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27051.

The octane numbers of 2,2,4-trimethylpentane (isooctane) and n-pentane are respectively-

Answer»

50,50
100,0
0,100
50,0

Answer :B
27052.

The octet rule is not valid for which one of the following molecules ?

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`CO_(2)`
`H_(2)S`
`NH_(3)`
`BF_(3)`

SOLUTION :Octate rule is INVALID for `BF_(3)`
27053.

Theodourofamineis

Answer»

ODOURLESS
PUNGENT
fishy
garlic LIKE

ANSWER :C
27054.

The number of ions formed when cupra ammonium sulphate is dissolved in water

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1
2
4
zero

Answer :B
27055.

The octahedral complex of a metal ion M^(3+) with four monodentate ligands L_(1),L_(2),L_(3) and L_(4) absorb wavelength in the region of red, green, yellow and blue, respectively. The increasing order of ligand strength of the four ligands is :

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`L_(4) lt L_(3) lt L_(2) lt L_(1)`
`L_(1) lt L_(3) lt L_(2) lt L_(4)`
`L_(3) lt L_(2) lt L_(4) lt L_(1)`
`L_(1) lt L_(2) lt L_(4) lt L_(3)`

ANSWER :B
27056.

The number of ions formed when copper ammonium sulphate, is dissolved in water is:

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1
2
4
Zero

Answer :B
27057.

The octahedral complex of a metal ion M^(3+)with four monodentate ligands L_(1), L_(2), L_(3) and L_(4) absorb wavelengthin the region of red, green, yellow and blue respectively. The increasing order of ligand strengths of the four ligands is :

Answer»

`L_(4) lt L_(3) lt L_(2) lt L_(1)`
`L_(1) lt L_(3) lt L_(2) lt L_(4)`
`L_(3) lt L_(2) lt L_(4) lt L_(1)`
`L_(1) lt L_(2) lt L_(4) lt L_(3)`

Solution :The ENERGY increases in the order (VIBGYOR)
RED < yellow < green < BLUE
The complex absorbing LOWER energy will be of lower strength of ligand and therefore, the correct order of ligand strength is
`L_(1) lt L_(3) lt L_(2) lt L_(4)`
27058.

The number of ions formed on dissolving one molecule of FeSO_(4)(NH_(4))_(2)SO_(4).6H_(2)O is

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4
5
3
6

Answer :B
27059.

The octahedral shape is associated with :

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`PF_(5)`
`SF_(4)`
`TeF_(6)`
`ClF_(3)`

ANSWER :C
27060.

The number of octahedral voids in a unit cell of cubic close packed structure is

Answer»


SOLUTION :`[FE(CN)_(6)]^(4-)` (after REARRANGEMENT)
27061.

The number of ions formed on dissolving one molecule of FeSO_4(NH_4)_2SO_46H_2O in water is:

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4
5
3
6

Answer :B
27062.

The occurrence of reaction is impossible if

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`DELTAH" is"+"ve",DELTAS" is"+"ve"`
`DeltaH" is"-"ve",DeltaS" is"-"ve"`
`DeltaH" is"-"ve",DeltaS" is"+"ve"`
`DeltaH" is"+"ve",DeltaS" is"-"ve"`

ANSWER :D
27063.

The obsrved osmotic pressure foe a 0.10 M solution of Fe (NH_(4))_(2) (SO_(4)) _(2) at 25^(@)C is 10.8 atm. The expected and experimental (observed) values of van't Hoff factor (i) will be respectively .

Answer»

5 and `4.42`
`4 and 4.00`
5 and `3.42`
`3 and 5.42`

Solution :GIVEN `PI _(AB)= 10.8` atm
`pi _(nw) =CST =0.10 xx0.0821 xx298 =2.446`
now experimental value of (i)
`= ("Observed osmotic pressure")/("Normal osmotic pressure") = (10.8)/(2.445) =4.42`
27064.

The number of ions formed in aqueous solution by the compound [Co(NH_3)_4CI_2]CI is:

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2
3
4
7

Answer :A
27065.

The occurrence of a reaction is impossible if

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`DeltaH" is "+ve, DeltaS " is ALSO + ve but "DELTAHLT TDeltaS`
`DeltaH" is- ve , "DeltaS " is also - ve but "DeltaH GT TDeltaS`
`DeltaH" is - ve ,"DeltaS " is "+ve`
`DeltaH" is + ve ,"DeltaS " is "- ve`

SOLUTION :`+ ve DeltaH and - ve DeltaS` both oppose the reaction.
27066.

The number of iodine atoms (N) present in 1 "cm"^(3) of its 0.1 M solution is

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`6.02 xx 10^(23)`
`6.02 xx 10^(2)`
`6.02 xx 10^(19)`
`1.204 xx 10^(20)`

Solution :1 cc of 0.1 M `I_(2)` SOL. `=0.1/1000=10^(-4)` mol
`N = n xx N_(A)`
`=10^(-4) xx 6.02 xx 10^(23)` molecules
`=6.02 xx 10^(19)` molecules
`=2 xx 6.02 xx 10^(19)` atoms `=1.204 xx 10^(20)`
27067.

The observed rat of acemical reaction is substantially lower than the collision frequecy. One or more of the following statements is/are true to account for this fact. A. The reactants to do not have the required energy B. The partnes do not collide in the proper orientation (C) Collision complex eists for a very short time. D. Collision frequecny over estimates the number of effective collisions.

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A,B and C
A,B and D
B,C and D
A,C and D

ANSWER :A
27068.

The number of iodine atoms present in 1cm^(3) of its 0.1 M solutions is :

Answer»

`6.02 xx 10^(23)`
`6.02xx10^(22)`
`6.02xx10^(19)`
`1.204xx10^(20)`

Solution :`1000cm^(3)` of 0.1 M solution CONTAIN
`=0.1` MOLE
`1cm^(3)` of 0.1 M solution contain `=(0.1)/(1000)`
`=1XX10^(-4)`mole
`1xx10^(-4)` mole of
`I_(2)=6.02xx10^(23)xx2xx1xx10^(-4)`
`=1.204xx10^(20)` atoms
27069.

The observed osmotic pressure of a solution of benzoic acid in benzene is less than its expected value because

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Benzene is a non-polar SOLVENT
Benzoic acid MOLECULES are assocuated in benzene
Benzoic acid molecules are dissociated in benzene
Benzoic acid is an organic COMPOUND

Answer :B
27070.

Theobserveddepressionin thefreezingpointof waterfor aperticularsolutionis 0.087 K . Calculatethe molalityof thesolutionif molaldepressionconstantfor wateris 1.86 K kg mol^(-1)

Answer»

SOLUTION :0.0467 MOL `KG^(-1)`
27071.

The number of intermediates + transition states passible during the following reaction is ________

Answer»


ANSWER :9
27072.

The observed angle of rotation of 2.0 gm sucrose containing 20 ml of aqueous solution in a polarimeter tube of 15 cm length is +15.3^(@). What is the specific rotation of this solution of sucrose ?

Answer»

`104^(@)`
`102^(@)`
`96.5^(@)`
`102.9^(@)`

SOLUTION :`102^(@)`
27073.

The number of incomplete orbitals in inner transition elements is:

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3
4
2
1

Answer :A
27074.

The observed and calculated E^0 values for M^(+2)/M are same for

Answer»

Fe
Co
Ni
Cu

Answer :A
27075.

The no.of hydroxyl ions produced by one molecule of Na_(2)CO_(3) on hydrolysis is

Answer»

2
1
3
4

Solution :`Na_(2)CO_(3)+2H_(2)Oto2NaOH+H_(2)CO_(3) and 2NAOH underset(("IONISATION"))hArr2Na^(+)+2OH^(-)`
Hence, it is clear that `2OH^(-)` ions will be formed on hydrolysis of ONE molecule of sodium CARBONATE.
27076.

The observed angle of rotation of 0.2 gm of Sucrose in 1 ml of aqueous solution in a Polarimeter tube 1 dm long is + 13.3^(@). What is the specific rotation of the solution of Sucrose ?

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`- 66.5^(@)`
`+66.5^(@)`
`+ 6.65^(@)`
1

Solution :`+ 66.5^(@)`
27077.

The observed and calculated E' values for M^(+2)//M are same for

Answer»

Fe
Co
Ni
Cu

Answer :A
27078.

The number of hydroxyl groups in pyrophosphoric acid is

Answer»

3
4
5
7

Answer :B
27079.

The observation by Louis Pasteur that crystals of certain compounds exist in the form of mirror images laid the foundation of modern stereochemistry.

Answer»


ANSWER :1
27080.

The O-O -H bond angle in H_(2)O_(2) is

Answer»

`106^(@)`
`109^(@),28`
`120^(@)`
NONE of these

Answer :D
27081.

The number of hydrogen bonds present in the sequence of a stretch of a double helical DNA 51 ATGCCTAA 3' is

Answer»

16
19
24
20

Answer :B
27082.

The number of hydrogen atoms required to convert 1 mole of nitrobenzene to hydrazobenzene is

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5
10
4
8

Answer :A
27083.

The O-O bond length in O_(3) is equal to that of

Answer»

SINGLE bond
DOUBLE bond
between single and double bond
between double and TRIPLE bond

Answer :C
27084.

The number of hydrogen atoms present in 25.6 g of sucrose (C_(12)H_(22)O_(11)) which has a molar mass of 342.3 g is:

Answer»

`22xx10^(23)`
`9.91xx10^(23)`
`11xx10^(23)`
`44xx10^(23)`

SOLUTION :Number of moles of SUCROSE
`=("MASS")/("Molar mass")=(25.6)/(342.3)`
Number of moles of hydrogen ATOM `=0.075xx22`
Number of atoms of hydrogen `=0.075xx22xx6.023xx10^(23)`
`=9.9xx10^(23)`.
27085.

The number of hydrogen atoms attached to phsophorus atom in hypophosphorus acid is :

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ZERO
TWO
one
THREE

ANSWER :C
27086.

The number of H^(+) ions present in 1 ml of a solution whose pH is 13 is :

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`6.022xx10^(10)`
`6.022xx10^(7)`
`6.022xx10^(20)`
`6.022xx10^(23)`

ANSWER :B
27087.

The number of [H] required to reduce with Sn-HCl to give aniline.

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SOLUTION :
27088.

The O-O bond length in Ozone is

Answer»

`1.33 A ^(0)`
`1.28A ^(0)`
`1.48A ^(0)`
`1.39A ^(0)`

ANSWER :B
27089.

The number of gram molecules of a substance present in unit volume is termed as

Answer»

Activity
NORMAL SOLUTION
MOLAR CONCENTRATION
Active mass

27090.

The O - O bond length in H_2O_2 is :

Answer»

`1.54 OVERSET@A`
`1.48overset@A`
`1.34 overset@A`
`1.01overset@A`

ANSWER :B
27091.

The number of grams of H_(2)SO_(4) present in 0.25 mole of H_(2)SO_(4) is:

Answer»

0.245
2.45
24.5
`49.0`

ANSWER :C
27092.

The nylon salt (Hexamethylenediammonium adipate) is

Answer»

`OVERSET(+)(N)H_3(CH_2)_4overset(+)(N)H_3overset(-)(O)OC(CH_2)_4 COO^(-)`
`overset(+)(N)H_3(CH_2)_6overset(+)(N)H_3overset(-)(O)OC (CH_2)_4COO^(-)`
`""^(-)OOC(CH_2)_6COO^(-)overset(+)(N)H_3(CH_2)_4 overset(+)(N)H_3`
`H_2N (CH_2)_6 NH_2 overset(-)(O)OC(CH_2)_4COO^(-)`

Solution :`overset(+)(N)H_3(CH_2)_6overset(+)(N)H_3overset(-)(O)OC (CH_2)_4COO^(-)`
27093.

The number of gram molecules of oxygen in6.02 xx 10^24 CO molecules is :

Answer»

10 gram molecules
5 gram molecules
1 gram molecule
0.5 gram molecule

Solution :No. of ATOMS of OXYGEN in`6.02 xx 10^24` CO molecules ` = 6.02 xx 10^24` (each CO contains 1 oxygen ATOM)
No. of oxygen molecules
` = (6.02 xx 10^24)/(2) = 3.01 xx 10^24`
Gram molecules of oxygen
` = (3.01 xx 10^24)/(6.02 xx 10^23) = 5`
27094.

The nutrient used in the body as a source of energy as a raw material for growthand repair is

Answer»

Fat
Carbohydrates
Proteins
Vitamins

Answer :C
27095.

The number of gram molecules of chlorine in6.02 xx 10^(25) hydrogen chloride molecules is

Answer»

10
100
50
5

Solution :Number of HCI MOLECULES = `6.02 xx 10^(25)` molecules
Number of moles of HCI = `(6.02 xx 10^(25))/(6.02 xx 10^(23)) = 100` moles.
`therefore` Gram MOLECULAR mass of chlorine in ` 6.023 xx 10^(25)` molecules of HCI = 100/2 = 50.
27096.

The number of gram atoms of oxygen in 0.2xx10^(24)CO molecules is

Answer»

`1`
`0.5`
`5`
`9`

Solution :`6.02xx10^(23)` molecules of CO = 1 mole of CO
`6.02xx10^(24)` MOLES of CO = 10 moles of CO
1 mole of CO CONTAIN 1 g atom of oxygen
10 moles of CO contain 10 g ATOMS of oxygen
27097.

The number of protons, electrons and neutrons in ""_(17)^(35)Cl^(-) are respectively

Answer»

17,18,18
17,17,18
17,18,17
17,18,38

Answer :A
27098.

The numerical value RT/PV for a gas at critical conditions is ………. Times of RT/PV at normal conditions:

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4
3/8
8/3
1/4

Answer :C
27099.

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

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8.314
`6.02xx10^(23)`
`0.0821`
`1.66xx10^(-19)`

SOLUTION :N/n represents the number of molecules per MILE of the gas i.e. Avogadro.s number.
27100.

The number of gram atoms of oxygen present in 0.25 g mole of (COOH)_2 2H_2Ois :

Answer»

0.125
0.5
1
1.5

Solution :1 MOLE of `(COOH)_2 . 2H_2O` CONTAINS = 6 gatom of oxygen
0.25 mole of `(COOH)_2 . 2H_2O` contains
` = 6 XX 0.25 = 1.5 ` grams atoms