datavorous/entrance-exam-dataset huggingface dataset:
Calculate standard free energy change for the reaction 2Ag+2H+→H2+2Ag+
Given: EAg+/Ago=+0.80V

Options:
A -308.08 kJ
B 154.4 kJ
C 77.2 kJ
D -154.4 kJ

The correct answer is:
154.4 kJ

Solution:
Ecello=EH2o-EAgo
=0-0.80=-0.80
ÎGÂ°=-nFEcello
=-2Ã96500Ã(-0.8)
=+154.4 kJ

Topic: Chemistry, Electrochemistry, NEET
Calculate standard potential of a cell having electrode reaction as
$\begin{aligned}
& \mathrm{Cd}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cd}_{\mathrm{s}} \mathrm{E}^{\circ}=-0.403 \mathrm{~V} \\
& \mathrm{Zn}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Zn}_{(\mathrm{s})} \mathrm{E}^{\circ}=-0.763 \mathrm{~V}
\end{aligned}$

Options:
A 0.201 V
B 0.360 V
C 0.481 V
D 1.166 V

The correct answer is:
0.360 V

Solution:
$\begin{aligned} & \mathrm{E}_{\text {cell }}^{\circ}=\mathrm{E}_{\text {Cathode }}^{\circ}-\mathrm{E}_{\text {Anode }}^{\circ} \\ & =-0.403-(-0.763) \\ & =0.763-0.403 \\ & =0.36 \mathrm{~V}\end{aligned}$

Topic: Chemistry, Electrochemistry, MHT CET, MHT CET 2022 (11 Aug Shift 1)
Calculate the atomic mass (average) of chlorine using the following data :
\(\begin{array}{lcl} & \text { % Natural Abundance } & \text { Molar Mass } \\ { }^{35} \mathrm{Cl} & 75.77 & 34.9689 \\ { }^{37} \mathrm{Cl} & 24.23 & 36.9659\end{array}\)

Options:

None

Solution:
Fractional abundance of \({ }^{35} \mathrm{Cl}=0.7577\)
Molar mass \(=34.9689\)
Fractional abundance of \({ }^{37} \mathrm{Cl}=0.2423\)
Molar mass \(=36.9659\)
Average atomic mass
\(=(0.7577)(34.9689 \mathrm{amu})+(0.2423)(36.9659 \mathrm{amu})\) \(=26.4959+8.9568=35.4527\)

Topic: Chemistry, Some Basic Concepts of Chemistry
Calculate the entropy change in surroundings when \(1.00 \mathrm{~mol}\) of \(\mathrm{H}_2 \mathrm{O}(l)\) is formed under standard conditions: \(\Delta_{\mathrm{f}} \mathrm{H}^{\circ}=-286 \mathrm{~kJ} \mathrm{~mol}^{-1}\).

Options:

None

Solution:
\(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{H}_2 \mathrm{O}(\ell) \Delta_{\mathrm{f}} \mathrm{H}^{\circ}=-286 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
This means that when \(1 \mathrm{~mol}_2 \mathrm{H}_2 \mathrm{O}(\mathrm{I})\), is formed, \(286 \mathrm{~kJ}\) of heat is released. This heat is absorbed by the surroundings, i.e., \(\mathrm{q}_{\text {surr }}=+286 \mathrm{kJmol}^{-1}\).
\(\begin{aligned}
\Delta \mathrm{S}=& \mathrm{q}_{\text {surr }} / \mathrm{T}=\frac{286 \mathrm{~kJ} \mathrm{~mol}^{-1}}{298 \mathrm{~K}} \\
&=0.9597 \mathrm{kJK}^{-1} \mathrm{~mol}^{-1}=959.7 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} .
\end{aligned}\)

Topic: Chemistry, Thermodynamics (C)
Calculate the entropy change in surroundings when 1.00 mol of H2O(l) is formed under standard conditions. (∆fH⊖= –286 kJ mol–1)

Options:
A ∆Ssurr= 959.73 J K–1
B ∆Ssurr= 910.42 J K–1
C ∆Ssurr= 456.84 J K–1
D ∆Ssurr= 494.15 J K–1

The correct answer is:
∆Ssurr= 959.73 J K–1

Solution:
Correct Option is : (A)
∆Ssurr= 959.73 J K–1

Topic: Chemistry, Thermodynamics (C), NEET
Calculate the standard cell potentials of galvanic cell in which the following reactions take place
(i) \(2 \mathrm{Cr}(s)+3 \mathrm{Cd}^{2+}(a q) \longrightarrow 2 \mathrm{Cr}^{3+}(a q)+3 \mathrm{Cd}(s)\)
(ii) \(\mathrm{Fe}^{2+}(a q)+\mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Fe}^{3+}(a q)+\mathrm{Ag}(s)\)
\(\mathrm{E}_{\mathrm{Cd}^{2+}, \mathrm{Cd}}^{\circ}=-0.40 \mathrm{~V}\),
\(\mathrm{E}_{\mathrm{Ag}^{+}, \mathrm{Ag}}^{\circ}=\mathbf{0 . 8 0 \mathrm { V } ,} \mathrm{E}_{\mathrm{Fe}^{3+}}^{\circ}, \mathrm{Fe}^{2+}=0.77 \mathrm{~V}\).
Calculate the \(\Delta_r G^{\circ}\) and equilibrium constant of the reactions.

Options:

None

Solution:
(i) \(\quad \mathrm{E}_{\text {cell }}^0=\mathrm{E}_{\text {cathode }}^{\circ}-\mathrm{E}_{\text {anode }}^{\circ}\)
\(\begin{aligned}
&=-0.40 \mathrm{~V}-(-0.74 \mathrm{~V})=+0.34 \mathrm{~V} \\
&\Delta_{\mathrm{r}} \mathrm{G}^{\circ}=-\mathrm{nFE}_{\text {cell }}^{\circ} \\
&=-6 \times 96500 \mathrm{C} \mathrm{mol}^{-1} \times 0.34 \mathrm{~V} \\
&=-196860 \mathrm{C} \mathrm{V} \mathrm{mol}^{-1} \\
&=-196860 \mathrm{~J} \mathrm{~mol}^{-1}=-196.86 \mathrm{~kJ} \mathrm{~mol}{ }^{-1} \\
&-\Delta_{\mathrm{r}} \mathrm{G}^{\circ}=2.303 \mathrm{RT} \operatorname{log~K} \\
&196860=2.303 \times 8.314 \times 298 \operatorname{log~K} \\
&\text { or } \quad \log \mathrm{K}=34.5014 \\
&\mathrm{~K}=\text { Antilog } 34.5014=3.172 \times 10^{34}
\end{aligned}\)
(ii) \(\mathrm{E}_{\text {cell }}^0=+0.80 \mathrm{~V}-0.77 \mathrm{~V}=+0.03 \mathrm{~V}\)
\(\begin{aligned}
&\Delta_r G^{\circ}=-\mathrm{nFE}_{\text {cell }}^{\circ} \\
&=-1 \times\left(96500 \mathrm{C} \mathrm{mol}^{-1}\right) \times(0.03 \mathrm{~V}) \\
&=-2895 \mathrm{C} \mathrm{V} \mathrm{mol}^{-1}=-2895 \mathrm{~J} \mathrm{~mol}^{-1} \\
&=-2.895 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
&\Delta_{\mathrm{r}} \mathrm{G}^{\circ}=-2.303 \mathrm{RT} \operatorname{log~K} \\
&-2895=-2.303 \times 8.314 \times 298 \times \operatorname{log~K} \\
&\text { or } \operatorname{log~K}=0.5074 \\
&\text { or } \mathrm{K}=\text { Antilog }(0.5074)=3.22 .
\end{aligned}\)

Topic: Chemistry, Electrochemistry
Calculate the standard enthalpy change (in kJ mol-1 ) for the reaction, H2(g)+O2(g)→H2O2(g) , given, that bond enthalpies of H–H,O=O,O–H and O–O (in kJ mol-1 ) are respectively 438, 498, 464 and 138

Options:
A -334
B -130
C +334
D +130

The correct answer is:
-130

Solution:
H2(g)+O2(g)→H2O2(g)
ΔHreaction=B.E.Reactants-B.E.products
=[B.E.(H-H)+B.E.(O=O)]-[2B.E.(O-H)+B.E.(O-O)]
=438+498–2×464+138=936–1066=-130 kJmol-1

Topic: Chemistry, Solutions, NEET
Calculate the standard enthalpy of formation of CH3OH(l) from the following data: CH3OH (l) + 3/2 O2(g) → CO2(g) + 2H2O(l) ; ∆rH⊖= –726 kJ mol–1C(graphite) + O2(g) → CO2(g) ; ∆cH⊖= –393 kJ mol–1H2(g) + 1/2 O2(g) → H2O(l) ; ∆fH⊖= –286 kJ mol–1

Options:
A –390 kJ mol–1
B –265 kJ mol–1
C –324 kJ mol–1
D –239 kJ mol–1

The correct answer is:
–239 kJ mol–1

Solution:
Correct Option is : (D)
–239 kJ mol–1

Topic: Chemistry, Thermodynamics (C), NEET
Calculate the standard enthalpy of formation of \(\mathrm{CH}_3 \mathrm{OH}(l)\) from the following data :
(i) \(\mathrm{CH}_3 \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_2\) (g) \(\longrightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(l)\); \(\Delta_{\mathrm{r}} \mathbf{H}^{\circ}=-726 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
(ii) \(\mathrm{C}\) (graphite) \(+\mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}_2(\mathrm{~g})\); \(\Delta_{\mathrm{c}} \mathbf{H}^{\circ}=-393 \mathbf{k J ~ m o l}^{-1}\)
(iii) \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2\) (g) \(\longrightarrow \mathrm{H}_2 \mathrm{O}(l)\); \(\Delta_{\mathrm{f}} \mathrm{H}^{\circ}=-286 \mathrm{~kJ} \mathrm{~mol}^{-1}\).

Options:

None

Solution:
Aim : \(\mathrm{C}(\mathrm{s})+2 \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CH}_3 \mathrm{OH}(\ell), \Delta_{\mathrm{f}} \mathrm{H}^{\circ}=\) ? Eqn. (ii) \(+2 \times\) Eqn. (iii) \(-\) Eqn. (i) gives the required eqn. with
\(\Delta \mathrm{H}=-393+2(-286)-(-726) \mathrm{kJ} \mathrm{mol}^{-1}\) \(=-239 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Topic: Chemistry, Thermodynamics (C)
Calculate the standard Gibbs energy (ΔrG) for the reaction: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s). The standard electrode potential is 1.1V.

Options:
A -100.15 kJ mol-1
B -212.27 kJ mol-1
C 55.34 kJ mol-1
D 0.85 kJ mol-1

The correct answer is:
-212.27 kJ mol-1

Solution:
The standard Gibbs energy (ΔrG) can be calculated using the equation ΔrG = -nFEo, where 'n' is the number of moles of electrons transferred, 'F' is the Faraday constant(96487 C mol-1), and 'Eo' is the standard electrode potential. In this case, 'n' is 2 (since two moles of electrons are involved in the balanced reaction), and 'Eo' is 1.1 V (as provided in the information). Therefore, ΔrG ≈ -212.27 kJmol-1.

Topic: Chemistry, Electrochemistry, NEET
Calculate the standard potential of the cell, if the standard electrode potentials of Zn2+/Zn and Ag+/Ag are - 0.763 V and + 0.799 V respectively.

Options:
A 0.036V
B 1.56V
C -1.562V
D 0.799V

The correct answer is:
1.56V

Solution:
Ecello=Ecathodeo-Eanodeo
= 0.799-(-0.763)=1.562 V

Topic: Chemistry, Electrochemistry, NEET
CAM helps the plants in:

Options:
A Conserving water
B Secondary growth
C Disease resistance
D Reproduction

The correct answer is:
Conserving water

Solution:
CAM plants are most common in arid environments. Being able to keep stomata closed during the hottest and driest part of the day reduces the loss of water through evapotranspiration.

Topic: Biology, Photosynthesis in Higher Plants, NEET
Can a body possess inertia without having weight?

Options:
A No, as weight is a fundamental property of matter.
B Yes, only in a vacuum environment.
C Yes, in conditions where the gravitational force acting on it is zero.
D No, inertia and weight are directly proportional properties.

The correct answer is:
Yes, in conditions where the gravitational force acting on it is zero.

Solution:
Correct Option is : (C)
Yes, in conditions where the gravitational force acting on it is zero.

Topic: Chemistry, Gravitation, NEET
Can a given organism occupy more than one trophic level simultaneously?

Options:
A Yes
B No
C Sometimes
D It depends on the ecosystem

The correct answer is:
Yes

Solution:
A given organism may occupy more than one trophic level simultaneously, for example, a sparrow may be a primary consumer when it eats seeds, fruits and peas and a secondary consumer when it eats insects and worms.

Topic: Biology, Ecosystem, NEET
Can competition occur between unrelated species?

Options:
A No
B Only if resources are abundant
C Yes
D Only if resources are limiting

The correct answer is:
Yes

Solution:
Competition can occur between unrelated species, where visiting flamingoes and resident fishes compete for their common food in shallow South American lakes.

Topic: Biology, Organisms and Populations, NEET
Can competition occur when resources are abundant?

Options:
A No, competition can only occur when resources are limiting.
B Yes, competition can occur even if resources are abundant.
C Competition only occurs in laboratory experiments.
D Competition occurs only when resources are limited in nature.

The correct answer is:
Yes, competition can occur even if resources are abundant.

Solution:
Resources need not be limiting for competition to occur; in interference competition, the feeding efficiency of one species might be reduced due to the interfering and inhibitory presence of the other species, even if resources (food and space) are abundant.

Topic: Biology, Organisms and Populations, NEET
Carbohydrates are the most abundant biomolecule present on earth. They are produced by

Options:
A some bacteria, algae and green plant cells.
B fungi, algae and green plant cells.
C all bacteria, fungi and algae.
D viruses, fungi and bacteria.

The correct answer is:
some bacteria, algae and green plant cells.

Solution:
Carbohydrates are organic compounds synthesized bychlorophyll containing photosynthetic cells of some bacteria, algae and green plant cells. Certain photoautotrophic bacteria e.g., Green sulphur bacteria and purple sulphur bacteria contain pigments like chlorobium chlorophyll and acteriochlorophyll respectively that helps them in photosynthesis.

Topic: Chemistry, Biomolecules, NEET
Carbohydrates, the most abundant biomolecules on earth, arc produced by

Options:
A some bacteria, algae and green plant cells
B fungi, algae and green plant cells
C all bacteria, fungi and algae
D viruses, fungi and bacteria.

The correct answer is:
some bacteria, algae and green plant cells

Solution:
Carbohydrate is the most abundant biomolecule on earth. Carbohydrate is organic compound synthesized in the chlorophyll-containing cells of some bacteria, algae and green plants, during photosynthesis. Certain photoautotrophic bacteria, e.g. Green sulphur bacteria and purple sulphur bacteria contain pigments like Chlorophyll and Bacteriochlorophyll, respectively that helps them in photosynthesis. During photosynthesis, carbon dioxide reduced into carbohydrates by water and oxygen is liberated.

Topic: Biology, Biomolecules (B), NEET, NEET 2005
Carbohydrates, the most abundant biomolecules on earth, are produced by:

Options:
A all bacteria, fungi and algae
B fungi, algae and green plant cells
C some bacteria, algae and green plant cells
D viruses, fungi and bacteria

The correct answer is:
some bacteria, algae and green plant cells

Solution:
Some photosynthetic bacteria such as Rhodopseudomonas can prepare carbohydrates. But during this type of food synthesis $\mathrm{O}_2$ does not evolved because in this case hydrogen donor is other than $\mathrm{H}_2 \mathrm{O}$. Algae (green and blue green) and all green plant cells prepare their food (carbohydrate) through photosynthesis.
Here, hydrogen ions are donated by water molecules by the process of photolysis of water i.e., $\mathrm{O}_2$ is released during this type of food synthesis.

Topic: Biology, Biomolecules (B), JIPMER, JIPMER 2007
Carbylamine reaction is given by aliphatic

Options:
A primary amine
B secondary amine
C tertiary amine
D quaternary ammonium salt

The correct answer is:
primary amine

Solution:
Carbylamine test is given by aliphatic primary amine.
Aliphatic (or aromatic) $1^{\circ}$ amine $+\mathrm{CHCl}_{3}+$ $\mathrm{KOH} \longrightarrow$ Bad smelling isocyanide.

Topic: Chemistry, Amines, VITEEE, VITEEE 2007
Catalytic dehydrogenation of a primary alcohol gives a

Options:
A secondary alcohol
B aldehyde
C ketone
D ester

The correct answer is:
aldehyde

Solution:
Primary alcohols get dehydrogenated with reduced copper at $573 \mathrm{~K}$, to give corresponding aldehydes.

Topic: Chemistry, Alcohols Phenols and Ethers, KCET, KCET 2008
Catalytic dehydrogenation of a primary alcohol gives a/an

Options:
A secondary alcohol
B aldehyde
C ketone
D ester

The correct answer is:
aldehyde

Solution:
Catalytic dehydration of a primary alcohol gives aldehyde i.e.,

Topic: Chemistry, Aldehydes and Ketones, COMEDK, COMEDK 2017
Cell is basic unit of life. Discuss in brief.

Options:

None

Solution:
Cell : The Basic Unit of life : All the living organism are composed of small, tiny structure or compartments called cells. These cells are called the 'building blocks' of life. The cells in true sense are considered as the basic unit of life because all the life processes i.e., metabolism, responsiveness, reproduction are carried out by the cells. Respiration, nutrition, release of energy for the body are carried out within the cells only. Even the animals and plants reproduce because the cells reproduce individually. Growth occurs because cell grow and multiply.
In Amoeba all the life processes are performed within the boundaries of the single cell. This is true of all other multicellular organisms. The only difference in the multicellular organisms is that the body of these organisms is made up of many cells. In these organisms, the cell do not behave independently, but get organized into tissues. Each tissue is specialized to perform specific functions. Different tissues then get organised into tissues. Each tissue is specialized to perform specific functions. Different tissues then get organised into organs which perform certain specific functions. Different organs are finally organised to form organ systems. Now it must be very clear that the basic structure to tissues, organs and organ system are the cells only. These tissues, organs and organ system of the organisms work because the cells work. Thus "the cells are structural and functional unit of the living beings" hence it is the basic unit of life.

Topic: Biology, Cell - The Unit of Life
Cell wall of fungi is composed of

Options:
A chitin
B pectin
C cellulose
D mannans

The correct answer is:
chitin

Solution:
The cell wall of fungi is composed of chitin. It is the second most abundant carbohydrate. It is a homopolymer of N -Acetyl Glucosamine (NAG) joined with $\beta$ 1-4 linkages. NAG is a modification of glucose molecule.

Topic: Biology, Biological Classification, JIPMER, JIPMER 2018
Cellulose does not form blue colour with Iodine because

Options:
A It breaks down when iodine reacts with it.
B It is a disaccharide.
C It is a helical molecule.
D It does not contain complex helices and hence cannot hold iodine molecules.

The correct answer is:
It does not contain complex helices and hence cannot hold iodine molecules.

Solution:
Cellulose is a polymeric polysaccharide consisting of only one type of monosaccharide, i.e., glucose. Cellulose is a homopolymer. Starch is a variant of this but is present as a storehouse of energy in plant tissues. Starch forms helical secondary structures, and it can hold iodine molecules in the helical portion. The starch-iodine is blue. Cellulose does not contain complex helices and hence cannot hold iodine.

Topic: Biology, Biomolecules (B), NEET, NEET 2023 (All India)
Cellulose, the most important constituent of the plant cell wall is made up of

Options:
A Branched-chain of glucose molecules linked by Î²-1, 4 glycosidic bonds in a straight chain and Î±-1, 6 glycosidic bonds at the site of branching
B Unbranched chain of glucose molecules linked by Î²-1, 4 glycosidic bond
C Branched-chain of glucose molecules linked by Î±-1, 6 glycosidic bonds at the site of branching
D Unbranched chain of glucose molecules linked by α-1, 4 glycosidic bond

The correct answer is:
Unbranched chain of glucose molecules linked by β-1, 4 glycosidic bond

Solution:
Cellulose is the most abundant carbohydrate. Cellulose molecule is composed of 1600 to 6000 glucose molecules joined together. Those polymers form long twisting macromolecules of cellulose. The chains are unbranched and linear. The successive glucose residues are joined together by β-1-4, linkages.

Topic: Chemistry, Biomolecules, NEET
Certainly, here's a single correct question with four options: Question: What is the meaning of the statement 'electric charge of a body is quantised'?

Options:
A Electric charge can take on any continuous value.
B Electric charge is a fundamental constant.
C Electric charge is only found in whole number multiples of elementary charge.
D Electric charge is proportional to the size of the body.

The correct answer is:
Electric charge is only found in whole number multiples of elementary charge.

Solution:
Correct Option is : (C)
Electric charge is only found in whole number multiples of elementary charge.

Topic: Chemistry, Electrostatics, NEET
$\left(\mathrm{CH}_3\right)_3 \mathrm{COCH}_3$ and $\mathrm{CH}_3 \mathrm{OC}_2 \mathrm{H}_5$ are treated with hydroiodic acid. The fragments obtained after reactions are respectively

Options:
A $\left(\mathrm{CH}_3\right)_3 \mathrm{CI}+\mathrm{CH}_3 \mathrm{OH} ; \mathrm{CH}_3 \mathrm{I}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$
B $\left(\mathrm{CH}_3\right)_3 \mathrm{CI}+\mathrm{CH}_3 \mathrm{OH} ; \mathrm{CH}_3 \mathrm{OH}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{I}$
C $\left(\mathrm{CH}_3\right)_3 \mathrm{COH}+\mathrm{CH}_3 \mathrm{I} ; \mathrm{CH}_3 \mathrm{OH}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{I}$
D $\mathrm{CH}_3 \mathrm{I}+\left(\mathrm{CH}_3\right)_3 \mathrm{COH} ; \mathrm{CH}_3 \mathrm{I}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$

The correct answer is:
$\left(\mathrm{CH}_3\right)_3 \mathrm{CI}+\mathrm{CH}_3 \mathrm{OH} ; \mathrm{CH}_3 \mathrm{I}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$

Solution:
When mixed ethersare used, the formation of alkyl iodide depends on the nature of alkyl groups. Methyl iodide is formed when one group is methyl and the other a primary or secondary alkyl group. Here reaction follows $\mathrm{S}_{\mathrm{N}} 2$ mechanism and because of the steric effect of the larger group, $\mathrm{I}^{-}$attacks the smaller (Me) group.
$\mathrm{CH}_3 \mathrm{OC}_2 \mathrm{H}_5+\mathrm{HI} \longrightarrow \mathrm{CH}_3 \mathrm{I}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$
When the substrate is a methyl $t$-alkyl ether, the products are $t-R \mathrm{I}$ and $\mathrm{MeOH}$. Here reaction follows $\mathrm{S}_{\mathrm{N}} 1$ mechanism and formation of products is controlled by the stability of carbocation. Since carbocation stability order is $3^{\circ}>2^{\circ}>1^{\circ}>\stackrel{+}{\mathrm{C}} \mathrm{H}_3$, therefore alkyl halide is always derived from tert-alkyl group.

Topic: Chemistry, Alcohols Phenols and Ethers, AIIMS, AIIMS 2017
Chargaff's rule states that in an organism

Options:
A amount of adenine (A) is equal to that of thymine $(\mathrm{T})$ and the amount of guanine $(\mathrm{G})$ is equal to that of cytosine (C)
B amount of adenine (A) is equal to that of guanine (G) and the amount of thymine $(T)$ is equal to that of cytosine (C)
C amount of adenine (A) is equal to that of cytosine $(\mathrm{C})$ and the amount of thymine $(T)$ is equal to that of guanine (G)
D amounts of all bases are equal

The correct answer is:
amount of adenine (A) is equal to that of thymine $(\mathrm{T})$ and the amount of guanine $(\mathrm{G})$ is equal to that of cytosine (C)

Solution:
Chargaff rule states that in DNA there is always equality in quantity between the bases $\mathrm{A}$ and $\mathrm{T}$ and between the bases $\mathrm{G}$ and $\mathrm{C}$.
Related Theory
In all the species the molar ratios $[A]=[T]$ and $[C]=[G]$, and that ratio $[C+G] /[A+T]$ was typically less than the unity with $[C+G]$ is less abundant.

Topic: Chemistry, Biomolecules, NEET, NEET 2003
Options:
A
B
C CH3-CH=CH-CH3
D

The correct answer is:

Solution:
The reaction between 3-methylbutan-2-ol and HBr is explained below.
In the first step, −OH group is protonated. In the second step, a molecule of water is lost to form secondary carbocation. Third step is the rearrangement of less stable secondary carbocation to more stable tertiary carbocation through 1,2-hydride ion shift. Final step is the nucleophilic attack of bromide ion on tertiary carbocation to form 2-bromo-2-methylbutane as shown below.

Topic: Chemistry, Alcohols Phenols and Ethers, NEET, NEET 2023 (All India)
China rose has

Options:
A Parietal placentation, twisted aestivation, monoadelphous, perigynous flower.
B Axile placentation, twisted aestivation, monoadelphous, hypogynous flower.
C Parietal placentation, valvate aestivation, diadelphous, perigynous flower.
D Axile placentation, valvate aestivation, diadelphous, hypogynous flower.

The correct answer is:
Axile placentation, twisted aestivation, monoadelphous, hypogynous flower.

Solution:
When the placenta is axial and the ovules are attached to it in a multilocular ovary, the placentation is said to be axile, as in china rose, tomato, and lemon. If one margin of the appendage overlaps that of the next one and so on as in china rose, lady's finger and cotton, it is called twisted aestivation. In the hypogynous flower the gynoecium occupies the highest position while the other parts are situated below it. The ovary in such flowers is said to be superior, e.g., mustard, china rose and brinjal. The stamens are united into one bunch or one bundle (monadelphous) as seen in in china rose.

Topic: Biology, Morphology of Flowering Plants, NEET
Chlorenchyma is known to develop in the

Options:
A cytoplasm of Chlorella
B mycelium of a green mould such as Aspergillus
C spore capsule of a moss
D pollen tube of Pinus.

The correct answer is:
spore capsule of a moss

Solution:
Chlorenchyma or assimilatory parenchyma are parenchymatous cells that possess abundant
chloroplasts in them. They are capable of photosynthesis. A spore capsule of moss can perform photosynthesis because of the presence of chlorenchyma cells in them.

Topic: Biology, Anatomy of Flowering Plants, AIIMS, AIIMS 2008
Chlorine has two naturally occurring isotopes, ${ }^{35} \mathrm{Cl}$ and ${ }^{37} \mathrm{Cl}$. If the atomic mass of $\mathrm{Cl}$ is $35.45$ the ratio of natural abundance of ${ }^{35} \mathrm{Cl}$ and ${ }^{37} \mathrm{Cl}$ is closest to

Options:
A $3.5: 1$
B $3: 1$
C $2.5: 1$
D $4: 1$

The correct answer is:
$3: 1$

Solution:
Mav $=\frac{\mathrm{M}_{1} \mathrm{n}_{1}+\mathrm{M}_{2} \mathrm{n}_{2}}{\mathrm{n}_{1}+\mathrm{n}_{2}}$
$35.45=\frac{35 \mathrm{n}_{1}+37 \mathrm{n}_{2}}{\mathrm{n}_{1}+\mathrm{n}_{2}}$
$\therefore \mathrm{n}_{1}: \mathrm{n}_{2}=3: 1$

Topic: Chemistry, Structure of Atom, KVPY, KVPY 2017 (5 Nov SB/SX)