_(3)CN`
`C_(2)H_(5)OH`
`C_(2)H_(5)NH_(2)`
`CH_(3)OH`

SOLUTION :`CH_(3)CN overset(H_(2)O //H^(+))(RARR) CH_(3)COOH`
Discussion
1704.

Which set contains only covalently bonded molecules ?

Answer»

`BCl_(3),SiCl_(4),PCl_(3)`
`NH_(4)Br,N_(2)H_(4),HBR`
`I_(2),H_(2)S,NAI`
`Al,O_(3),As_(4)`

Solution :The set of compounds `BCl_(3),SiCl_(4),PCl_(3)` are predominatly COVALENT compounds. `NH_(4)Br` and NaI ionic compounds and Al contains METALLIC lattice.
Discussion
1705.

Under which conditions, do we use magnetic separation for concentration of ore?

Answer»

SOLUTION :When either the METAL or the impurities are magnetic in PROPERTIES.
Discussion
1706.

Which of the following is the correct relation of the first law of thermodynamics?

Answer»

`DELTA = Q - W`
`Delta E = Delta q +Delta W`
`Delta E = q + W`
`Delta E = Delta q +W`

SOLUTION :`Delta E = q + W`
Discussion
1707.

Write equation for the preparation of teflon.

Answer»

SOLUTION :`N F_(2)C=CF_(2)`
Discussion
1708.

What are the oxidation of Cr in CrO_(3) and Ti in H_(2)TiO_(4) ?

Answer»


ANSWER :`+6, +4`
Discussion
1709.

Which is correct about electron sea model ?

Answer»

It was PROPOSED by Lorentz
It explains the lattice energies of ionic COMPOUNDS
It can explain the electrical CONDUCTIVITY of METALS
all 

ANSWER :C
Discussion
1710.

Which of the following is aromatic-

Answer»





Solution :Any planar cyclic system containing `(4N + 2)pi` ELECTRONS and having a single cyclic `pi`-electron cloud encompassing all the carbon ATOMS in the RING is aromatic.
Discussion
1711.

Which of the following properties increases on going down from F to I in Group VII-A of the periodic table

Answer»

Electronegativity
VOLATILE NATURE
ionic radius
OXIDIZING power

Solution :Ionic radius INCREASES on GOING down the group because no. of shells increases.
Discussion
1712.

Which is used as a painkiller ?

Answer»

ANTIBIOTIC
ANALGESIC
Antipyretic
Antioxidants

ANSWER :B
Discussion
1713.

What is the number of donar atoms in dimethylglyoximato ligand?

Answer»

1
2
3
4

Answer :B
Discussion
1714.

The relative basicity of amino follows the order as ……………… .

Answer»

`"Alkyl amines"gt"Aralkyl amines"gt "Ammonia"gt"N - aralkylamine"gt"Arylamine"`
`"Aralkyl amines"gt"Ammonia" gt "Arylamine" gt "Alkyl amine" gt "N - aralkylamine"`
`"Arylamine"LT "Alkyl amines"gt"Alkyl amine" lt"Ammonia" gt "N - aralkylamine"`
`"N - aralkylamine"lt "Arylamine" lt "Ammonia" lt "Alkyl amine" lt "Aralkyl amine"`

SOLUTION :`"Alkyl amines"gt"Aralkyl amines"gt "Ammonia"gt"N - aralkylamine"gt"Arylamine"`
Discussion
1715.

When limestone is heated, CO_(2) is given off. The metallurgical operation is __________

Answer»

SMELTING
reduction
calcination
roasting

Answer :C
Discussion
1716.

Which of the following have identical bond order?

Answer»

`CN^(-)`
`O_(2)^(-)`
`NO^(+)`
`CN^(+)`

Solution :Both `CN^(-)` and `NO^(+)` each have 14 electrons and their electronic configuration are same.
`(SIGMA1S^(2))(sigma(1s^(2))(sigma2s^(2))(sigma*2s^(2))(sigma2p_(z)^(2))(pi2p_(x)^(2))(pi2p_(y)^(2))`
and so their bond is same i.e. 3
The NUMBER of electrons in `CN^(+)` is 12 and in `O_(2)^(-)` is 17.
Discussion
1717.

Two containers A and B have the some volume. Container A contains 5 mole of O_(2) gas. Containter B contains 3 moles of He and 2 moles of N_(2). Both the containers are separately kept in vacuum at the same temperature. Both the containers have very small orifices of the same area through which the gases leak out. Compare the rate of effusion of O2 with that of He gas mixture.

Answer»

Solution :Since both the containers are in the same conditions of P, V and T,
`r_(O_(2))/(r_("mix"))=sqrt((M_("mix"))/(M_(O_(2)))`
As the mixture contains three moles of He and 2 moles of `N_(2)`, the effective molecular weight of the mixture would be.
`(3)/(5)xx4+(2)/(5)xx28=13.6`
`therefore (r_(O_(2)))/(r_("mix"))=sqrt((13.6)/(32))=0.652`
Though this solution looks OK, there is one big flaw in it.
The error is that we have assumed that `He and N_(2)` from vessel B would effuse out with the same rate. This assumption was made in because we have taken the composition of the gas mixture coming out of the vessel to be same as that of the mixture that was inside the vessel. It should be duly noted that the two mixtures (inside and the are that effused out) have different compositions. Therefore first we must find the composition of the gas mixture coming out of the vessel B.
`(r_(N_(2)))/(r_("He"))=(2)/(3) sqrt((4)/(28))=(2)/(3) sqrt((1)/(7))=0.252`
This means that initially the ratio of moles of `N_(2)` to the moles of He coming out of the vessel are in the MOLAR ratio of 0.252 and not `(2)/(3)`
Let moles of the He coming out to be x
`therefore` Moles of `N_(2)` coming out is 0.252 x
`therefore (n_(N_(2)))/(n_("total"))=(0.252 x)/(1.252x)=0.2`
`(n_("He"))/(n_("total"))=0.8 rArr M_("mix") =0.2xx28+0.8xx4=8.8`
`therefore (r_(O_(2)))/(r_("mix"))=sqrt((8.8)/(32))=0.52`
Discussion
1718.

Which of the following methods are used determine rate of the reaction containing more than one type of reactants? (P)Intergrated rate equation method. (Q)Half life method. (R )Ostwald.s isolation method.

Answer»

Only R
Q and R
P and Q
P.Q and R

Answer :A
Discussion
1719.

Value of CFSE, in tetrahedral complex having 3d^4 configuration of metal lon, surrounded by weak field ligands. will be

Answer»

`-2/5 triangle_(t)`
`+2/5 triangle_(t)`
`-4/5 triangle_(t)`
`+3/5 triangle_(t)`

Solution :For TETRAHEDRAL complexes, the CFSE is LOWER than pairing energy
Discussion
1720.

What happen when borax treated with ammonium chloride?

Answer»

Solution :When borax treated with ammonium chloride, it forms BORON nitride.
`underset(("Borax"))(Na_(2)B_(4)O_(7))+2NH_(4)Cl to 2NaCl+underset(("Boronnitride"))(2BN)+B_(2)O_(3)+2H_(2)O`
Discussion
1721.

Which of the following is not suitable for use in a desiccator to dry substances

Answer»

Conc. `H_(2)SO_(4)`
`Na_(2)SO_(4)`
`CaCl_(2)`
`P_(4)O_(10)`

Answer :A
Discussion
1722.

When a transition metal ion (usually) is involved in octahedral complex formation, the five degenerate d-orbitals split into two set of degenerate orbitals (3+2). Three degenerate orbitals of lower energy (d_(xy),d_(yz),d_(zx)) and a set of degenerate orbitals of higher energy (d_(x^(2)-y^(2)" and "d_(z^(2)). The orbitals with lower energy are called t_(2g) orbitals and those with higher energy are called e_(g) orbitals. In octahedral complexes, positive metal ion may be considered to be present at the centre and negative ligands at the corner of a regular octahedron. As lobes of d_(x^(2)-y^(2)" and "d_(z^(2) lie along the axis, i.e., along the ligands, the repulsions are more and so, high is the energy. The lobes of the remaining three d-orbitals lie between the Axis i.e., between the ligands. The repulsions between them are less, so lesser the energy. In the octahedral complexes, if metal ion has electrons more than 3, then for pairing them, the options are (i) Pairing may start with 4th electron in t_(2g) orbitals. (ii) Pairing may start normally with 6th electron when t_(2g)" and "e_(g) orbitals are singly filled. In which of the following configurations, hybridisation and magnetic moment of octahedral complexes are independent of nature of ligands. (P) d^(3) configuration of any metal cation (Q) d^(6) configuration of IIIrd transition series metal cation (R ) d^(8) configuration of Ist transition series metal cation (S) d^(7) configuration of any metal cation Select the correct code:

Answer»

R, S
P, R, S
P, Q, S
P, Q, R

Answer :D
Discussion
1723.

When a transition metal ion (usually) is involved in octahedral complex formation, the five degenerate d-orbitals split into two set of degenerate orbitals (3+2). Three degenerate orbitals of lower energy (d_(xy),d_(yz),d_(zx)) and a set of degenerate orbitals of higher energy (d_(x^(2)-y^(2)" and "d_(z^(2)). The orbitals with lower energy are called t_(2g) orbitals and those with higher energy are called e_(g) orbitals. In octahedral complexes, positive metal ion may be considered to be present at the centre and negative ligands at the corner of a regular octahedron. As lobes of d_(x^(2)-y^(2)" and "d_(z^(2) lie along the axis, i.e., along the ligands, the repulsions are more and so, high is the energy. The lobes of the remaining three d-orbitals lie between the Axis i.e., between the ligands. The repulsions between them are less, so lesser the energy. In the octahedral complexes, if metal ion has electrons more than 3, then for pairing them, the options are (i) Pairing may start with 4th electron in t_(2g) orbitals. (ii) Pairing may start normally with 6th electron when t_(2g)" and "e_(g) orbitals are singly filled. Which of the following electronic arrangement is/are possible for inner orbital octahedral complex. (P) t_(2g)^(3)e_(g)^(2)""(Q) t_(2g)^(6)e_(g)^(1) (R ) t_(2g)^(3)e_(g)^(0)""(S) t_(2g)^(4)e_(g)^(2) Select the correct code :

Answer»

<P>P, S
Q, R
R only
R, S

Answer :C
Discussion
1724.

When a transition metal ion (usually) is involved in octahedral complex formation, the five degenerate d-orbitals split into two set of degenerate orbitals (3+2). Three degenerate orbitals of lower energy (d_(xy),d_(yz),d_(zx)) and a set of degenerate orbitals of higher energy (d_(x^(2)-y^(2)" and "d_(z^(2)). The orbitals with lower energy are called t_(2g) orbitals and those with higher energy are called e_(g) orbitals. In octahedral complexes, positive metal ion may be considered to be present at the centre and negative ligands at the corner of a regular octahedron. As lobes of d_(x^(2)-y^(2)" and "d_(z^(2) lie along the axis, i.e., along the ligands, the repulsions are more and so, high is the energy. The lobes of the remaining three d-orbitals lie between the Axis i.e., between the ligands. The repulsions between them are less, so lesser the energy. In the octahedral complexes, if metal ion has electrons more than 3, then for pairing them, the options are (i) Pairing may start with 4th electron in t_(2g) orbitals. (ii) Pairing may start normally with 6th electron when t_(2g)" and "e_(g) orbitals are singly filled. Select incorrect match for the following complexes.

Answer»

`[IrF_(6)]^(3-)(Delta gt P)`
`[Co(H_(2)O)_(6)]^(3+)(Delta LT P)`
`FE(CO)_(5)(Delta gt P)`
`[PdCl_(2)(SCN)_(2)]^(2-)(Delta gt P)`

Answer :B
Discussion
1725.

The value of any colligative property of any electrolyte is determined experimentally, it is found to be higher than the theoretically calculated vallue using the usual expressions. This is because the electrolytes undergo dissociation in the solution. Similarly, in case of some substances association takes place, e.g., acetic acid in benzene and the experimental value of colligative value. The experimentally observed molar masses come out to be different than the theoretical values. These are called abnormal molar masses. In such cases, a correction factor, i, called van't Hoff factor is introduced which is a ratio of observed value of colligative property to calculated value when the solution behaves ideally. Knowing i, the degree of dissociation or association of the solute can be calculated. Which of the following aqueous solution will have the highest freezing point

Answer»

0.1 M urea
0.1 M SUCROSE
`0.1 M AlCl_(3)`
`0.1 M K_(4)[Fe(CN)_(6)]`

Solution :All except sucrose dissociate to GIVE ions in the solution. Thus, conc. of particles in sucrose solution is minimum. HENCE, depression in freezing point is minimum or actualfreezing point is highest.
Discussion
1726.

The value of any colligative property of any electrolyte is determined experimentally, it is found to be higher than the theoretically calculated vallue using the usual expressions. This is because the electrolytes undergo dissociation in the solution. Similarly, in case of some substances association takes place, e.g., acetic acid in benzene and the experimental value of colligative value. The experimentally observed molar masses come out to be different than the theoretical values. These are called abnormal molar masses. In such cases, a correction factor, i, called van't Hoff factor is introduced which is a ratio of observed value of colligative property to calculated value when the solution behaves ideally. Knowing i, the degree of dissociation or association of the solute can be calculated. The van't Hoff factor for 0.1 M Ba(NO_(3))_(2) solution is 2.74. The degree of dissociation is

Answer»

`91.3%`
`87%`
`100%`
`74%`

SOLUTION :`{:(,Ba(NO_(3))_(2),hArr,Ba^(2+),+,2NO_(3)^(-)),("Initial","1 MOLE",,,,),("After disso.",1-ALPHA,,alpha,,2alpha):}`
Total `=1+2alpha`
`therefore i=1+2alpha`
or `alpha=(i-1)/(2)=(2.74-1)/(2)=0.87=87%`
Discussion
1727.

The value of any colligative property of any electrolyte is determined experimentally, it is found to be higher than the theoretically calculated vallue using the usual expressions. This is because the electrolytes undergo dissociation in the solution. Similarly, in case of some substances association takes place, e.g., acetic acid in benzene and the experimental value of colligative value. The experimentally observed molar masses come out to be different than the theoretical values. These are called abnormal molar masses. In such cases, a correction factor, i, called van't Hoff factor is introduced which is a ratio of observed value of colligative property to calculated value when the solution behaves ideally. Knowing i, the degree of dissociation or association of the solute can be calculated. The boiling point of 0.1 molal K_(4)[Fe(CN)_(6)] solution will be (Given K_(b) for water = 0.52^(@)"C kg mol"^(-1))

Answer»

`100.52^(@)C`
`100.104^(@)C`
`100.26^(@)C`
`102.6^(@)C`

Solution :`DELTA T_(B)=imk_(b)`
`THEREFORE T_(b)=5xx0.1xx52=0.26`
`therefore T_(b)=100+0.26=100.26^(@)C`
Discussion
1728.

Vapour density of a substance is dependant of :

Answer»

Volume
Temperature
Pressure
Mol. Weight

Answer :D
Discussion
1729.

This diagram represents the behaviour of a pure solvent upon cooling. Which of the diagrams below best represents the cooling curve of a solution in that solvent upon cooling ? (Assume that all diagrams are drawn to the same scale.)

Answer»




ANSWER :C
Discussion
1730.

The smog is essentially caused by the presence of :

Answer»

`O_3` and `N_2`
`O_2` and `N_2`
OXIDES of N and S
`O_2` and `O_3`

ANSWER :C
Discussion
1731.

Which of the following chemicals are used to manufacture methyl isocyanate that caused Bhopal tragedy (i) Methyl amine (ii) Chloroform/KOH (iii) Phosphine (iv) Dimethylamine

Answer»

(i) and (III)
(iii) and (IV)
(i) and (II)
(ii) and (iv)

ANSWER :C
Discussion
1732.

Which of the following is expected to have highest boiling point ?

Answer»

`(CH_3)_2CHCl`
`(CH_3)_2CHOH`
`CH_3CH_2CH_2OH`
`CH_3CH_2CH_2Cl`

Solution :ALCOHOLS have HIGHEST boiling point because of presence of hydrogen bonding. Normal alcohols have higher boiling point than ISOMERIC branched alcohols. Compound (C) is normal PROPYL alcohol and compound (B) is branched propyl alcohol.
Discussion
1733.

The sealed containers of the same capacity and at the same temperature are filled with 44 g of H_(2) in one and 44 g of CO_(2) in the other. If the pressre of carbon dioxide in the second container is 1 atm. That of hydrogen in the first container would be.

Answer»

<P>1 ATM
10 atm
22 atm
44 atm

Solution :`(P_(1)V_(1))/(P_(2)V_(2))=(n_(1)RT_(1))/(n_(2)RT_(2))=(n_(1)T_(1))/(n_(2)T_(2))`
As, `V_(1) =V_(2) & T_(1) =T_(2)`
`(P_(1))/(P_(2))=(n_(1))/(n_(2))`
`(P_(H_(2)))/(P_(co_(2)))=(n_(H_(2)))/(n_(co_(2)))`
`(P_(H_(2)))/(1)=(44//2)/(44//44)=22` atm
Hence, (C ) is the correct answer.
Discussion
1734.

Which of the following statement is/are correct with the extraction of silver?

Answer»

Zinc is used to extract silver by solvent extraction from molten lead in Parke's process.
Silver is obtained as a by-product in the extraction of copper , lead and zinc.
Gold can ALSO be used to extract silver from Sodium argentocyanide in cyanidation.
Silver is obtained from the anode waste FORMED in the ELECTROLYTIC REFINING of copper and zinc.

Answer :A::B::D
Discussion
1735.

What type of devation is shown by a mixture of ethnol and acetone ? What type of azeotrope is formed on mixing the two ?

Answer»

SOLUTION :`+ve` DEVIATION, MAXIMUM BOILING AZEOTROPE.
Discussion
1736.

The role of addition of Me_(3)SiCl during the hydrolysis followed by conduction of Me_(3)SiCl_(2) is.

Answer»

To catalyte the reaction.
To TERMINATE the chain and hence CONTROLLING the MOLECULAR weight.
For obtaiming a proper cross linking.
All of the above

Solution :Being containing one HALOGENS atom it is ultilized for terminating the chain.
Discussion
1737.

Which one of the following does not give tetrathionate as the product, when treated with an aq solution of hypo?

Answer»

`C1_(2)`
`I_(2)`
`CuC1_(2)`
`FeC1_(3)`

ANSWER :A
Discussion
1738.

What will be the emf for the given cell Pt|H_(2)(P_(1))|H_((aq))^(+)||H_(2)(P_(2))|Pt

Answer»

`(RT)/(F)ln(P_(1))/(P_(2))`
`(RT)/(2f)ln(P_(1))/(P_(2))`
`(RT)/(f)ln(P_(2))/(P_(1))`
None of these

Solution :Anodic reaction: `H_(2)(P_(1))to2H^(+)`
Cathodic reaction: `2H^(+)toH_(2)(P_(2))`
`E_("cathode")=-(RT)/(2F)ln (P_(2))/([H^(+)]^(2)),E_("anode")=-(RT)/(2F)ln([H^(+)]^(2))/(P_(1))`
`E_("INF")=E_("anode")+E_("cathode")` ltBrgt `=-(RT)/(2F)ln((H^(+))^(2))/(P_(1))-(RT)/(2F)ln(P_(2))/((H^(+))^(2))`
`=-(RT)/(2F)ln(P_(2))/(P_(1))=(RT)/(2F)ln(P_(1))/(P_(2))`
Discussion
1739.

Whch of the following is an example of Freidel- Crafts reaction?

Answer»




NONE of the above

Answer :A, B, C
Discussion
1740.

The stability of ferric ion is due to:

Answer»

COMPLETELY FILLED d-orbitals
Half filled d-orbitals
Half filled F orbitals
Completely filled f-orbitals

Answer :B
Discussion
1741.

Which of the following amino acid has pH greater than 7 ?

Answer»

GLUTAMIC ACID
LYSINE
GLYCINE
Alanine

Solution :Lysine
Discussion
1742.

Which of the following has dodechahedral geometry

Answer»

`[ReH_(9)]^(2-)`
`[ZF_(7)]^(3-)`
`[MO(CN)_(8)]^(4+)`
`[UF_(8))]^(3-)`

Solution :`[Mo(NC)_(8)]^(4-)` has octagonal bipyramidal or dodecahedral (12 faces) SHAPE.
Discussion
1743.

Whichof the followingreactswithnot answerCarbylaminereaction ?

Answer»

`CH_(3)CH_(2)NH_(2)`
`(CH_(3))_(2)NH`
`(CH_(3))_(3)N`
`C_(2)H_(5)X`

ANSWER :A
Discussion
1744.

Which one is the first antibiotic produced by microorganism?

Answer»

PENICILLIN
Tetracycline
Chloramphenicol
All of these 

SOLUTION :Penicillin
Discussion
1745.

What are copolymers ? Give one example of a copolymer.

Answer»

Solution :Polymers whose repeating structural UNITS are derived from two or more TYPES of monomer units are CALLED copolymers. For example , nylons, polyester , MELMAC, Buna-N, etc.
Discussion
1746.

When two atoms of hydrogen combine to form a molecule of hydrogen gas, the energy of the molecule is:

Answer»

GREATER than that of SEPARATE atoms
Equal to that of separate atoms
LOWER than that of separate atoms
SOMETIMES lower than sometimes higher

Answer :C
Discussion
1747.

Which of the two is functional isomer? Glucose and lactose or xylose and galactose.

Answer»

SOLUTION :The FUNCTIONAL ISOMER : GLUCOSE and LACTOSE.
Discussion
1748.

Which of the following statements regarding the manufacture of H_(2)SO_(4) by Contact process is not true?

Answer»

Sulphur is burnt in air to form `SO_(2)`
`SO_(2)` is catalytically oxidised to `SO_(3)`
`SO_(3)` is dissolved in WATER to get `100%` sulphuric acid
`H_(2)SO_(4)` OBTAINED by CONTACT PROCESS is of higher purity than that obtained by lead chamber process.

Answer :C
Discussion
1749.

When 9.65 Coulomb of electricity is passed through a solution of silver nitrate (Atomic weigth of Ag = 107.85g), the amount of silver deposited is ………………….. .

Answer»

`10.8 mg`
`5.4 mg `
`16.2 mg`
`21.2 mg`

SOLUTION :`W_(Ag) = (E_(Ag) xx Q)/(96500) = (108 xx 9.65)/(96500)`
`= 1.08 xx 10^(-2)g = 10.8 mg`.
Discussion
1750.

Which is not present in clear hard water ?

Answer»

`MG(HCO_3)_2`
`CaCI_2`
`MgSO_4`
`MgCO_3`

ANSWER :D
Discussion