datavorous/entrance-exam-dataset huggingface dataset:
Evaluate: \(\int x \tan ^{-1} x d x\)

Options:
A \(\frac{1}{2}\left(x^2+1\right) \tan ^{-1} x-\frac{1}{2} x+c\)
B \(\frac{1}{2}\left(x^2+1\right) \tan ^{-1} x+\frac{1}{2} x+c\)
C \(\frac{1}{2}\left(\mathrm{x}^2-1\right) \tan ^{-1} \mathrm{x}-\frac{1}{2} \mathrm{x}+\mathrm{c}\)
D None of these

The correct answer is:
\(\frac{1}{2}\left(x^2+1\right) \tan ^{-1} x-\frac{1}{2} x+c\)

Solution:
\(\begin{aligned}
& \int x \tan ^{-1} x d x=\frac{x^2}{2} \tan ^{-1} x-\frac{1}{2} \int \frac{x^2}{1+x^2} d x \\
& =\frac{x^2}{2} \tan ^{-1} x-\frac{1}{2} \int\left[1-\frac{1}{1+x^2}\right] d x \\
& =\frac{1}{2}\left(x^2+1\right) \tan ^{-1} x-\frac{1}{2} x+c
\end{aligned}\)

Topic: Mathematics, Definite Integration, JEE Main
Evaluate: \( \int\left\{\frac{(\log x-1)}{1+(\log x)^{2}}\right\}^{2} d x \)

Options:
A \( \frac{x}{x^{2}+2}+C \)
B \( \frac{\log x}{(\log x)^{2}+1}+C \)
C \( \frac{x}{(\log x)^{2}+1}+C \)
D \( \frac{x e^{x}}{1+x^{2}}+C \)

The correct answer is:
\( \frac{x}{(\log x)^{2}+1}+C \)

Solution:
Method1: by cross checking the options

Consider fx=xlogx2+1

∴ f'x=1+log x2-2 xlog xx1+log x22
∴ f'x=1+log x2-2log x1+log2x2=log x-11+log x22
∴ ∫log x-121+log x2dx=∫f'xdx=fx+C
∴ ∫logx-11+log x22dx=x1+log x2+C
Hence option 3 is the correct answer and we can check the other choices by the similar argument.

Alternate solution

∫log x-11+log x22dx
Put logx=t⇒x=et⇒dx=etdt
=∫ett-12t2+12dt
∫ett2+1-2tt2+12dt
=∫et1t2+1+-2tt2+12dt
=ett2+1+C ∵ ∫exfx+f'xdx=exfx+C
=xlog x2+1

Topic: Mathematics, Indefinite Integration, JEE Main
Evaluate: \( \int_{-\pi}^{199 \pi} \sqrt{\left(\frac{1-\cos 2 x}{2}\right)} d x \)

Options:
A \( 200 \)
B \( 400 \)
C \( -200 \)
D \( -400 \)

The correct answer is:
\( 400 \)

Solution:
Let, I=∫-π199π1-cos2x 2 dx
=∫-π199πsinxdx
=199--1∫0πsinxdx
(∵sinx is periodic with period π and ∫mTnTfxdx=n-m∫0Tfxdx if T is the period of the function fx ).
=200 ∫0πsinxdx
=200-cosx0π
=2001--1=400.

Topic: Mathematics, Definite Integration, JEE Main
Evaluate the correct matching order of the following terms.

i	Narcotic drugs	a	Artificial sweetner

ii.	Saponification	b	Heroin

iii	Alitame	c	Hydrolysis of fats and oils

Options:
A (i)-(b),(ii)-(c),(iii)-(a)
B (i)-(a),(ii)-(c),(iii)-(b)
C (i)-(b),(ii)-(a),(iii)-(c)
D None of these

The correct answer is:
(i)-(b),(ii)-(c),(iii)-(a)

Solution:
(i) Narcotic drugs: Certain narcotics are also used as analgesics. For example, morphine and its derivatives codeine, heroin, marijuana are used in server pain as analgesics. These are known to be habit-forming.

(ii) Saponification: When oil and fats are heated with a solution of sodium hydroxide, they break down to sodium salt of the respective fatty acid and glycerol. This process of making soap by the hydrolysis of fats and oils with alkalies is called saponification.

(iii) Alitame: It is more stable than aspartame during cooking. It is high potency sweetener and one main difficulty with alitame and similar type of high potency sweetener is the difficulty in controlling the sweeteners of the food.

Topic: Chemistry, Chemistry in Everyday Life, NEET
Evaluate the following statements for their correctness.

A. The elevation in boiling point temperature of water will be same for 0.1 M NaCl and 0.1 M urea.
B. Azeotropic mixture boil without change in their composition.
C. Osmosis always takes place from hypertonic to hypotonic solution.
D. The density of 32% H2SO4 solution having molarity 4.09 M is approximately 1.26 g mL–1.
E. A negatively charged sol is obtained when KI solution is added to silver nitrate solution.

Choose the correct answer from the options given below:

Options:
A B and D only
B B, D and E only
C A and C only
D A, B and C only

The correct answer is:
B and D only

Solution:
(A) Elevation in boiling point (∆Tb) temperature of the water is not same, as ‘i’ value is different for two solutions.
(B) Azeotropic mixture: Liquid mixtures that distill over without changes in composition are called constant boiling mixtures or Azeotropes or Azeotropic mixtures.
(C) Osmosis always takes place from hypotonic (low concentration of solute) solution to hypertonic (high concentration of solute) solution.
(D) Let the mass of H2SO4 (32%) is 100.

∴ Weight of H2SO4 = 32

Moles of H2SO4 = 3298

Now, 4.09 = 3298 × V

⇒ V = 79 mL

Density = 10079 = 1.265
(E) A negative charged sol is obtained when KI solution is added to silver nitrate solution whereas positive charged sol is obtained when AgNO3 solution is added to KI solution.

So, the correct answer is B and D only.

Topic: Chemistry, Solutions, JEE Main, JEE Main 2023 (31 Jan Shift 2)
Evaluate the following statements related to group 14 elements for their correctness.
(A) Covalent radius decreases down the group from \(\mathrm{C}\) to \(\mathrm{Pb}\) in a regular manner.
(B) Electronegativity decreases from \(\mathrm{C}\) to \(\mathrm{Pb}\) down the group gradually.
(C) Maximum covalance of \(\mathrm{C}\) is 4 whereas other elements can expand their covalance due to presence of d orbitals.
(D) Heavier elements do not form \(\mathrm{p} \pi-\mathrm{p} \pi\) bonds.
(E) Carbon can exhibit negative oxidation states.
Choose the correct answer from the options given below :

Options:
A (A), (B) and (C) Only
B (C), (D) and (E) Only
C (C) and (D) Only
D (A) and (B) Only

The correct answer is:
(C), (D) and (E) Only

Solution:
(A) Down the group; radius increases
(B) EN does not decrease gradually from $\mathrm{C}$ to $\mathrm{Pb}$.
(C) Correct.
(D) Correct.
(E) Range of oxidation state of carbon ; -4 to +4

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main, JEE Main 2024 (06 Apr Shift 2)
Even Carnot engine cannot give 100% efficiency because we cannot

Options:
A
prevent radiation

B
find ideal sources

C
reach absolute zero temperature

D
eliminate friction

The correct answer is:
reach absolute zero temperature

Solution:
Absolute zero temperature is practically not reachable

Topic: Physics, Thermodynamics, JEE Main
Even Carnot engine cannot give 100% efficiency because we cannot

Options:
A
prevent radiation

B
find ideal sources

C
reach absolute zero temperature

D
eliminate friction

The correct answer is:
reach absolute zero temperature

Solution:
Absolute zero temperature is practically not reachable

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2002
Even in absence of pollinating agents seed-setting is assured in

Options:
A Commellina
B Zostera
C Salvia
D fig

The correct answer is:
Commellina

Solution:
Commellina benghalensis or kankauoa bears aerial, chasmogamous (stigma and anthers exposed to pollinating agents), insect pollinated flowers and underground cleistogamous flowers. Cleistogamous flowers are bisexual flowers which never open, i.e., always remain closed. In such flowers the anthers and stigma lie very close to each other. When anthers dehisce in the flower buds, pollen grains come in contact with the stigma of the same flower, i.e., autogamy occurs. So, these flowers produce assured seed set, even in the absence of pollinators.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET, NEET 2012 (Screening)
Events $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are mutually exclusive events such that $\mathrm{P}(\mathrm{A})=\frac{3 \mathrm{x}+1}{3}, \mathrm{P}(\mathrm{B})=\frac{\mathrm{x}-1}{4}$ and $\mathrm{P}(\mathrm{C})=\frac{1-2 \mathrm{x}}{4}$. The set of possible values of $\mathrm{x}$ are in the interval.

Options:
A
$[0,1]$

B
$\left[\frac{1}{3}, \frac{1}{2}\right]$

C
$\left[\frac{1}{3}, \frac{2}{3}\right]$

D
$\left[\frac{1}{3}, \frac{13}{3}\right]$

The correct answer is:
$\left[\frac{1}{3}, \frac{13}{3}\right]$

Solution:
$\mathrm{P}(\mathrm{A})=\frac{3 \mathrm{x}+1}{3}, \mathrm{P}(\mathrm{B})=\frac{1-\mathrm{x}}{4}, \mathrm{P}(\mathrm{C})=\frac{1-2 \mathrm{x}}{2}$
These are mutually exclusive
$0 \leq \frac{3 x+1}{3} \leq 1, \quad 0 \leq \frac{1-x}{4} \leq 1$ and $0 \leq \frac{1-2 x}{2} \leq 1$
$-1 \leq 3 x \leq 2,-3 \leq x \leq 1$ and $-1 \leq 2 x \leq 1$
$-\frac{1}{3} \leq x \leq \frac{2}{3},-3 \leq x \leq 1$, and $-\frac{1}{2} \leq x \leq \frac{1}{2}$
Also $0 \leq \frac{1+3 x}{3}+\frac{1-x}{4}+\frac{1-2 x}{2} \leq 1$
$0 \leq 13-3 x \leq 12 \Rightarrow 1 \leq 3 x \leq 13 \Rightarrow \frac{1}{3} \leq x \leq \frac{13}{3}$
$\max \left\{-\frac{1}{3},-3,-\frac{1}{2}, \frac{1}{3}\right\} \leq \mathrm{x} \leq \min \left\{\frac{2}{3}, 1, \frac{1}{2}, \frac{13}{3}\right\}$
$\frac{1}{3} \leq x \leq \frac{1}{2} \Rightarrow x \in\left[\frac{1}{3}, \frac{1}{2}\right]$

Topic: Mathematics, Probability, JEE Main, JEE Main 2003
Every atom makes one free electron in copper. If 1.1 A Current is flowing in the wire of copper having 1 mm diameter, then the drift velocity (approx.) will be (density of copper =9Ã103 kg m-3 and atomic weight of copper = 63)

Options:
A 0.1 mm s-1
B 0.2 mm s-1
C 0.3 mm s-1
D 0.2 mm s-1

The correct answer is:
0.1 mm s-1

Solution:
vâ¡dâ¡=iâ¡nâ¡Ae;
where nâ¡=NAÏMâ¡
=6.023Ã1026Ã9Ã10363=0.860Ã1029
=8.6Ã1028
and A=ÏD24=227Ã10-324 m2;
=1114Ã10-6 m2
vd=1.18.6Ã1028Ã1114Ã10-6Ã1.6 Ã10-19
=19.6Ã10+3=100Ã10-496=1.0Ã10-4 m s-1
=0.1 mm s-1

Topic: Physics, Current Electricity, JEE Main
Every chromosome essentially has a primary constriction or the ________, on the sides of which disc shaped structures called _____ are present

Options:
A centrosome, centromere
B chromocenter, centromere
C centromere, kinetochores
D centromere, centromere

The correct answer is:
centromere, kinetochores

Solution:
Each chromosome consists of two chromatids that remain attached at a point called centromere or primary constriction. At centromere two disc-shaped structures are presently known as kinetochore. During cell division spindle fibers attaches to the kinetochore.

Topic: Biology, Cell - The Unit of Life, NEET
Every planet revolves around the sun in an elliptical orbit :

A. The force acting on a planet is inversely proportional to square of distance from sun.

B. Force acting on planet is inversely proportional to product of the masses of the planet and the sun.

C. The centripetal force acting on the planet is directed away from the sun.

D. The square of time period of revolution of planet around sun is directly proportional to cube of semi-major axis of elliptical orbit.

Choose the correct answer from the options given below :

Options:
A A and D only
B C and D only
C B and C only
D A and C only

The correct answer is:
A and D only

Solution:
Newton's law of gravitation state that gravitational force between two point masses is directly proportional to the product of their masses m1 and m2 and inversely proportional to the square of the distance r between them. It is stated as F=Gm1m2r2

A planet moves around the sun in a circular path for which the gravitational force of attraction on the planet by the sun provides the necessary centripetal force. This centripetal force is always directed towards the centre of the sun at each point of its path which is responsible for circular motion of planet.

From Kepler's Third Law, we know that the squares of the orbital periods of the planets are directly proportional to the cubes of the semi-major axes of their orbits.

⇒T2∝a3

Thus, A and D only are correct.

Topic: Physics, Gravitation, JEE Main, JEE Main 2023 (25 Jan Shift 2)
"Evil Quartet" is related to

Options:
A causes of biodiversity loss.
B importance of species diversity to ecosystem.
C species area relationship.
D nutrient recycling.

The correct answer is:
causes of biodiversity loss.

Solution:
There are four main reason or causes of biodiversity loss which are referred as "Evil quartet". It is habitat loss and fragmentation, over- exploitation, alien species invasion and co-exitinction.

Topic: Biology, Biodiversity and its Conservation, NEET
Evolutionary history of an organism is known as:

Options:
A ontogeny
B phylogeny
C ancestry
D paleontology

The correct answer is:
phylogeny

Solution:
Phylogeny pertains to the evolutionary history of a taxonomic group of organisms. Thus, phylogenetics is mainly concerned with the relationships of an organism to other organisms according to evolutionary similarities and differences. Ontogeny is the development of a single individual, or a system within the individual, from the fertilized egg to maturation and death. Ancestry represents persons initiating or comprising a line of descent. Palaeontology is the study of the history of life on Earth as based on fossils.
Related Theory
The biogenelic law was proposed by Ernst Haeckel which states that each embryo's developmental stage represents an adult form of an evolutionary ancestor i.e., an organism, in the course of its development, goes through all the stages of those forms oflife from which it has evolved.

Topic: Biology, Evolution, NEET, NEET 2006
Examine the figure below and select the right option giving all the four parts (A, B, C and (D) correctly identified.
\(\begin{array}{|l|l|}
\hline & \text{A} & \text{B} & \text{C} & \text{D} \\
\hline \text{(a)} & \text{Archegoniophore} & \text{Female thallus} & \text{Gemma
cup} & \text{Rhizoids} \\
\hline \text{(b)} & \text{Archegoniophore} & \text{Female thallus} & \text{Bud} & \text{Foot} \\
\hline \text{(c)} & \text{Seta} & \text{Sporophyte} & \text{Protonema} & \text{Rhizoids} \\
\hline \text{(d)} & \text{Antheridiophore} & \text{Male thallus} & \text{Globule} & \text{Roots} \\
\hline
\end{array}\)

Options:
A (a)
B (b)
C (c)
D (d)

The correct answer is:
(a)

Solution:
Marchantia, a liverwort, is a dioecious. The female thallus bears female sex organs called archegonia on special branches called archegoniophores. The plants body is anchored by unicellular rhizoids.

Topic: Biology, Plant Kingdom, NEET, NEET 2011 (Mains)
Examine the figures $A, B, C$ and D. In which one of the four options, all the items A, B, C and D are correct?
(A)
(B)
(C)
(D)

Options:
A (a)
B (b)
C (c)
D (d)

The correct answer is:
(c)

Solution:
A-Selaginella
B-Equisetum
C-Salvinia
D-Ginkgo

Topic: Biology, Plant Kingdom, NEET, NEET 2010 (Mains)
Examine the figures (A-D) given below and select the right OPTION out of a-d, in which all the four structures A, B, C and D are identified correctly.

Options:
A (a)
B (b)
C (c)
D (d)

The correct answer is:
(c)

Solution:
A-Offset of Eichhornia B-Antheridiophore in Marchantia C-Antipodals in embryo sac D-Oogonium (nucule) in Chara.

Topic: Biology, Plant Kingdom, NEET, NEET 2010 (Mains)
Examine the following two structures for the anilinium ion and choose the correct statement from the ones given below

Options:
A II is not an acceptable canonical structure, because carbonium ions are less stable than ammonium ions
B II is not an acceptable canonical structure, because it is non-aromatic
C II is not an acceptable canonical structure, because the nitrogen has 10 valence electrons
D II is an acceptable canonical structure

The correct answer is:
II is not an acceptable canonical structure, because the nitrogen has 10 valence electrons

Solution:
II is not acceptable canonical form, because the nitrogen atom has 10 valence electron. Since N does not have d-orbitals in its valence shell, therefore, its octet cannot be expanded.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Examine the food web for a particular terrestrial ecosystem. Which species is autotrophic and decomposer respectively?

Options:
A A and E
B B and E
C C and A
D E and A

The correct answer is:
A and E

Solution:
A is producer as it is source of food and energy for all organism in the ecosystem, Dead remains or detritus of all organism is degraded by decomposer and which causes recycling of nutrients.

Topic: Biology, Ecosystem, NEET
Examine the structural formulas of the following compounds and find how many compounds will produce \( \mathrm{CO}_{2} \) on oxidative ozonolysis:
\( \mathrm{H}_{3} \mathrm{C}-\mathrm{C} \equiv \mathrm{CH}, \mathrm{Ph}-\mathrm{CH} \equiv \mathrm{CH}_{2} \),

Options:

The correct answer is:
5

Solution:
Ozonolysis of alkene and alkyne gives different compounds based on reductive ozonolysis and oxidative ozonolysis.

Oxidative ozonolysis of non-terminal alkenes gives carboxylic acid.

Oxidative ozonolysis of non-terminal alkenes gives ketone.

Oxidative ozonolysis of terminal alkyne gives carboxylic acid.

Similarly, oxidative cleavage of terminal alkyne gives carboxylic acid and carbon dioxide.

.

Topic: Chemistry, Hydrocarbons, JEE Main
Example of a three-dimensional silicate is :

Options:
A Ultramarines
B Feldspars
C Zeolites
D All of these

The correct answer is:
All of these

Solution:
Three-dimensional silicates are formed by sharing all four oxygen atom of SiO44-. Ultramarines, feldspars and zeolites are examples of three- dimensional silicates. Molecular formula of ultramarine is Na8(SiAlO4)6 feldspar is KAlSi3O8 and zeolite is Na2Al2Si2O8. xH2O.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main, JEE Main 2014 (19 Apr Online)
Example of unit cell with crystallographic dimensions $a \neq b \neq c, \alpha=\gamma=90^{\circ}, \beta \neq 90^{\circ}$ is

Options:
A Calcite
B Graphite
C Rhombic sulphur
D Monoclinic sulphur

The correct answer is:
Monoclinic sulphur

Solution:
Monoclinic sulphur is an example of Monoclinic crystal system.

Topic: Chemistry, Solid State, JEE Main
Example of vinylic halide is

Options:
A
B
C
D

The correct answer is:

Solution:
The key difference between allylic and vinylic carbon is that allylic carbon is the carbon atom adjacent to the double-bonded carbon atom, whereas the vinylic carbon atom is one of the two atoms that share the double bond.

Haloarene or aryl halide is a chemical compound containing arenas, where one or more hydrogen atoms are bonded to the aromatic ring and are replaced with halogens.

Vinyl carbon is sp2 hybridized aliphatic carbon

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (30 Jan Shift 1)
Examples of In-situ conservation method are national parks, sanctuaries, biosphere reserves, and hotspots. The number of national parks (A), wildlife sanctuaries (B), and biosphere reserves (C), of India, are

Options:
A (A – 90), (B – 448), (C – 20)
B (A – 78), (B – 212), (C – 18)
C (A – 90), (B – 448), (C – 14)
D (A – 58), (B – 412), (C – 14)

The correct answer is:
(A – 90), (B – 448), (C – 14)

Solution:
In situ conservation refers to the conservation of wild species in their natural habitats and environment. The national parks, wildlife sanctuaries, and biosphere reserve are some of the examples.
National parks are areas maintained by the government and reserved for the betterment of wildlife (both flora and fauna). Here, grazing, the felling of trees, habitat manipulation, and cultivation are not allowed. India has 90 National parks.
Wildlife sanctuaries are tracts of land where animals (fauna) are protected from all types of exploitation and habitat disturbance. Collection of forest products, harvesting of timber, tilling of land, private ownership, etc., are allowed. India has 448 wildlife sanctuaries.
Biosphere reserves are special categories of protected land and/or coastal environments, wherein tribal people are an integral component of the system. India has 14 biosphere reserves.

Topic: Biology, Biodiversity and its Conservation, NEET
Examples of synthetic polymer $(X)$ and semi-synthetic polymer $(Y)$ are $x$

Options:
A

B Rayon $\quad$ Nylon 6, 6
C

D

The correct answer is:

Solution:
Synthetic polymer
Synthetic polymers are human-made polymer they can be classified into four main categories
(i) thermoplastics
(ii) thermosets
(iii) elastomers
(iv) synthetic fibers
e.g. polythene
Semi synthetic polymer
A semi synthetic polymer is a polymer made by chemically treating a natural polymer.
e.g. rayon
Rayon is derived from naturally occurring cellulose.

Topic: Chemistry, Polymers, AP EAMCET, AP EAMCET 2018 (24 Apr Shift 1)
excess

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Excess $\mathrm{KI}$ reacts with $\mathrm{CuSO}_4$ solution, and then $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$ solution is added to it. Which of the statements is incorrect for this reaction?

Options:
A $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$ is oxidised
B $\mathrm{Cu} \mathrm{l}_2$ is formed
C $\mathrm{Cu}_2 \mathrm{I}_2$ is formed
D Evolved $\mathrm{I}_2$ is reduced

The correct answer is:
$\mathrm{Cu} \mathrm{l}_2$ is formed

Solution:
Copper sulphate reacts with potassium iodide to form cuprous iodide and iodine.
$2 \mathrm{CuSO}_4+4 \mathrm{KI} \rightarrow \mathrm{Cu}_2 \mathrm{I}_2 \downarrow+\mathrm{I}_2+2 \mathrm{~K}_2 \mathrm{SO}_4$
Thus, $\mathrm{CuI}_2$ is not formed in this reaction.
Hence, the option $\mathrm{B}$ is incorrect and option $\mathrm{A}$ is correct.
The liberated iodine is titrated with sodium thiosulphate to form sodium tetrathionate.
$2 \mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3+\mathrm{I}_2 \rightarrow \mathrm{Na}_2 \mathrm{~S}_4 \mathrm{O}_6+2 \mathrm{NaI}$
Iodine is reduced and sodium thiosulphate is oxidized.

Topic: Chemistry, Redox Reactions, JEE Main
Excess of isobutane on reaction with Br2 in presence of light at 125°C gives which one of the following, as the major product?

Options:
A
B
C
D

The correct answer is:

Solution:
This reaction follows the free radical mechanism as follows:

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2021 (26 Aug Shift 1)
Excess of KI reacts with CuSO4 solution and then Na2S2O3 solution is added to it. Which of the statements is incorrect in this reaction?

Options:
A Evolved I2 is reduced
B CuI2 is formed
C Na2S2O3 is oxidised
D Cu2I2 is formed

The correct answer is:
CuI2 is formed

Solution:
When excess of KI reacts with Copper sulphate followed by addition of hypo the reactions are as follows
2CuSO4+4KI→Cu2I2+2K2SO4+I2
I2+2Na2S2O3→Na2S4O6+2NaI
This liberated I2 is reduced and Na2S2O3 is oxidised.

Topic: Chemistry, d and f Block Elements, JEE Main
Excess of $\mathrm{KI}$ reacts with $\mathrm{CuSO}_4$ solution and then $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$ solution is added to it. Which of the statements is incorrect for this reaction?

Options:
A
$\mathrm{Cu}_2 \mathrm{I}_2$ is reduced

B
Evolved $\mathrm{I}_2$ is reduced

C
$\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$ is oxidized

D
$\mathrm{Cul}_2$ is formed

The correct answer is:
$\mathrm{Cul}_2$ is formed

Solution:
$2 \mathrm{CuSO}_4+4 \mathrm{KI}$ (excess) $\longrightarrow 2 \mathrm{~K}_2 \mathrm{SO}_4+\mathrm{Cu}_2 \mathrm{I}_2+\mathrm{I}_2 \uparrow$
$\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3+\mathrm{I}_2 \longrightarrow \mathrm{Na}_2 \mathrm{~S}_4 \mathrm{O}_6+2 \mathrm{Nal}$

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2004
Excess of NaOH (aq) was added to 100 mL of FeCl3 aq resulting into 2.14 g of FeOH3. The molarity of FeCl3aq is:
(Given the molar mass of Fe=56 g mol-1 and molar mass of Cl=35.5 g mol-1)

Options:
A 0.3 M
B 0.2 M
C 0.6 M
D 1.8 M

The correct answer is:
0.2 M

Solution:
Moles of FeOH3 =2.14107=2×10-2mol
Moles FeCl3=100 ml × 10−3 × M
According to stoichiometry, mole of FeCl3=Mole of Fe(OH)3
100 × 10−3 ×M=2 × 10−2
M=2×10-2100×1000=0.2 M

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2017 (08 Apr Online)
Excess of $\mathrm{PCl}_{5}$ reacts with conc $\mathrm{H}_{2} \mathrm{SO}_{4}$ giving

Options:
A chlorosulphonic acid
B thionyl chloride
C sulphuryl chloride
D sulphurous acid

The correct answer is:
sulphuryl chloride

Solution:
$\mathrm{PCl}_{5}$ reacts with conc $\mathrm{H}_{2} \mathrm{SO}_{4}$ to give sulphuryl chloride by replacing its hydroxyl groups with chlorine atoms.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Excessive release of CO2 into the atmosphere results in

Options:
A global warming.
B polar vortex.
C depletion of ozone.
D formation of smog.

The correct answer is:
global warming.

Solution:
Carbon dioxide is very important for maintaining the climate which is suitable for our survival.
CO2 has the property of trapping heat from the sun. So higher the level of carbon dioxide, the higher will be the heat trapped. This leads to the rise in the temperature of the earth, resulting in global warming.Therefore, we can say excessive release of CO2 into the atmosphere results in global warming.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2019 (09 Apr Shift 1)