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26851.

The number of moles of KMnO_(4) that will be needed to react completely with one mole of ferrus oxalate in acidic solution is

Answer»

`3//5`
`2//5`
`4//5`
1

Answer :a
Discussion
26852.

The orbitals of same energy level providing the most efficient overlapping are:

Answer»

`SP^3 -sp^3`
sp - sp
`sp^2-sp^2`
All

Answer :B
Discussion
26853.

The number ofmoles of KMnO_(4) that will be needed to react completely with one mole of ferrious oxalate in acidic solution is

Answer»

`(3)/(5)`
`(2)/(5)`
`(4)/(5)`
1

Answer :A
Discussion
26854.

The order for the reaction, H_2 +Cl_2 overset(hv) to 2HCl over water is

Answer»

0
1
2
3

Solution :The order of the reaction over water is ZERO and in general CASE it is two. This is an EXPERIMENTAL FACT.
Discussion
26855.

The number of moles of KMnO_(4) reduced by one mole of Kl in alkaline medium is :

Answer»

one
two
five
one-fifth.

Solution :`overset(+7)KMnO_(4) overset(OH^(-))to overset(+6)K_(2)MnO_(4)`
Change in oxidation NUMBER of Mn in basic medium is 1.
`therefore` Moles of Kl is EQUAL to moles of `KMnO_(4)`.
Discussion
26856.

The orbitals having lower energy in tetrahedral complexes according to CFT are

Answer»

`d_(XY),d_(XZ),d_(z^2)`
`d_(xy), d_(YZ),d_(x^2-y^2)`
`d_(xy),d_(yz),d_(ZX)`
`d_(x^2-y^2),d_(z^2)`

ANSWER :D
Discussion
26857.

The number of moles of KMnO_4 reduced by one mole of KI in alkaline medium is

Answer»

ONE
two
five
one fifth

Answer :B
Discussion
26858.

The orbital picture of a triplet carbene can be drown as :

Answer»




NONE of these

Solution :(c ) Multiplicity `RARR 2s+1rArr2(1)+1=3rArr`TRIPLET
Discussion
26859.

The number of moles of KMnO_(4) in acidic medium that will be needed to react with one mole of sulphide ion is:

Answer»

`(2)/(5)`
`(3)/(5)`
`(4)/(5)`
`(1)/(5)`

ANSWER :A
Discussion
26860.

The orbitalpicture of a singlet carbene ( :CH_(2)) can be drown as :

Answer»




None of these

Solution :(a) Singlet CARBENE name by multiplicity = 2s + 1
Here SPIN QUANTUM numberzero,s = 0 then = 2(0) + 1 = 1
Discussion
26861.

The orbital nearest to the nucleus is

Answer»

4f
5d
4S
7P

ANSWER :C
Discussion
26862.

The number of moles of ions produced per mole of K_4[Fe(CN)6] in aqueous solution will be _____

Answer»

5
2
11
10

Answer :A
Discussion
26863.

The orbital diagram in which the Aufbau principle is violated

Answer»




ANSWER :B
Discussion
26864.

The number of moles of hydroxide (HO^(-))ion in 0.3 litre of 0.005 M solution of Ba(OH)_(2) is

Answer»

0.0075
0.0015
0.003
0.005

Solution :0.3L of 0.005 M `Ba(OH)_(2)` SOL as`M=n/V(L)`
`:. N = 0.005 M Ba(OH)_(2)` sol as `M=n/(V(L))`
`:.n = 0.005 XX 0.3` mol = 0.0015 mol `Ba(OH)_(2)` `(Ba^(2+)2OH^(-))`
mol of `OH^(-)` = `2 xx 0.0015 = 0.0030` moles
Discussion
26865.

The orbital diagram in which both the Pauli 's exclusion principle and Hund's rule are violated is

Answer»





Solution :Electrons in an orbital cannot have spin in the same direction (Pauli .s exclusion principle). p-orbitals MUST contain `1 E^(-)` each before pafring STARTS (Hund.s rule).
Discussion
26866.

The number of moles of hydrogen that can be added to 1 mole of an oil is the highest in

Answer»

LINSEED oil
Groundanut oil
Sunflower SEED oil
Mustard oil

Answer :A
Discussion
26867.

The number of moles of hydroxide (OH^(-)) ion in 0.3 litre of 0.005 M solution of Ba(OH)_(2) is

Answer»

0.005
0.003
0.0015
0.0075

Answer :B
Discussion
26868.

The orbital diagram in which both Pauli's exclusion principle and Hund's rule are violated is:

Answer»




ANSWER :A
Discussion
26869.

The number of moles of H_(2)O_(2) required to completely react with 400 ml of 0.5 N KMnO_(4) in acidic medium are

Answer»

0.1
0.2
`1.0`
0.5

Answer :A
Discussion
26870.

the orbital diagram in which Aufbau principle is violated is

Answer»




ANSWER :B
Discussion
26871.

The number of moles of hydrogen molecules required to produce 20 moles of ammonia through Haber's proccess is :

Answer»

40
10
20
30

Solution :`N_2(g) + 3H_2(g) to 2NH_3(g)`
The required `H_2` moles for the MANUFACTURING of 2 MOLE `NH_3 = 3` mole
The required `H_2` moles for the manufacturing of 20 moles `NH_(3) = (20 xx 3)/2`=30 mole `H_2` is required
Discussion
26872.

The number of moles of HIO_(3) formed during periodic oxidation of cyclohexa (1,2,3) triol____

Answer»


ANSWER :2
Discussion
26873.

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

Answer»

`+1/2.h/(2pi)`
ZERO
`h/(2pi)`
`sqrt2.h/(2pi)`

Solution :For 2S ORBITAL, L= 0,
angular MOMENTUM= `sqrt(l(l+1))h/(2pi)=0`
Discussion
26874.

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

Answer»

`h/4pi`
ZERO
`h/2PI`
sqrt2.h/(2pi)

ANSWER :B
Discussion
26875.

The number of moles of electrone required to deposits 36 g of Al from an aqueous solution of Al(NO_(3))_(3) is (atomic mass of Al = 27)

Answer»

4
2
3
1

Solution :`Al^(3+) + 3e^(-) to Al` .
`1 F-= 96500 C -= 1 g ` EQ. of Al `= (27)/(3) = 9g`
`therefore` 1 MOLE of electrons i.e. 36 g of Al = n
`thereforen = (36 xx 1)/(9) = 4 `moles
Discussion
26876.

the orbital angular momentum of 4f electron is

Answer»

`4(H/(2π))`
`sqrt12(h/(2π))`
`sqrt6π(h/(2π)`
`sqrt2×h/(2π)`

ANSWER :B
Discussion
26877.

The number of moles of Ca(OH)_(2) that must be dissolved to make 250 mL solution in water of pH = 10.65 is

Answer»

`5.6 xx 10^(-5)`
`6.5 xx 10^(-5)`
`4.5 xx 10^(-5)`
`5.4 xx 10^(-5)`

Solution :We know that pH + POH = 3.35
pOH = (14-10.65) = 3.35
`[OH^(-)] = 10^(-3.35)= 4.47 xx 10^(-4) M`
ONE molecule of `Ca(OH)_(2) = (4.47 xx 10^(-4))/(2) = 2.235 xx 10^(-4) M`
No. of moles in `250 ML = (2.235 xx 10^(-4))/(4) = 5.6 xx 10^(-5)`
Discussion
26878.

The orbital angular momentum for a 2p -electron is

Answer»

`SQRT(3)h`
`sqrt(6)h`
`0`
`sqrt(2)h/(2pi)`

ANSWER :D
Discussion
26879.

The orbital angular momentum of an electron in 2s orbital

Answer»

`+1/2h/(2PI)`
zero
`h/(2pi)`
`sqrt(2).h/(2pi)`

Solution :ORBITAL angular MOMENTUM
`=sqrt(l(l+1))h/(2pi)=0"" ( :. l=0)`
Discussion
26880.

The number of moles of electron required to reduce 0.2 mole ofCr_(2)O_(7)^(2-) to Cr^(3+)

Answer»

6
1.2
0.6
12

Solution :`Cr_(-2)O_(7)^(2-)+14H^(+)+6e^(-)to2Cr^(3+)+7H_(2)O`
1-6
0.2 - X`implies` 1.2
Discussion
26881.

The orange solid on heating gives a colourless gas and a greensolid which can be reduced to metal by aluminium powder. The orange and the green solids are respectively

Answer»

`(NH_(4))_(2)Cr_(2)O_(7) and Cr_(2)O_(3)`
`Na_(2)Cr_(2)O_(7) and Cr_(2)O_(3)`
`K_(2)Cr_(2)O_(7) and CrO_(3)`
`(NH_(4))_(2) Cr_(2)O_(4) and CrO_(3)`

SOLUTION :
Discussion
26882.

The number of moles of BaCO_3 Which contains 1.5 moles of oxygen atom is :

Answer»

0.5
1
3
`6.02 XX 10^23`

ANSWER :A
Discussion
26883.

thenumberof molesof acidified KMnO_(4) requiredto oxidize 1 moleof ferrousoxalate (FeC_(2) O_(4)) is

Answer»

5
3
0.6
1.5

Answer :C
Discussion
26884.

The orange solid on heating gives a colourless gas and a green solid which can be reduced to metal by aluminium powder. The orange and the green solids are, respectively.

Answer»

`(NH_4)_2 Cr_2 O_7 and Cr_2O_3`
`Na_2Cr_2O_7` and `Cr_2O_3`
`K_2Cr_2O_7` and `CrO_3`
`(NH_4)_(2)CrO_(4)` and `CrO_(3)`

Solution :`UNDERSET("ORANGE SOLID")((NH_4)_(2)Cr_2O_7) + underset("Colourlessgas")(N_2) + Cr_2O_3 + 4H_2O`
Discussion
26885.

The number of moles of a solute in its solution is 20 and total number of moles are 80. The mole fraction of solute is

Answer»

`2.5`
`0.25`
1
`0.75`

SOLUTION :MOLE fraction of SOLUTE `= (20)/(80)=0.25`.
Discussion
26886.

The orange solid on heating gives a colourless gas and a green solid which can be reduced to metal by aluminium powder. The orange and the green solids are, respectively

Answer»

`(NH_(4))_(2)Cr_(2)O_(7) and Cr_(2)O_(3)`
`Na_(2)Cr_(2)O_(7) and Cr_(2)O_(3)`
`K_(2)Cr_(2)O_(7) and CrO_(3)`
`(NH_(4))_(2)CrO_(4) and CrO_(3)`

SOLUTION :`underset("Orange solid")((NH_(4))_(2)Cr_(2)O_(7)) overset(Delta)rarr underset("Colouress gas")(N_(2)) + underset("Green solid")(Cr_(2)O_(3))+4H_(2)O`
`Cr_(2)O_(3) + 2Al overset(Delta)rarr 2Cr + Al_(2)O_(3)`
Discussion
26887.

The number of moles of acidified KMnO, required to oxidize 1 mole of ferrous oxalate (FeC_(2)O_(2)) is ....

Answer»

5
3
0.6
1.5

Solution :`MnO_(4)^(-)+FeC_(2)O_(2) to Mn^(2+)+ FE^(3+)2CO_(2)`
`5 "MOLES of" FeC_(2)O_(4)4 = 3 "moles of" KMnO_(4)`
`5 "moles of" FeC_(2)O_(4)4 = (3)/(5) "moles of" KMnO_(4)`
`5 "moles of" FeC_(2)O_(4)4 = 0.6 "moles of" KMnO_(4)`
Discussion
26888.

The orange colour of K_(2)Cr_(2)O_(7) is due to….

Answer»

`SIGMA RARR sigma^(**)` transition
`d-d` transition
LIGAND to metal charge transfer
Metal to ligand charge transfer

Answer :C
Discussion
26889.

The number of moles of a gas 1 m^(3) of volume at NTP is:

Answer»

4.46
0.446
1.464
44.6

Solution :`1 m^(3)=1000L`
NUMBER of MOLES `=(1000)/(22.4)=44.6`
Discussion
26890.

The optium temoperature range for enzymatic activity is ………and optimum pH range is ……….

Answer»

SOLUTION :`298-310 K, 5-7`
Discussion
26891.

The number of moles in 0.64 g of SO_2is :

Answer»

100
10
0.1
0.01

Solution :MOLES of `SO_2 = ("MASS")/("MOL WT")= (0.64)/(64) = 0.01`
Discussion
26892.

The optically inactive amino acid is

Answer»

LYSINE
Glysine
ARGININE
ALANINE

ANSWER :B
Discussion
26893.

The number of molecules present in 1cm^(3) of an ideal gas at STP is called …………..and its value is…………… .

Answer»

SOLUTION :Loschmidt NUMBER, `(6.022xx10^(23))/22400=2.69xx10^(19)`
Discussion
26894.

The optically active molecule is:

Answer»




ANSWER :B
Discussion
26895.

The number of molecules present in 1 cm^(3) of water is :-

Answer»

`2.7xx10^(19)`
`3.3xx10^(22)`
`6.02xx10^(20)`
1000

Solution :Density water=1G/mL
`therefore`Mass of `1cm^(3)` water=density`xx`volume
`=1xx1`
=1g
`therefore`No. of mole of water MOLECULE in `1cm^(3)=`
`therefore`No. of molecule of `H_(2)O` in `1cm^(3)=(1)/(18)xxN_(A)`
`=(1)/(18)xx6xx10^(23)`
`=3.3xx10^(22)`
Discussion
26896.

The optically active tartaric acid is named as D-(+)-tartaric acid because it has a positive

Answer»

Optical rotation and is derived from D-glucose
PH in organic solvent
Optical rotation and is derived from D-(+)- glyceraldehyde
Optical rotation when substituted by deuterium

Solution :D-word is used to represent the ARRANGEMENT of -OH group in right side at SECOND LAST carbon atoms as in glyceraldehyde.
`{:(""CHO),("|"),(H-C-OH),("|"),(""CH_(2)OH):}`
and (+) sign is used to represent the rotation in right side. Hence, in D-(+)-tartaric acid.

Hence, it has a positive optical rotation and it is devided with glyceraldehyde.
Discussion
26897.

The number of molecules of water needed to convert one molecule of P_2O_5 into orthophosphoric acid is:

Answer»

2
3
4
5

Answer :B
Discussion
26898.

The optically active molecule is

Answer»




Solution :Others are meso COMPOUND DUE to presence of plane of symmetry.
Discussion
26899.

The optically active product obtained from S_(N)2 reaction of dextro rotatory compound will be ________.

Answer»

DEXTRO rotatory
laevo rotatory
racemic mixture
partially OPTICALLY active

Answer :B
Discussion
26900.

The number of molecules of NaCI in an unit cell of its crystal is:

Answer»

2
4
6
8

Answer :B
Discussion