datavorous/entrance-exam-dataset huggingface dataset:
Nitration of aniline in strong acidic medium also gives m-nitroaniline because

Options:
A In absence of substituents nitro group always goes to m-position.
B In electrophilic substitution reactions amino group is meta directive.
C Inspite of substituents nitro group always goes to only m-position.
D In acidic (strong) medium aniline is present as anilinium ion.

The correct answer is:
In acidic (strong) medium aniline is present as anilinium ion.

Solution:
As -NH3⊕ is m-directing and strongly ring deactivating, hence besides para (51%) and ortho (2%), meta product (47%) is also formed in significant yield here.

Topic: Chemistry, Amines, JEE Main
Nitration of phenyl benzoate yields the product

Options:
A

B

C

D

The correct answer is:

Solution:
It is an electrophilic aromatic substitutions reaction. Because the $\mathrm{COO}^{-}$acts as an electron withdrawiry group to one phenyl group ard a moderate electron donating group to the other pheryl group the nitration will occur on the phenyl group where $\mathrm{COO}^{-}$is acting as ars electron donating group.
The major product of the nitration will have the
$\mathrm{NO}_2$ in the para position because the or tho position is partially blocked by the large COPh grơup.

Topic: Chemistry, General Organic Chemistry, JEE Main
Nitric oxide is prepared by the action of $\mathrm{HNO}_3$ on

Options:
A $\mathrm{Fe}$
B $\mathrm{Cu}$
C $\mathrm{Zn}$
D $\mathrm{Sn}$

The correct answer is:
$\mathrm{Cu}$

Solution:
$3 \mathrm{Cu}+8 \mathrm{HNO}_3 \rightarrow 3 \mathrm{Cu}\left(\mathrm{NO}_3\right)_2+4 \mathrm{H}_2 \mathrm{O}+2 \mathrm{NO}$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Nitrobenzene can be prepared from benzene by using a mixture of conc. $\mathrm{HNO}_3$ and conc. $\mathrm{H}_2 \mathrm{SO}_4$. In the mixture, nitric acid acts as a/an -

Options:
A catalyst
B reducing agent
C acid
D base

The correct answer is:
base

Solution:
$\mathrm{HNO}_3$ on nitrating mixture acts as a base.

Topic: Chemistry, Haloalkanes and Haloarenes, NEET, NEET 2009 (Mains)
Nitrogen can be estimated by Kjeldahl's method for which of the following compound?

Options:
A
B
C
D

The correct answer is:

Solution:
Kjeldahl method is not applicable to compounds containing nitrogen in nitrogroup, Azo groups and nitrogen present in the ring (e.g Pyridine) as nitrogen of these compounds does not change to Ammonium sulphate under these conditions.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2021 (17 Mar Shift 2)
Nitrogen detection in an organic compound is carried out by Lassaigne's test. The blue colour formed corresponds to which of the following formulae?

Options:
A $\mathrm{Fe}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]_2$
B $\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3$
C $\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_2$
D $\mathrm{Fe}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3$

The correct answer is:
$\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3$

Solution:
$\underset{\text{Sodium ferrocyanide}}{3 \mathrm{Na}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]}+2 \mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3 \rightarrow \underset{\substack{\text{Ferric ferrocyanide} \ \text{(Prussian blue)} }}{\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3}+\mathrm{Na}_2 \mathrm{SO}_4$

Topic: Chemistry, General Organic Chemistry, JEE Main
Nitrogen dioxide cannot be obtained from

Options:
A Cu(NO3)2
B Hg(NO3)2
C NaNO3
D AgNO3

The correct answer is:
NaNO3

Solution:
Nitrogen dioxide cannot be obtained from NaNO3 .
CuNO32→ΔCuO+2NO2↑+12O2↑
HgNO32→ΔHg↓+2NO2↑+O2↑
AgNO3→ΔAg+NO2↑+12O2↑

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Nitrogen fixation in root nodules of Alnus is brought about by

Options:
A Azorhizobium
B Bradyhizobium
C Clostridium
D Frankia

The correct answer is:
Frankia

Solution:
Frankia is actinomycete.
Bradyrhizobium - Root nodules in soyabean.

Topic: Biology, Mineral Nutrition, NEET, NEET 2008 (Mains)
Nitrogen gas is at 300 °C temperature. The temperature (in K ) at which the rms speed of a H2 molecule would be equal to the rms speed of a nitrogen molecule, is ………. (Molar mass of N2 gas 28 g ).

Options:

The correct answer is:
41

Solution:
Given,

Temperature of Nitrogen gas is

T1=300 °C=300+273=573 K

Molar mass of H2 gas is MH2=2 g

Molar mass of N2 gas is MN2=28 g

The rms speed of gas is given by

Vrms=3RTM

Now rms speed of hydrogen gas is equal to rms speed of nitrogen gas, i.e.; VrmsH2=VrmsN2
⇒TH2TN2=MH2MN2

Substitute the given values i above equation, we get,

TH2=228×573

TH2=41 K

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2020 (05 Sep Shift 2)
Nitrogen gas is kept in an open beaker at 273 K and 1 atm pressure. If the pressure of the surrounding suddenly falls to 0.5 atm and the temperature increases to 546 K, then the percentage of niitrogen remaining in the beaker is mn % of the initial amount. Then the value of m+n is:

Options:

The correct answer is:
7.00

Solution:
Let the initial number of moles of nitrogen gas in the beaker =n1
Now, total volume of gas in the beaker will remain constant.
Then, n∝PT
Therefore,
⇒n1n2=P1P2×T1T2
⇒n1n2=10.5×546273=4
Since, the final moles of nitrogen in the beaker is one-fourth of the initial amount in 25 % of the initial amount.
Therefore, m=2,n=5 and m+n=7

Topic: Chemistry, States of Matter, JEE Main
Nitrogen gas is obtained by thermal decomposition of:

Options:
A NaNO2
B NaNO3
C BaN32
D BaNO32

The correct answer is:
BaN32

Solution:
NaNO3→∆NaNO2+12O2

(Thermally stable and does not undergo decomposition further)

BaN32→∆Ba+3N2

2BaNO32→∆2BaO+4NO2+O2

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2022 (28 Jun Shift 1)
No current flows between two charged bodies connected together when they have the same

Options:
A Capacitance or $\frac{Q}{V}$ ratio
B Charge
C Resistance
D Potential or $\frac{Q}{C}$ ratio

The correct answer is:
Potential or $\frac{Q}{C}$ ratio

Solution:
The correct option is D Potential or $\frac{Q}{C}$ ratio No current flows between two charged bodies connected together when they have the same potential. For the flow of current between two conductors, there must be a potential difference. Only then the electrons can flow from a lower potential to a higher potential.

Topic: Physics, Capacitance, JEE Main
No new follicles develop in the luteal phase of the menstrual cycle because

Options:
A follicles do not remain in the ovary after ovulation
B FSH levels are high in the luteal phase
C LH levels are high in the luteal phase
D both FSH and LH levels are low in the luteal phase

The correct answer is:
both FSH and LH levels are low in the luteal phase

Solution:
No new follicles develop in the luteal phase of menstrual cycle. It is because during this phase, Luteinising Hormone (LH) and Follicle Stimulating Hormone (FSH) levels decrease. Instead, the already ruptured follicle closes after releasing the egg and forms a corpus luteum during luteal phase, which produces progesterone.

Topic: Biology, Human Reproduction, NEET, NEET 2019 (Odisha)
No of allylic hydrogens in thermodynamic product is “X”
No of allylic hydrogens atom in kinetic product is “ Y ”
The value ofX+Y is ⋯

Options:

The correct answer is:
8

Solution:
Here, X = 1
Y = 7
Now, X+Y = 8

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
No. of atoms in one molecule of sulphur is

Options:
A 8
B 4
C 3
D None of these

The correct answer is:
8

Solution:
Sulphur molecules have 8 atoms in it because of its high catenation property.
$S_8$ :Crown Sulphur

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
No. of Î±-H atoms in the product (alkene) formed when CH33CCHOHCH3 reacts with conc.H2SO4

Options:

The correct answer is:
12

Solution:
The major product of the reaction of (CH3)3CCH(OH)CH3 with conc. H2SO4 is (CH3)2C=C(CH3)2
It is having 12 α -H-atoms.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
"No square of a real number is less than zero" is not equivalent to

Options:
A For every real number \( a, a^{2} \) is non-negative
B \( \forall a \in R, a^{2} \geq 0 \)
C Both (For every real number \( a, a^{2} \) is non-negative) or \( \left(\forall a \in R, a^{2} \geq 0\right) \)
D None of these

The correct answer is:
None of these

Solution:
Square of a real number is always non-negative so both of these statements are true.

Topic: Mathematics, Mathematical Reasoning, JEE Main
Noble gases do not react with other elements because

Options:
A None of these
B They have completely paired up and stable electron shells
C The sizes of their atoms are very small
D Are monoatomic

The correct answer is:
They have completely paired up and stable electron shells

Solution:
Noble gas has completely filled valence shell and all the electrons are paired up in each orbit.
Hence, noble gas are very stable because it cannot accept electron or denote electrons (doesn't react with other elements) due to its completely filled and paired up electronic configuration.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Non-albuminous seed is produced in:

Options:
A Maize
B Castor
C Wheat
D Pea

The correct answer is:
Pea

Solution:
Albuminous seeds have stored food in a special type of nourishing tissue, endosperm. Endosperm remains persistent till maturity (germination) in these seeds. Examples - maize, wheat, and castor.

Non-albuminous seeds (also known as exalbuminous seeds) have stored food and cotyledons. In these seeds, the endosperm is completely used during the development of the embryo. Example - pea.

Topic: Biology, Morphology of Flowering Plants, NEET, NEET 2014
Non-stoichiometric cuprous oxide, Cu2O can be prepared in laboratory. In this oxide, copper to oxygen ratio is slightly less than 2 : 1. This substance is a?

Options:
A p-type semiconductor
B n-type semiconductor
C Intrinsic semiconductor
D Insufficient information

The correct answer is:
p-type semiconductor

Solution:
The ratio of copper to oxygen is slightly less than 2 : 1. This shows that some cuprous (Cu+) ions have been replaced by cupric (Cu2+) ions. In order to maintain the electrical neutrality of the molecule, every two Cu+ ions will be replaced by one Cu2+ ion which results in creating cation vacancies leading to positive holes. Since the conduction is due to positive holes, it is a p-type semiconductor.

Topic: Chemistry, Solid State, NEET
Non-stoichiometric cuprous oxide, \(\mathrm{Cu}_2 \mathrm{O}\) can be prepared in the laboratory. In this oxide, copper to oxygen ratio is slightly less than 2 : 1. Can you account for the fact that this substance is a p-type semiconductor?

Options:

None

Solution:
The stoichiometric ratio slightly less than \(2: 1\) in \(\mathrm{Cu}_2 \mathrm{O}\) shows that some cuprous \(\left(\mathrm{Cu}^{+}\right)\)ions have been replaced by cupric \(\left(\mathrm{Cu}^{2+}\right)\) ions. In order to maintain electroneutrality, every two \(\left(\mathrm{Cu}^{+}\right)\)ions will be replaced by one \(\mathrm{Cu}^{2+}\) ion thereby, creating a hole. As the conduction will be due to the presence of these positive holes, hence it is a \(p\)-type semiconductor.

Topic: Chemistry, Solid State
Noradrenaline is a/an

Options:
A Antihistamine
B Neurotransmitter
C Antacid
D Antidepressant

The correct answer is:
Neurotransmitter

Solution:
Noradrenaline is an organic compound that functions in the brain and body as a hormone and neurotransmitter.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2019 (09 Apr Shift 2)
Normal butane convert into isobutane by

Options:
A $\mathrm{LiAlH}_4$
B $\mathrm{AlCl}_3$
C $\mathrm{NaBH}_4$
D $\mathrm{Zn} / \mathrm{HCl}$

The correct answer is:
$\mathrm{AlCl}_3$

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
n-pentane and iso pentane can be distinguished by

Options:
A $\mathrm{Br}_2$
B $\mathrm{O}_3$
C conc. $\mathrm{H}_2 \mathrm{SO}_4$
D $\mathrm{KMnO}_4$

The correct answer is:
$\mathrm{KMnO}_4$

Solution:
On oxidation, with $\mathrm{KMnO}_4$, they give different alcohols

Topic: Chemistry, Hydrocarbons, JEE Main
n-propyl bromide on treatment with ethanolic potassium hydroxide produces

Options:
A Propane
B Propene
C Propyne
D Propanol

The correct answer is:
Propene

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
Nuclear energy is released in fission since binding energy per nucleon is

Options:
A sometimes larger and sometimes smaller
B larger for fission fragments than for parent nucleus
C same for fission fragments and nucleus
D smaller for fission fragments than for parent nucleus

The correct answer is:
larger for fission fragments than for parent nucleus

Solution:
Nuclear energy is relased in fission because $\mathrm{BE} /$ nucleon is larger for fission fragments than for parent nucleus.

Topic: Physics, Nuclear Physics, JEE Main
Nuclear fusion is possible in which of the following cases?

Options:
A Only between light nuclei
B Only between heavy nuclei
C Between both light and heavy nuclei
D Only between nuclei, which are stable against decay

The correct answer is:
Only between light nuclei

Solution:
Nuclear fusion is possible between light nuclei. High temperature is essential for initiating the fusion reaction. Therefore, nuclear fusion is possible only between light nuclei and not between heavy nuclei.

Topic: Physics, Nuclear Physics, JEE Main
Nuclear pores are formed when

Options:
A both inner and outer nuclear membranes remain parallel.
B both inner and outer nuclear membranes get fused.
C Both inner and outer nuclear membranes get degenerated.
D Both inner and outer nuclear membranes break by chromosomes.

The correct answer is:
both inner and outer nuclear membranes get fused.

Solution:
Nuclear pores of various diameters are present in the nuclear membrane, through which the bidirectional movement of the RNAs and proteins occurs between the cytoplasm and nucleoplasm. The nuclear pores are formed when the outer nuclear membrane is fused with the inner nuclear membrane. The nuclear pores are enclosed by circular structures called annuli. The pore and annuli together are called pore complex or pore basket.

Topic: Biology, Cell - The Unit of Life, NEET
Nucleic acids exhibit secondary structure, justify with example.

Options:

None

Solution:
Nucleic acids are large biological molecules, essential for all known forms of life. The secondary structure of a nucleic acid molecule refers to the base pairing interactions within a single molecule or set of interacting molecules. DNA and RNA represent two main nucleic acids, their secondary structures however differ the secondary structure of DNA comprises of two complementary strands of polydeoxyribonucleotide, spirally coiled on a common axis forming a helical structure.
This double helical structure of DNA is stabilized by phosphodiester bonds (between 5' of sugar of one nucleotide and 3 sugar of another nucleotide), hydrogen bonds (between bases, and ionic interactions.

Topic: Biology, Biomolecules
Nucleopolyhedroviruses

Options:
A Cannot be used in ecologically sensitive areas
B Have negative impacts on non-target organisms
C Is the main genus of baculoviruses, that are used as biocontrol agents.
D All of these

The correct answer is:
Is the main genus of baculoviruses, that are used as biocontrol agents.

Solution:
The nuclear polyhedrosis virus, part of the family of baculoviruses, is a virus affecting insects, predominantly moths and butterflies. It has been used as a pesticide.

Topic: Biology, Microbes in Human Welfare, NEET
Nucleus A decays into B with a decay constant Î»1 and B further decays into C with a decay constant Î»2. Initially, at t=0, the number of nuclei of A and B were 3N0 and N0 respectively. If at t = t0 number of nuclei of B becomes constant and equal to 2N0, then

Options:
A t0=1λ1ln3λ12λ2
B t0=1λ1lnλ1λ2
C t0=1λ1ln2λ13λ2
D t0=1λ2ln3λ12λ2

The correct answer is:
t0=1λ1ln3λ12λ2

Solution:
NA=3N0e–λ1t

dNAdt=-3λ1N0e-λ1t
dNBdt=3λ1N0e-λ1t-λ2NB

At t=t0, dNBdt=0

⇒3λ1N0e-λ1t-λ2NB=0

⇒3λ1N0e-λ1t=λ22N0

eλ1t0=3λ12λ2

⇒t0=1λ1ln3λ12λ2

Topic: Physics, Nuclear Physics, JEE Main
Nucleus a having Z=17 and equal number of protons and neutrons has 1.2 MeV binding energy per nucleon. Another nucleus B of Z=12 has total 26 nucleons and 1.8 MeV binding energy per nucleons. The difference of binding energy of B and A will be ______ MeV.

Options:

The correct answer is:
6

Solution:
In A, Z=17 and have equal numbers of protons and neutrons i.e. nucleons=2×17=34

So, for A,

Total binding energy BEA=1.2×34=40.8 MeV

Another nucleus B mass number =26

So, total binding energy of B is BEB=1.8×26 MeV=46.8 MeV

Difference of BE=BEB-BEA=6 MeV

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2023 (01 Feb Shift 2)
Number greater than 1000 but less than 4000 is formed using the digits 0, 1, 2, 3, 4 (repetition allowed) is

Options:
A
125

B
105

C
375

D
625

The correct answer is:
375

Solution:
Required number of numbers $=3 \times 5 \times 5 \times 5=375$

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2002
Number of 4-digit numbers that are less than or equal to 2800 and either divisible by 3 or by 11 , is equal to _____ .

Options:

The correct answer is:
710

Solution:
Four-digit numbers that are less than or equal to 2800 are 1000,1001,1002,....,2800

Numbers divisible by 3 are 1002, 1005, ....,2796,2799

Let total numbers be n, then

1002+n-1 3=2799

⇒n=600

Numbers divisible by 11 are 1001, 1012, ..., 2794

Let total numbers be m, then

2794=1001+m-111

⇒m=164

Numbers divisible by 33 are

1023,1056,...,2772

Let total numbers be r, then

2772=1023+r-133

⇒r=54

Number of 4-digit numbers that are less than or equal to 2800 and either divisible by 3 or by 11 , is equal to

=600+164-54=710

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2023 (31 Jan Shift 1)
Number of 6-digit telephone numbers, which can be constructed with digits \(0,1,2,3,4,5,6,7,8,9\), if each number starts with 35 and no digit appears more than once is

Options:
A 1680
B \(8 !\)
C 6 !
D 6.6 !

The correct answer is:
1680

Solution:
Since each number consisting of 6 digits starts with 35,3 and 5 are fixed in the first and second places.
The other four places can be filled up with remaining 8 digits in \({ }^8 \mathrm{P}_4\) ways \(=8.7 .6 .5=1680\) ways.
Hence the required number of telephone numbers is 1680 .

Topic: Mathematics, Permutation Combination, VITEEE, VITEEE 2022
Number of acidic hydrogen atoms in butyne- 1 are

Options:
A 2
B 3
C 1
D 4

The correct answer is:
1

Solution:
The hydrogen atom which is attached to triple bond is acidic.

Topic: Chemistry, Hydrocarbons, JEE Main
Number of acidic hydrogen atoms in butyne-1 are

Options:
A 2
B 3
C 1
D 4

The correct answer is:
1

Solution:
The hydrogen atom which is attached to triple bond is acidic.

Topic: Chemistry, Hydrocarbons, JEE Main
Number of alkanes obtained on electrolysis of a mixture of CH3COONa and C2H5COONa is_____.

Options:

The correct answer is:
3

Solution:
An aqueous solution of sodium or potassium salt of carboxylic acid is electrolysed in this reaction, resulting in the dissociation of the salt into a carboxylate ion and sodium or potassium ions.

The total number of different alkanes are formed when the following mixture is subjected to electrolysis:

CH3COONaaq and C2H5COONaaq

In this ethane, butane and propane are formed.

CH3COONa→C∙H3

C2H5COONa→C∙2H5

2C∙2H5→CH3-CH2-CH2-CH3

2CH∙3→CH3-CH3

C∙H3+C∙2H5→CH3-CH2-CH3

So, the total number of alkanes formed here will be 3

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2024 (31 Jan Shift 1)
Number of ambidentate ligands among the following is ______
\(\mathrm{NO}_2^{-}, \mathrm{SCN}^{-}, \mathrm{C}_2 \mathrm{O}_4^{2-}, \mathrm{NH}_3, \mathrm{CN}^{-}, \mathrm{SO}_4^{2-}, \mathrm{H}_2 \mathrm{O}\).

Options:

The correct answer is:
3

Solution:
Ligands which have two different donor sites but at a time connects with only one donor site to central metal are ambidentate ligands.
Ambidentate ligands are $\mathrm{NO}_2^{-} ; \mathrm{SCN}^{-} ; \mathrm{CN}^{-}$

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2024 (09 Apr Shift 1)
Number of ambidentate ligands in a representative metal complex M(en)(SCN)4 is _________. [en=ethylenediamine]

Options:

The correct answer is:
4

Solution:
Ambidentate ligand has two or more donor atoms, but during complex formation, only one donor atom is attached to central metal. SCN− ion has donor atoms N and S.

Examples of ambidentate ligands include NO2- and SO32-. So,4 SCN- ligands are ambidentate.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2023 (06 Apr Shift 1)
Number of amine compounds from the following giving solids which are soluble in \(\mathrm{NaOH}\) upon reaction with Hinsberg's reagent is _______

Options:

The correct answer is:
5

Solution:
Primary amine give an ionic solid upon reaction with Hinsberg reagent which is soluble in $\mathrm{NaOH}$.

Topic: Chemistry, Amines, JEE Main, JEE Main 2024 (08 Apr Shift 1)
Number of amphoteric compounds among the following is

(A) BeO

(B) BaO

(C) BeOH2

(D) SrOH2

Options:

The correct answer is:
2

Solution:
Both compounds BeO and BeOH2 are amphoteric in nature.
and both compounds BaO and SrOH2 are basic in nature

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2021 (24 Feb Shift 1)
Number of atoms in $558.5$ gram Fe (at.wt. of $\mathrm{Fe}=55.85 \mathrm{~g} \mathrm{~mol}^{-1}$ ) is

Options:
A
twice that in 60 g carbon

B
$6.023 \times 10^{22}$

C
half that in $8 \mathrm{g} \mathrm{~He}$

D
$558.5 \times 6.023 \times 10^{23}$

The correct answer is:
twice that in 60 g carbon

Solution:
$\mathrm{Fe}($ no. of moles $)=\frac{558.5}{55.85}=10$ moles
$C$ (no. of moles $)=60 / 12=5$ moles.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
Number of atoms in $558.5$ gram Fe (at.wt. of $\mathrm{Fe}=55.85 \mathrm{~g} \mathrm{~mol}^{-1}$ ) is

Options:
A
twice that in 60 g carbon

B
$6.023 \times 10^{22}$

C
half that in $8 \mathrm{g} \mathrm{~He}$

D
$558.5 \times 6.023 \times 10^{23}$

The correct answer is:
twice that in 60 g carbon

Solution:
$\mathrm{Fe}($ no. of moles $)=\frac{558.5}{55.85}=10$ moles
$C$ (no. of moles $)=60 / 12=5$ moles.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2002
Number of atoms in the following samples of substances is largest in:

Options:
A
$4.0 \mathrm{~g}$ of hydrogen

B
$71.0 \mathrm{~g}$ of chlorine

C
$127.0 \mathrm{~g}$ of iodine

D
$48.0 \mathrm{~g}$ of magnesium

The correct answer is:
$4.0 \mathrm{~g}$ of hydrogen

Solution:
$4 \mathrm{~g}$ of hydrogen $=4$ mole of hydrogen $=4 \times 6.023 \times 10^{23}$ atoms
$$
\begin{aligned}
& 71.0 \mathrm{gm} \text { of chlorine }=\frac{71.0}{71.0}=1 \text { moles of } \\
& \text { chlorine } \\
& 127 \mathrm{gm} \text { of iodine }=\frac{1}{2} \text { mole of }_2 \\
& =\frac{127}{254}=6.023 \times 10^{23} \text { atoms } \\
& 48.0 \mathrm{gm} \text { of magnesium } \\
& =\frac{48.0}{24.0}=2 \times 6.023 \times 10^{23} \times \frac{1}{2} \text { atoms } \\
& \therefore 4.0 \mathrm{gm} \mathrm{H}_2 \text { has largest number of atoms. }
\end{aligned}
$$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2013 (23 Apr Online)
Number of bridging CO ligands in Mn2CO10 is

Options:

The correct answer is:
0

Solution:
Mn2CO10 structure is

Zero bridging CO ligands are present

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2021 (26 Feb Shift 1)
Number of bridging \( \mathrm{CO} \) ligands in \( \left[\mathrm{Mn}_{2}(\mathrm{CO})_{10}\right] \) is

Options:

The correct answer is:
0

Solution:
Mn2CO10 structure is

Zero bridging CO ligands are present

Topic: Chemistry, Coordination Compounds, JEE Main