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27101.

The numerical value of the equilibrium constant or any chemical change is affected by changing the

Answer»

concentration of product
catalyst
concentration of reacting substance
TEMPERATURE

Solution :The equilibrium constant is unaffected by CHANGING the concentration of products, catalyst and conentration of REACTANTS. It is AFFECTED by changing the temperature.
Discussion
27102.

The numerical value of N//n ( where N is the number of molecules in a given sample of gas and n is the number of moles of the gas ) is :

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8.314
`6.02 xx 10^(23)`
0.0821
`1.66 xx 10^(-19)`

ANSWER :B
Discussion
27103.

The numerical value of K_p and K_c for the equilibrium 2NH_3 hArr N_2 + 3H_2 are related as :

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`K_p = K_c xx(RT)^3`
`K_p = K_c xx(RT)^(-2)`
`K_p = K_c xx(RT)^2`
NONE of these

Answer :C
Discussion
27104.

The numerical value of energy involved in the given process , S rarr S^(-) is less than, which of the following process

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`S^(-) rarr S`
`SE rarr Se^(-)`
`S rarr S^(+)`
(2) and (3) both

Solution :DATA based
Discussion
27105.

The number of geometrical isomers that can exist for the square planar complexis :

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4
6
2
3

Solution :The complex is also MABCD type and is the square planar in NATURE. THREE GEOMETRICAL ISOMERS also possible.
Discussion
27106.

The numerical value of a solubility product K_(sp) is measured by experiment.For example, we could determine K_(sp) for CaSO_(4) by adding on excess of solid CaSO_(4) to water, stirring the mixture to give a saturated solution of CaSO_(4) and then measuring the concentration of Ca^(2+) and SO_(4)^(-) in the saturated solution. Value of K_(sp) are unaffected by the presence of other ions in soluiotn. At 27^(@)C experiemntal value of K_(sp) is 2.4 xx 10^(-5) M^(2). Solubility of CaSO_(4) solution in 0.02M aqueous Ca(NO_(3))_(2) solution

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`1.2 xx 10^(-3)M //` litre
`1.02 xx 10^(-3)G m //` litre
`3.6 xx 10^(-3)` M `//` litre
`4.9 xx 10^(-3)` M `//` litre

ANSWER :A
Discussion
27107.

The number of geometrical isomers that can exist for square planar [Pt(Cl)(py)(NH_(3))(NH_(2)OH)]^(+) is (py = pyridine)

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4
6
2
3

Solution :A square planar COMPLEX of the type MABCD forms THREE isomers.

Two of these are CIS and one is trans.
Discussion
27108.

The numerical value of 'a' the van der Waals' constant is maximum for :

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`NH_3`
`H_2`
`O_2`
He

ANSWER :A
Discussion
27109.

The number of geometrical isomers possible for the complex [CoL_(2)Cl_(2)]^(-)(L=H_(2)NCH_(2)CH_(2)O^(-)) is

Answer»


Solution :
Alternatively, it is a complex of the TYPE `[M(AB)_(2)C_(2)]`. POSSIBLE GEOMETRICAL isomers are
Discussion
27110.

The numerical value of C_p -C_v is equal to :

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R
R/M
M/R
None

Answer :B
Discussion
27111.

The numerical value of a solubility product K_(sp) is measured by experiment.For example, we could determine K_(sp) for CaSO_(4) by adding on excess of solid CaSO_(4) to water, string the mixture to give a saturated solution of CaSO_(4) and then measuring the concentration of Ca^(2+) and SO_(4)^(-) in the saturated solution. Value of K_(sp) are unaffected by the presence of other ions in solution. At 27^(@)C experiemntal value of K_(sp) is 2.4 xx 10^(-5) M^(2). CaSO_(4) obtained as ppt. when 0.02M when 0.02 M CaCl_(2) mixed with "................"Na_(2)SO_(4) solution

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0.0004M
0.004 M
0.0008 M
0.0006M

ANSWER :B
Discussion
27112.

The numerical value of a solubility product K_(sp) is measured by experiment.For example, we could determine K_(sp) for CaSO_(4) by adding on excess of solid CaSO_(4) to water, string the mixture to give a saturated solution of CaSO_(4) and then measuring the concentration of Ca^(2+) and SO_(4)^(-) in the saturated solution. Value of K_(sp) are unaffected by the presence of other ions in solution. At 27^(@)C experiemntal value of K_(sp) is 2.4 xx 10^(-5) M^(2). Will a precipitate for if equal volumes of 0.08 M CaCl_(2) and 0.02 M Na_(2)SO_(4)are mixed ?

Answer»

YEAS
No
Can not predict
Reaction is not possible

Answer :A
Discussion
27113.

The number of geometrical isomers possible for the square planar complex MABCDis :

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2
3
4
6

Solution :THREE GEOMETRICAL ISOMERS are POSSIBLE.
Discussion
27114.

The numer of optically active stereoisomers of tartaric acid, (HOOC.CHOH.CHOH.COOH) is

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4
2
1
3

Answer :B
Discussion
27115.

The numer of quaternary and chiral carbon atoms present in elatol, isolated from an algae are respectively

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2, 3
4, 2
3, 2
1, 3

Answer :A
Discussion
27116.

The number of geometrical isomers in case of a compound with the structure, CH_3 - CH = CH -CH =CH -C_2H_5 are:

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Four
Three
Two
Five

Answer :A
Discussion
27117.

The numbers of tetrahedral and Octahedral holes per atom in cubical closest packings of atoms respectively are-

Answer»

1,1
1,2
2,1
2,2

Solution :2,1
Discussion
27118.

The number of electrons in the highest principal energy level for an element(Z=26) is

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8
6
2
16

Solution :The CONFIGURATION is `[Ar]3d^(5)4S^(2).` No of electrosn in n=4 is 2.
Discussion
27119.

The numberof possible enantiomeric pairs that can be produced during monochlorination of 2-methyl butane is:

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2
3
4
1

Answer :A
Discussion
27120.

The numberof all possible products excluding stereoisomers obtained on monochlorination of n-butane and iso-butane are respectively

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2 and 3
3 and 2
2 and 1
2 and 2

Answer :D
Discussion
27121.

The number of which subtomic particle is same is case of chlorine atoms and chloride in ?

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Electron
Proton
NEUTRONS
All of the above

Answer :B::C
Discussion
27122.

The number P = O bonds present in tetrabasic H_(4)P_(2)O_(7) is…………….

Answer»

Three
two
one
FOUR

SOLUTION :
Discussion
27123.

The number of geometrical isomers in CH_(3)CH=CHCH_(2)CH=CH_(2) is:

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TWO
three
four
FIVE.

ANSWER :A
Discussion
27124.

The number of geometrical isomers of,[Co(NH_3)_3(NO_2)_3] are:

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Zero
2
3
4

Answer :B
Discussion
27125.

The numbers of ions formed on dissolving one molecule of FeSO_4(NH_4)_2SO_4.6H_2O is :

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4
5
3
6

Answer :B
Discussion
27126.

The number of waves made by an electron moving in an orbit having maximum quantum number (m) +3 is

Answer»

3
4
5
6

Solution :l=3 and n=4
Discussion
27127.

The number of geometrical isomers of [Co(NH_(3))_(3)(NO_(3))_(3)] are :

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0
2
3
4

Solution :`Ma_(3)b_(3) to [M (a a)(B b)(AB)],[M(ab)(ab)(ab)]`
Discussion
27128.

The number of water molecules present in a drop of water weighing 0.18g is :

Answer»

`6.022 xx 10^26`
`6.022 xx 10^23`
`6.022 xx 10^19`
`6.022 xx 10^21`

Solution :MOLES of water in a DROP
` = (0.018)/(18) = 1 xx 10^(-3) ` moles
1MOL of water ` = 6.022 xx 10^23` molecules
` 1 xx 10^(-3) ` mole ofwater
` = 6.022 xx 10^23 xx 1 xx 10^(-3)`
` = 6.022 xx 10^20` molecules
Discussion
27129.

The number of geometrical isomers in the following compound CH_(3)-CH-CH-CH=CH-C_(2)H_(5) is:

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4
3
2
5

Answer :A
Discussion
27130.

The number of geometrical isomers in case of a compound with the structure CH_(3)-CH=CH-CH-C_(2)H_(5) is

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4
3
2
5

Solution :
Discussion
27131.

The number of water molecules present in a drop of water weighing 0.018g is

Answer»

`6.022 xx 10^(26)`
`6.022 xx 10^(23)`
`6.022 xx 10^(19)`
`6.022 xx 10^(20)`

Solution :No. of moles = `("Moles")/("MOL. Mass") = (0.018)/(18) = 1 xx 10^(-3)`
No. of moles = `("No. of molecules")/(N_(0))`
`1 xx 10^(-3) = ("No. of molecules")/(6.022 xx 10^(23)) = 6.022 xx 10^(20)`
Discussion
27132.

The number of geometrical isomers for the compound with molecular formula C_(2)BrClFI is

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3
4
5
6

Solution :There are SIX GEOMETRICAL ISOMERS (E and Z).
.
Discussion
27133.

The number of water molecules is maximum in

Answer»

18 gram of water
18 MOLES of water
18 MOLECULES of water
1.8 gram of water

Solution :`"18 G of water "=(18)/(18)" mole = 1 mole "=6.022xx10^(23)" molecules"`
`"18 moles of water"=18xx6.022xx10^(23)" molecules"`
`"1.8 g of water "=(1.8)/(18)" mole = 0.1 mole"`
`=0.1xx6.022xx10^(23)=6.022xx10^(22)" molecules"`
Thus, 18 molecules of `H_(2)O` contain maximum NUMBER of molecules.
Discussion
27134.

The number of geometrical isomers for [Pt(NH_(3))_(2)Cl_(2)] is :

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1
2
3
4

Answer :B
Discussion
27135.

The number of water molecules in 1 litre of water is

Answer»

18
`18xx1000`
`N_(A)`
`55.55 N_(A)`

Solution :`d=M/V` (d = density, M = mass, V = volume)
Since `d=1`
So, `M=V`
18 GM = 18 ml
18 ml `=N_(A)` MOLECULES (`N_(A)=` AVOGADRO's no.)
`1000 ml=N_(A)/18xxx1000=55.555 N_(A)`
Discussion
27136.

The number of geometrical isomers for octahedral [Co(NH_(3))_(2)Cl_(4)]^(-) , square planar [AuCl_(2)Br_(2)]^(-) respectviely are :-

Answer»

4,2
2,2
3,2
2,3

ANSWER :B
Discussion
27137.

The number of geometrical isomers for [P(NH_(3))_(2)Cl_(2)] is

Answer»

2
1
3
4

Answer :A
Discussion
27138.

The number of water molecules directly bonded to the metal centre is CuSO_4.5H_2O is

Answer»


SOLUTION :`[CU(H_2O)_4]SO_4 . H_2O`
Discussion
27139.

The number of geometrical isomers for octahedral [Co(NH_(3))_(2)Cl_(4)]^(-) and square planar [AuCl_(2)Br_(2)]^(-) respectively are :

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1, 2
2, 2
3, 2
2, 3

Solution :`[M a_(2) b_(4)] implies (a a)(B b)(b b)(ab)(ab)(b b)`
`[M a_(2)b_(2)] implies (a a)(b b)(ab)(ab)`
Discussion
27140.

The number of geometric isomers that can exist for square planar[Pt(Cl)(py)(NH_(3))(NH_(2)OH)]^(+) is (py=pyridine)

Answer»

2
3
4
6

Answer :B
Discussion
27141.

The number of water molecule(s) directly bonded to the metal centre is CuSO_(4).5H_(2)O is

Answer»


SOLUTION :
Discussion
27142.

The number of geometrical isomers and optical isomers of [Pt(NH_(3))(Br)(Cl)(Py)] are respectively.

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2, 0
3, 2
3, 0
2, 2

Solution :M ABCD `IMPLIES` G.I.= 3, O.I.=0
( because P.O.S.= Molecular plane)
Discussion
27143.

The number of geometric isomers that can exist for square planar [Pt(Cl)(py) (NH_(3))(NH_(2))OH]^(+) is (py = pyridine) :

Answer»

2
3
4
6

Solution :THREE geometrical isomers are possible.

These are written by fixing the position of one LIGAND (say CL) and placing the other ligands `(NH_(2)OH, NH_(3), py)` trans to it. OTHERS structures will be identical.
Discussion
27144.

The number of visible lines when an electrons returns from the 5^(th) orbit to ground state in the hydrogen spectrum is P. Then value of P is:

Answer»


SOLUTION :No of VISIBLE LINES MEANS Balmer seriesltbgt HENCE n-2=5-2=3
Discussion
27145.

The number of g-atom of oxygen in 6.02xx10^(24) CO molecule is

Answer»

1
0.5
5
10

Solution :One molecula of `CO` CONTAINS one oxygen atom.
`:. 6.02xx10^(24)` MOLECULES of `CO` contain `6.02xx10^(24)` oxygen atoms.
`6.02xx10^(23)` atoms of oxygen `=1` g-atom of oxygen
`:. 6.02xx10^(24)` atoms of oxygen `=10` g-atom of oxygen.
Discussion
27146.

The number of formula units of calcium fluoride CaF_(2) present in 146.4 g of CaF_(2) (The molar mass of CaF_(2) is 78.08 g/mol) is

Answer»

`1.129xx10^(24) CaF_(2)`
`1.146xx10^(24) CaF_(2)`
`7.808xx1-^(24) CaF_(2)`
`1.877xx10^(24) CaF_(2)`

Solution :NUMBER of moles of `CaF_(2)=146.4/78.08=1.875`
`:.` No. of FORMULA UNITS `=1.875xxN_(A)`
`=1.875xx6.023xx10^(23)=11.29xx10^(23)=1.129xx10^(24)`
Discussion
27147.

The number of geometrical isomers of [Co(NH_(3))_(3)(NO_(3))_(3)] is

Answer»

4
2
6
3

Answer :A
Discussion
27148.

The number of free valencies available for adsorption if four Pt atoms are linked together by convalent bonds is

Answer»

SOLUTION :
THUS, there are 8 FREE VALENCIES.
Discussion
27149.

The number of valence electrons for each member of halogen family is equal to six.

Answer»

SOLUTION :The number of valence electrons for each MEMBER of HALOGENN family is seven
Discussion
27150.

The number of Faradays required to deposit 1 g equivalent of aluminium (At. Wt. = 27) from a solution of aluminium chloride will be

Answer»

1
2
3
4

Solution :To deposit 1 G EQ. of any ELEMENT, 1 F is required
Discussion