Explore topic-wise InterviewSolutions in .

This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.

27401.

The number of atoms in 0.004 g of magnesium is close to (atomic mass of Mg = 24)

Answer»

24
`2 xx 10^20`
`10^20`
`6.02 xx 10^23`

Solution :24 G of `MG = 6 xx 10^23` atoms
0.004 g of Mg ` = (6 xx 10^23)/(24) xx 0.004= 1 xx 10^20`
Discussion
27402.

The number of atoms belongs to bcc unit cell is ...............................

Answer»

2
4
6
12

Answer :A
Discussion
27403.

The number of atom is an mole gas can be given by :

Answer»

`N XX "Av.no" xx "ATOMICITY"`
`(n xx Av.no)/"Atomicity"``
`(Av.no xx "Atomicity")/n`
NONE of these

Answer :A
Discussion
27404.

The number of asymmetric carbon atoms in tartaric acid is:

Answer»

ONE
TWO
four
zero

Answer :B
Discussion
27405.

The number of asymmetric carbon atoms in fructose are :

Answer»

2
3
4
5

Answer :B
Discussion
27406.

The number of asymmetric carbon atoms and the number of optical isomers inCH_(3)(CHOH)_(2)COOH are respectively :–

Answer»

3 and 4
1 and 3
2 and 4
2 and 3

Solution :`CH_(3)- underset(OH) underset(|)overset("*")CH- underset(OH) underset(| )overset("*")CH-COOH`
`2^(2)=4`
Discussion
27407.

The number of asymmetric carbon atoms in a molecule of glucose is:

Answer»

6
4
5
3

Answer :B
Discussion
27408.

The number of asymmetric C-atom created and number of possible stereoisomers when benzil (Ph CO CO Ph) is reduced with LiAlH_(4).

Answer»

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

Solution :
The product CONTAINS two similar ASYMMETRIC carbon atoms and two OPTICALLY active and moreoptically inactive meso FORM.
Discussion
27409.

The number of asymmetric carbon atom in glucose are

Answer»

2
3
4
5

Answer :C::D
Discussion
27410.

The number of asymmetric carbon atom/sin lactic acid is

Answer»

1
3
2
4

Answer :A
Discussion
27411.

The number of assymmetric carbon atoms in a molecule of glucose is

Answer»

4
5
3
6

Answer :A
Discussion
27412.

The number of antibiotics among the following is ampicillin, sulphanilamide, veronal, equanil, serotonin, luminal, seconal.

Answer»


ANSWER :1
Discussion
27413.

The number of antibonding electron pairs in O_(2)^(2-) molecular ion on the basis of molecular orbital theory is(Atomic number of O is 8)

Answer»

5
4
3
2

Solution :`O_(2)^(2-)`, no. Of antibonding electron PAIRS = 4 pairs

MOLECULAR orbital diagram of `O_(2)^(2-)`(SUPEROXIDE ion).
Discussion
27414.

The number of anti - bonding electrons present in O_(2)^(-) molecular ion is :

Answer»

8
7
5
4

Solution :`O_(2)^(-):KK(SIGMA2S)^(2)(sigma^(**)2s)^(2)(sigma2p_(Z))^(2)(pi2p_(X))^(2) (pi2p_(y))^(2)(pi^(**)2p_(x))^(2)(pi^(**)2p_(y))^(1)`
No, of electrons in antibonding orbitals = 5 (excluding inner orbitals = KK).
Discussion
27415.

The number of anions presentin bleaching powder.

Answer»


SOLUTION :
Discussion
27416.

The number of Amino acids in the two protomers of insulin are

Answer»

21
30
29
27

Solution :Insulin contains two POLYPEPTIDE chains. One CHAIN consists of 21 and ANOTHER chain consists of 30 AMINO acids.
Discussion
27417.

The number of amino acids found in proteins that a human body can synthesize is

Answer»

20
10
5 
14 

ANSWER :B
Discussion
27418.

The number of amino acids in insulin is :

Answer»

51
61
81
101

Answer :A
Discussion
27419.

The number of amino acids and numberof peptidebonds in a linear tetrapeptide (made of different amino acids) are respectively.

Answer»

4 and 4
5 and 5
5 and 4
4 and 3

Solution :In general ,
No. of PEPTIDE BONDS = No. of amino acids - 1
Discussion
27420.

The number of aluminium ions present in 54g of aluminium (atomic weight27) is

Answer»

2
18
`1.1 xx 10^(24)`
`1.2 xx 10^(24)`

SOLUTION :NUMBER of aluminium ions present in 54 G of aluminium
`= (6.203 xx 10^(23) xx 54)/(27) = 1.2 xx 10^(24)`
Discussion
27421.

The number of amino acid that human body can synthesise is

Answer»

20
10
14
18

Answer :B
Discussion
27422.

The number of alpha-particles emitted per second by 1g of radium is 3.608 xx 10^(10). Calculate decay constant and t_((1)/(2))

Answer»

SOLUTION :`1.35 XX 10^(-11)s^(-1)`
`5.13 xx 10^(10)s`
Discussion
27423.

The number of alpha beta -particles emitted in the nucleus reaction ._(92)U^(238) rarr ._(90)Th^(234) rarr ._(91)Pa^(234) are respectively

Answer»

1 and 1
1 and 2
2 and 1
2 and 2

Solution :`._(92)U^(238) rarr ._(90)Th^(234) rarr ._(91)PA^(234)`
No. of `alpha" particle" = (238 - 234)/(4) = (4)/(4) = 1`
No. of `beta " particle " = 91 - 90 = 1`
Discussion
27424.

The number of alpha particle in the nuclear reaction ""_(90)^(238) Th to ""_(81)^(212) Bi are : ____

Answer»


Solution :`""_(92) Th^(228) to ""_(83) Bi^(212) + a (""_(2)He^(4)) + b (""_(-1) beta^(0)) , a = ((228 - 212)/(4)) = (16)/(4) = 4 ALPHA , b = ?`
Discussion
27425.

The number of alpha- and beta-particles emitted when a radioactive element ._(90)E^(322) changes into ._(86)G^(220) will be

Answer»

5 and 4
2 and 3
3 and 2
4 and 1

Solution :`._(90)E^(232) rarr ._(86)G^(220)`
No. of `ALPHA " particle " = (232 - 220)/(4) = 3`
No. of `beta" particle " = 86 - [90 - 2 xx 3] = 2`
Discussion
27426.

The number of alpha and pi-bonds present in the molecule of carbolic acid are respectively

Answer»

7, 3
2, 3
4, 3
13, 3

Answer :D
Discussion
27427.

The number of alpha and beta particles emitted in the nucleur disintegration series _90^228 Th to _83^213 Bi are:

Answer»

`8alpha,1beta`
`4alpha,7beta`
`3alpha,7beta`
`4alpha,1beta`

ANSWER :A
Discussion
27428.

The number of alpha and beta particles emitted in the nuclear reaction, ._(90)Th^(298) rarr ._(83)Bi^(212) are

Answer»

`4 alpha` and `7 BETA`
`4 alpha` and `1 beta`
`8 alpha` and `1//beta`
`3 alpha` and `7 beta`

Answer :B
Discussion
27429.

The number of alpha and beta-particles emitted in the nuclear reaction ._(90)Th^(228) rarr ._(83)Bi^(212) are respectively

Answer»

4 , 1
3 , 7
8, 1
4, 7

Solution :`._(90)Th^(228)rarr ._(83)Bi^(212)`
No. of `ALPHA-"particle" = (228 - 212)/(4) = (16)//(4) = 4`
No. of `BETA-"particle" = 2xx 4 - 90 + 83`
Discussion
27430.

The number ofalpha and beta- particles emitted in the nuclear reaction Th_90^228 to Th_90^228 are respectively

Answer»

4, 1
3, 7
8, 1
4, 7

ANSWER :A
Discussion
27431.

The number of alpha and beta-particles emitted during the transformation of ._(90)Th^(232) " to " ._(82)Pb^(208) are respectively

Answer»

4, 2
2, 2
8, 6
6, 4

Solution :`._(90)Th^(232) RARR ._(82)Pb^(208) + X ._(2)He^(4) + y ._(-1)BETA^(0)`
Equating mass no.
`232 = 208 + 4x + 0 y or 4x = 24 or x = 6`
Equating atomic no
`90 = 82 + 2x - y or 90 = 82 + 2 xx 6 - y or y = 4`
Hence `6 alpha and 4 beta` particles will be emitted
Discussion
27432.

The number of alpha and beta particles emitted during the transformation of ""_(90)Th^(232) to ""_(82)Pb^(208) are respectively

Answer»

4,2
2,2
8,6
6,4

Solution :`""_(90)Th^(232)to""_(82)Pb^(208)`
Loss in weight=232-208=24.
HENCE number of `alpha`-particles emitted=`(24)/(4)=6`
Therefore,
`""_(90)Th^(232) underset(-6a)to""_(78)X^(208) underset(-4beta)to""_(82)Pb^(208)`
For increasing atomic no. by four units `4beta` particle must be emitted.
Discussion
27433.

The number of alpha and beta emitted in the nuclear reaction ""_(92)^(238)U to ""_(82)^(214)Pb is

Answer»

`8 beta`
`6 alpha`
`2 beta`
`5 alpha`

Solution :LET x `alpha- and y beta`- particle be emitted in the REACTION.
`""_(92)^(238)U to x ""_(2)^(4)alpha + y ""_(-1)^(0)beta + ""_(82)^(214)Pb`
Equating the MASS numbers on both sides,
`238 = 4x + y xx 0 + 214 implies 4x = 24 :.x= 6`
Equating atomic numbers on both sides,
`92 = 2x + (-y) + 82`
`92 = 2 xx 6 + (-y) + 82 implies - y = - 2 :. y = 2`
HENCE in the NUCLEAR reaction, `""_(92)^(238)U to ""_(82)^(214)Pb, 6 alpha`- and `2 beta` - particles are emitted.
Discussion
27434.

The number of alkene(s) which can produce 2-butanol by the successive treatment of (i) B_(2)H_(6) in tetrahydrofuran solvent and (ii) alkaline H_(2)O_(2) solution is

Answer»

1
2
3
4

Solution :STRUCTURE of butane-2-ol is:
`CH_(3)- underset(OH) underset(|) CH-CH_(2)-CH_(3)`

Now, `CH_(3)-CH= CH-CH_(3)` can have both CIS and transforms.
Discussion
27435.

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

Answer»

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

Answer :D
Discussion
27436.

The number of alkene(s) which can produces 2-butanol by the successive treatment of (i) B_2H_6 in tetrahydrofuran solvent and (ii) alkaline H_2O_2 solution is-

Answer»

1
2
3
4

Solution :
Discussion
27437.

The number of alpha- and beta- particles emitted during the transformation of ""_(90)^(232) Th to ""_(82)^(208) Pb is respectively

Answer»

2,2
4,2
6,4
8,6

Solution :`""_(90) Th^(232) to ""_(82) Pb^(208) + a (""_(2) He^(4)) + b (""_(-1) BETA^(0)) , a = ((232- 208)/(4)) = 6 ALPHA`
` 8 = (2 XX 6) -b, b = 12 - 8 beta`
Discussion
27438.

The number of alkenes (including stereoisomers) which can produce 2 - butanol by the successive treatment of (i) B_(2)H_(6) in tetrahydrofuran solvent and (ii) alkaline H_(2)O_(2) solution is/are

Answer»


ANSWER :2
Discussion
27439.

Thenumberof alkanylgroupspossiblefromC_ 4H _ 7are

Answer»

7
5
3
8

Solution :Followingfive alkenyl groupsare possiblefrom` -C _4 H _ 7 `
(i)` CH _ 3 - CH _ 2-CH =CH - `
(ii)`CH _3 -CH = CH-CH _ 2-`
(iii )` CH _ 2=CH -CH _ 2-CH _ 2- `
(iv)` CH _ 3- underset (CH _ 3) underset(|) C -CH -`
(v)` CH _3 - underset ( CH _ 2- ) underset ( |)C =CH `
Discussion
27440.

The number of alkanols and ethers represented by the molecular formulaeC_(3) H_(8) O and C_(4)H_(10) Orespectively are gives by the set :

Answer»

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

Solution :
Discussion
27441.

The number of aldol reaction (s) that occurs in the given transformation is

Answer»

1
2
3
4

Answer :C
Discussion
27442.

The number of aldol reaction(s) that occurs in the given transformation is

Answer»

4
3
2
1

Answer :B
Discussion
27443.

The number of alcohols giving iodoform test among the following is CH_(3)CH_(2)OH,CH_(3)OH,CH_(3)CH_(2)CH_(2)OH,(CH_(3))_(2)CHOH CH_(3)CH_(2)CH_(2)CH_(2)OH,CH_(3)CH_(2)CH(OH)CH_(3) CH_(3)CH(OH)CH(CH_(3))_(2),(C_(2)H_(5))_(2)CHOH,(CH_(3))_(3)COH

Answer»


SOLUTION :Alcohols conaining `CH_(3)CHOH` -GROUP give iodofor test. These are `CH_(3)CH_(2)OH,CH_(3)CH_(2)CH(OH)CH_(3),CH_(3)CH(OH),CH(CH_(3))_(2),CH_(3)CH(OH)CH_(3)`.
Discussion
27444.

The number of alcohols giving iodoform test among the following is: CH_(3) CH_(2) OH, CH_(3) OH, CH_(3) CH_(2) CH_(2) OH, (CH_(3))_(2) (CHOH), CH_(3) CH_(3) CH_(2) CH_(2) OH CH_(3) CH_(2) CH (OH)CH_(3), CH_(3) CH (OH)CH (CH_(3))_(2) (C_(2) H_(5))_(2) CHOH, (CH_(3))_(3) COH

Answer»


ANSWER :4
Discussion
27445.

The number of aldehydes of molecular formula C_5H_10O is :

Answer»

2
3
4
5

Answer :C
Discussion
27446.

The number of acidic protons in H_(3)PO_(3) are

Answer»

0
1
2
3

Solution :`H_3PO_3` is DIBASIC ACID i.e, TWO HYDROGEN ATOMS are ionisable.
Discussion
27447.

The number of alcohol isomers for molecular formula C_(4)H_(10) is _______.

Answer»

7
10
3
4

Answer :D
Discussion
27448.

The number of acidic protons in H_3PO_3 are :

Answer»

0
1
2
3

Answer :C
Discussion
27449.

The number of a-particles emitted by ""^(218)Ra in changing to stable isotope ""^(206)Pb is :

Answer»

3
4
6
2

Solution :`_(84)^(218)Rato_(82)^(206)Pb++x""_(2)^(4)He+y""_(-1)overset(0)E`
Comparing MASS NUMBERS
`218=206+4x+0`
4x=12 or x=3
Discussion
27450.

The number of a bonds in the product formed in acid catlysed reaction of acetaldehyde ----

Answer»


SOLUTION :`2CH_3- OVERSET(O ) overerset(||)C-HtoCH_3 overset(OH) overset(|)CH-CH_2 - overset(O) overset(||)C-H`
Discussion