datavorous/entrance-exam-dataset huggingface dataset:
Compound (A), C8H9Cl, gives a white precipitate when warmed with alcoholic AgNO3. Oxidation of (A) gives an acid (B), C8H6O4. (B) easily forms anhydride on heating. Identify the compound (A).

Options:
A ​
B ​
C ​
D ​

The correct answer is:
​

Solution:
Compound (A) gives a white precipitate when warmed with alcoholic AgNO3means it contain a removable chloride atom(Benzylic position). Oxidation of (A) gives an acid (B), C8H6O4. (B) easily forms anhydride on heating. Means both acid should be ortho position of each other.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main, JEE Main 2013 (07 Apr)
Compound AC9H10O shows positive iodoform test. Oxidation of A with KMnO4/KOH gives acid BC8H6O4 . Anhydride of B is used for the preparation of phenolphthalein. Compound A is:

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2019 (10 Apr Shift 2)
Compound A contains 8.7% Hydrogen, 74% Carbon and 17.3% Nitrogen. The molecular formula of the compound is, Given : Atomic masses of C,H and N are 12,1 and 14 amu respectively.

The molar mass of the compound A is 162 g mol-1.

Options:
A C4H6N2
B C2H3N
C C5H7N
D C10H14N2

The correct answer is:
C10H14N2

Solution:
C	74%	7412=6.16	6.161.23=5

N	17.3%	17.314=1.23	1.231.23=1

H	8.7%	8.71=8.7	8.71.23=7

Emperical formula =C5NH7

Emperical weight =81

Multiplying factor =16281=2

Molecular formula=Empirical formulan

Molecular formula=C5NH72

Molecular formula =C10N2H14

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2022 (28 Jun Shift 2)
Compound A formed in the following reaction reacts with B gives the product C. Find out A and B.

Options:
A A=CH3-C≡C-Na+,B=CH3-CH2-CH2-Br
B A=CH3-CH=CH2, B=CH3-CH2-CH2-Br
C A=CH3-CH2-CH3, B=CH3-C≡CH
D A=CH3-C≡C-Na+,B=CH3-CH2-CH3

The correct answer is:
A=CH3-C≡C-Na+,B=CH3-CH2-CH2-Br

Solution:
When CH3C≡CH reacts with sodium an intermediate CH3C≡CNa is formed which on further reaction with propyl bromide gives CH3CH2CH2C≡CCH3 as the product. The second step of the reaction follows, SN2 mechanism.

The reaction is as follows,

Hence in the above reaction,

A=CH3-C≡C-Na+,B=CH3-CH2-CH2-Br

Hence option A is the answer.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2024 (30 Jan Shift 1)
Compound A from the following reaction sequence is :

Options:
A Benzoic Acid
B Aniline
C Salicylic Acid
D Phenol

The correct answer is:
Phenol

Solution:
When aniline treated with bromine, the reaction results in the formation of 2,4,6-tribromo phenylamine. This reaction is an example of bromination. The process of conversion of a primary aromatic amino compound into a diazonium salt is known as diazotization. This process is carried out by adding an aqueous solution of sodium nitrite to a solution of primary aromatic amine (e.g., aniline) in excess of HCl at a temperature below 5°C. Benzene diazonium chloride reacts with H3PO2 in aqueous medium to form benzene.

Topic: Chemistry, Amines, JEE Main, JEE Main 2023 (13 Apr Shift 2)
Compound A is converted to B on reaction with CHCl3 and KOH. The compound B is toxic and can be decomposed by C. A, B and C respectively are:

Options:
A primary amine, nitrile compound, conc. HCl
B secondary amine, isonitrile compound, conc. NaOH
C primary amine, isonitrile compound, conc. HCl
D secondary amine, nitrile compound, conc. NaOH

The correct answer is:
primary amine, isonitrile compound, conc. HCl

Solution:

Topic: Chemistry, Amines, JEE Main, JEE Main 2021 (20 Jul Shift 1)
Compound 'A' (molecular formula C3H8O) is treated with acidified potassium dichromate to form a product 'B' (molecular formula C3H6O). 'B' does not forms a shining silver mirror on warming with ammoniacal silver nitrate. 'B' when heated with concentrated sulphuric acid form 'C' which do not give test for unsaturation. Identify the sum of number of carbon and hydrogen atoms in compound 'C'

Options:

The correct answer is:
21

Solution:
Compound 'A' (molecular formula C3H8O) is isopropyl alcohol.
Compound 'B' (molecular formula C3H6O) is acetone. With concentrated sulphuric acid acetone form mesitylene.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Compound A of molecular formula C 4 H 10 O on treatment with Lucas reagent at room temperature gives compound B. When compound B is heated with alcoholic KOH, it gives isobutene. Compound A and B are respectively

Options:
A 2-Methyl-2-propanol and 2-Methyl-2-chloropropane
B 2-Methyl-1-propanol and 1-Chloro-2-methylpropane
C 2-Methyl-1-propanol and 2-Methyl-2-chloropropane
D Butan-2-ol and 2-Chlorobutane

The correct answer is:
2-Methyl-2-propanol and 2-Methyl-2-chloropropane

Solution:
Reaction involved is given as:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Compound A used as a strong oxidizing agent is amphoteric in nature. It is the part of lead storage batteries. Compound A is:

Options:
A PbO
B Pb3O4
C PbO2
D PbSO4

The correct answer is:
PbO2

Solution:
PbO2 is amphoteric and strong oxidizing agent and also a component of lead storage batteries.

In lead storage batteries PbO2 is used. In this Oxidation state of Pb is +4 so, it is always reduced and behaves as oxidizing agent

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main, JEE Main 2021 (26 Feb Shift 1)
Compound C in above reaction sequence is

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Compound (C) is

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Compound found by hydrolysis of BiCl3 is :-

Options:
A Bismuth hydroxide
B Bismuth oxychloride
C Bismuth oxide
D Oxo acid of bismuth

The correct answer is:
Bismuth oxychloride

Solution:
BiCl3+H2O→BiOCl+2HCl

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Compound I is heated with Conc. HI to give a hydroxy compound A which is further heated with Zn dust to give compound B. Identify A and B.

​​​​​​​

Options:
A

B

C

D

The correct answer is:

Solution:
Alkyl aryl ethers are cleaved at the alkyl-oxygen bond due to the more stable aryl-oxygen bond. The reaction yields phenol and alkyl halide.

And then Phenol is converted to benzene on heating with zinc dust.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2022 (28 Jul Shift 2)
Compound PdCl4.6H2O is a hydrated complex; 1 molal aqueous solution of it has freezing point 269.28 K. Assuming 100% ionization of complex, calculate the molecular formula of the complex
( Kf for water = 1.86 K kg mol-1 )

Options:
A [Pd(H2O)6]Cl4
B [Pd(H2O)4Cl2]Cl2.2H2O
C [Pd(H2O)3Cl3]Cl.3H2O
D [Pd(H2O)2Cl4].4H2O

The correct answer is:
[Pd(H2O)3Cl3]Cl.3H2O

Solution:
ΔT=i×Kf×m
(273-269.28)=i×1.86×1
3.72=i×1.86
i=2
α=i-1n-1
1=2-1n-1
or n=2
Thus, the complex should give two ions in the solution, i.e., the complex will be [Pd(H2O)3Cl3]Cl.3H2O .

Topic: Chemistry, Solutions, JEE Main
Compound 'P' on nitration with dil. HNO3 yields two isomers A and B. These isomers can be separated by steam distillations. Isomers A and B show the intramolecular and intermolecular hydrogen bonding respectively. Compound P on reaction with conc. HNO3 yields a yellow compound 'C', a strong acid. The number of oxygen atoms is present in compound 'C'

Options:

The correct answer is:
7

Solution:
So, as we can see that the final proudct C is picric acid and it contain 7 oxygen atoms.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2022 (26 Jun Shift 1)
Compound P will be

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
Compound squamous epithelium is found in

Options:
A stomach
B intestine
C trachea
D pharynx

The correct answer is:
pharynx

Solution:
In the structure of compound squamous epithelium, several layers of cells, deep layers are cuboidal to columnar, surface layers flat and scale-like. This epithelium founds in mouth,oesophagus, part of epiglottis (pharynx) and vagina. The main function of this epithelium is protection.

Topic: Biology, Plant - Growth and Development, JIPMER, JIPMER 2008
Compound that is both paramagnetic and coloured is

Options:
A K2Cr2O7
B (NH4)2 [TiCl6]
C VOSO4
D K3[Cu(CN)4]

The correct answer is:
VOSO4

Solution:
K2Cr2O7 contains Cr+6(3d0) which is diamagnetic but coloured due to charge transfer spectra.
(NH4)2[TiCl6] contains Ti4+(3d0), which is diamagnetic and colourless.
VOSO4 contains V4+(3d1), which is paramagnetic and coloured.
K3[Cu(CN)4] contains Cu+(3d10), which is diamagnetic and colourless.

Topic: Chemistry, d and f Block Elements, JEE Main
Compound that will give positive Lassaigne's test for both nitrogen and halogen is

Options:
A N2H4·HCl
B CH3NH2.HCl
C NH4Cl
D NH2OH.HCl

The correct answer is:
CH3NH2.HCl

Solution:
Lassaigne's test gives presence of chlorine and nitrogen in organic compound by fusion of organic compound in the presence of sodium. The fused products in this case are:
CH3NH2.HCl→fusionNaNaCN and NaCl

NaCN gives +ve test for nitrogen and NaCl gives +ve test for halogen.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2023 (29 Jan Shift 1)
Compound which gives acetone on ozonolysis

Options:
A $\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3$
B $\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{C}\left(\mathrm{CH}_3\right)_2$
C $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}=\mathrm{CH}_2$
D $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}_2$

The correct answer is:
$\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{C}\left(\mathrm{CH}_3\right)_2$

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Compound which gives acetone on ozonolysis

Options:
A $\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3$
B $\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{C}\left(\mathrm{CH}_3\right)_2$
C $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}=\mathrm{CH}_2$
D $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}_2$

The correct answer is:
$\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{C}\left(\mathrm{CH}_3\right)_2$

Solution:
$\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{C}\left(\mathrm{CH}_3\right)_2 \xrightarrow{\mathrm{O}_3} 2 \mathrm{CH}_3-\mathrm{CO}-\mathrm{CH}_3$

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Compound with molecular formula C3H6O can show:

Options:
A Positional isomerism
B Both positional isomerism and metamerism
C Metamerism
D Functional group isomerism

The correct answer is:
Functional group isomerism

Solution:
They are functional group isomerism.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2021 (18 Mar Shift 1)
Compound (X) on reduction with LiAlH4 gives a hydride (Y) containing 21.72% hydrogen along with other products. The compound (Y) reacts with air explosively resulting in boron trioxide. Compounds x and Y are respectively

Options:
A BCl3,B2H6
B B2H6,BCl3
C BF3,Al2O6
D B2H6,BF3

The correct answer is:
BCl3,B2H6

Solution:
4BCl3X+3LiAlH4→2B2H6Y+3AlCl3+3LiCl
Percentage of hydrogen in B2H6=627.62×100=21.72
B2H6+3O2→ Explosion B2O3+3H2O

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Compounds A and B are treated with dil. HCl separately. The gas liberated are Y and Z respectively. Y turns acidified K2Cr2O7 paper green while Z turns lead acetate paper black. The compounds A and B are respectively.

Options:
A NaCl and Na2CO3
B Na2SO3 and Na2S
C Na2S and Na2SO3
D Na2SO3 and Na2SO4

The correct answer is:
Na2SO3 and Na2S

Solution:
Here A & (B) are Na2SO3andNa2S respectively.
Na2SO3(A)+2HCl→2NaCl+SO2(Y)+H2O
K2Cr2O7+H2SO4+3SO2→K2SO4+Cr2(SO4)3Green+H2O
Na2S(B)+2HCl→2NaCl+H2S(Z)
PbCH3COO2+H2S→PbSBlack+2CH3COOH

Topic: Chemistry, d and f Block Elements, JEE Main
Compounds A is

Options:
A

B

C

D

The correct answer is:

Solution:
$\mathrm{As}^{-\mathrm{CCl}_3}$ is a $m$-directing group.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Compound(s) which will liberate carbon dioxide with sodium bicarbonate solution is/are :

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

The correct answer is:
B and C only

Solution:
Compounds which are more acidic then H2CO3, gives CO2 gas on reaction with NaHCO3. Compound B i.e. Benzoic acid and compound C i.e. picric acid both are more acidic than H2CO3.

equilibrium favours forward direction and CO2↑ is liberated.

Equilibrium favours backward direction and CO2↑ is not liberated.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2021 (25 Feb Shift 1)
Comprise the pressure at the point \( P \) in the three identical capillary tubes (shown in figure), which are dipped into \( 3 \) different liquids, and exposed to the atmosphere. The liquid meniscus within the tubes is shown in the diagram. Choose correct option:
(a)
(b)
(c)

Options:
A \( P_{2}>P_{1}>P_{3} \)
B \( P_{3}>P_{1}>P_{2} \)
C \( \mathrm{P}_{1}>\mathrm{P}_{3}>\mathrm{P}_{2} \)
D \( P_{2}>P_{3}>P_{1} \)

The correct answer is:
\( P_{2}>P_{3}>P_{1} \)

Solution:
In the case of high cohesive force the surface tension is greater, because the surface tension is directly related to cohesive force,

In first case the surface tension is minimum, because in this the attraction between liquid molecules and glass molecules say adhesive force is greater than cohesive force, while in second diagram the cohesive force is greater than adhesive force, so surface tension is maximum is second case.

While in third case cohesive force is equal to adhesive force,

it means,

T2>T3>T1

And excess pressure inside capillary is directly related with surface tension,

P=2Tr,

⇒P2>P3>P1

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Compute the LC product of a tuned circuit required to generate a carrier wave of \( 1 \mathrm{MHz} \) for amplitude modulation ?

Options:
A \( 2.54 \times 10^{-14} \mathrm{~s} \)
B \( 3.54 \times 10^{-14} \mathrm{~s} \)
C \( 2.34 \times 10^{-14} \mathrm{~S} \)
D \( 6.51 \times 10^{-2} \mathrm{~s} \)

The correct answer is:
\( 2.54 \times 10^{-14} \mathrm{~s} \)

Solution:
Given, the frequency of carrier wave is 1 MHz.
Formula for the frequency of tuned amplifier
1 2 π LC = 1 MHz
LC = 1 2 π × 1 0 6
LC = 1 ( 2 π × 1 0 6 ) 2 = 2 &CenterDot; 5 4 × 1 0 - 1 4 s
Thus, the product of LC is 2.54 × 10–14s.

Topic: Physics, Communication System, JEE Main
Conc. $\mathrm{HNO}_3$ is added before proceeding to test for group III members. This is to

Options:
A Oxidise any remaining $\mathrm{H}_2 \mathrm{~S}$
B Convert ferrous ion into ferric ion
C Form nitrates which give granular ppts
D Increase ionisation of ammonium hydroxide

The correct answer is:
Convert ferrous ion into ferric ion

Solution:
$\mathrm{HNO}_3$ oxidises $\mathrm{Fe}^{2+}$ to $\mathrm{Fe}^{3+}$ otherwise $\mathrm{Fe}^{2+}$ is not completely precipitated as hydroxide.

Topic: Chemistry, Practical Chemistry, JEE Main
Conc. $\mathrm{HNO}_3$

Options:
A Reacts with aluminium vigrously
B Reacts with aluminium to form aluminium nitrate
C Does not react with aluminium
D Reacts with platinum

The correct answer is:
Does not react with aluminium

Solution:
Aluminum metal is not attacked by nitric acid of any concentration because of the thin and unreactive protective layer of aluminum oxide formed on the metallic surface due to the reaction of aluminium metal with oxygen of air.
Hence option 3 is correct.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Concentrated aqueous sodium hydroxide can separate a mixture of

Options:
A Al3+ and Sn2+
B Al3+ and Fe3+
C Al3+ and Zn2+
D Zn2+ and Pb2+

The correct answer is:
Al3+ and Fe3+

Solution:
Concentrated aqueous sodium hydroxide can separate a mixture of Al3+ and Fe3+
Al3+ forms white gelatinous precipitate of AlOH3 , which is soluble in excess of NaOH.
AlOH3+NaOH→NaAlO2+2H2O
Fe3+ forms reddish brown precipitate of FeOH3 , which is insoluble in excess of NaOH.

Topic: Chemistry, s Block Elements, JEE Main
Concentrated $\mathrm{HNO}_3$ reacts with $I_2$ to give

Options:
A $\mathrm{HI}$
B $\mathrm{HOI}$
C $\mathrm{HIO}_3$
D $\mathrm{HOIO}_3$

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

Solution:
$I_2+10 \mathrm{HNO}_3 \rightarrow 2 \mathrm{HIO}_3+10 \mathrm{NO}_2+4 \mathrm{H}_2 \mathrm{O}$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Concentrated HNO3 reacts with Iodine to give

Options:
A HI,NO2 and H2O
B HIO2,N2O and H2O
C HIO3,NO2 and H2O
D HIO4,N2O and H2O

The correct answer is:
HIO3,NO2 and H2O

Solution:
Concentrated nitric acid is a strong oxidizing agent.

Concentrated nitric acid also oxidises non–metals and their compounds. Iodine is oxidised to iodic acid.

I2+10HNO3conc⇒2HIO3+10NO2+4H2O

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2022 (28 Jul Shift 2)
Concentrated hydrochloric acid when kept in open air sometimes produces a cloud of white fumes. The explanation for it is that

Options:
A
oxygen in air reacts with the emitted HCI gas to form a cloud of chlorine gas

B
strong affinity of HCI gas for miosture in air results in forming of droplets of liquid solution which appears like a cloudy smoke.

C
due to strong affinity for water, concentrated hydrochloric acid pulls moisture of air towards it self. This moisture forms droplets of water and hence the cloud.

D
concentrated hydrochloric acid emits strongly smelling HCI gas all the time.

The correct answer is:
oxygen in air reacts with the emitted HCI gas to form a cloud of chlorine gas

Solution:
$4 \mathrm{HCI}+\mathrm{O}_2 \rightarrow 2 \mathrm{Cl}_2+2 \mathrm{H}_2 \mathrm{O}$
Cloud of white fumes

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2003
Concentrated hydrochloric acid when kept in open air sometimes produces a cloud of white fumes. The explanation for it is that

Options:
A
oxygen in air reacts with the emitted HCI gas to form a cloud of chlorine gas

B
strong affinity of HCI gas for miosture in air results in forming of droplets of liquid solution which appears like a cloudy smoke.

C
due to strong affinity for water, concentrated hydrochloric acid pulls moisture of air towards it self. This moisture forms droplets of water and hence the cloud.

D
concentrated hydrochloric acid emits strongly smelling HCI gas all the time.

The correct answer is:
oxygen in air reacts with the emitted HCI gas to form a cloud of chlorine gas

Solution:
$4 \mathrm{HCI}+\mathrm{O}_2 \rightarrow 2 \mathrm{Cl}_2+2 \mathrm{H}_2 \mathrm{O}$
Cloud of white fumes

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Concentrated hydrochloric acid when kept in open air sometimes produces a cloud of white fumes. The explanation for it is that

Options:
A oxygen in air reacts with the emitted HCl gas to form a cloud of chlorine gas
B strong affinity of HCl gas for moisture in air results in forming of droplets of liquid solution which appears like a cloudy smoke
C due to strong affinity for water, concentrated hydrochloric acid pulls moisture of air towards itself. This moisture forms droplets of water and hence the cloud
D concentrated hydrochloric acid emits strongly smelling HCl gas all the time

The correct answer is:
oxygen in air reacts with the emitted HCl gas to form a cloud of chlorine gas

Solution:
4HCl+O2airâ2Cl2+2H2Ocloud of white fumes

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Concentrated nitric acid used in laboratory work is \( 68 \% \) nitric acid by aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is \( 1.504 \mathrm{~g} \mathrm{~mL}^{-1} \) ?

Options:

The correct answer is:
16.23

Solution:
68% HNO3 by mass means
Mass of nitric acid = 68 g
Mass of solution - 100 g
Molar mass of HNO3 = (1) + (14) + (16 x 3) = 63 g mol-1
Number of moles of HNO 3 n HNO 3 = W M = 68 g 63 g / mol = 1.079 mol
Density of solution = 1.504 g mL-1
Volume of solution = Mass Density = 100 g 1.504 g mL - 1
= 66.5 mL or 0.0665 L
Molarity = Moles of solute Volume of solution in litres
= 1.079 mol 0.0665 L
= 16.23 mol L-1 or 16.23 M

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
Concentrated of aqueous sulphuric acid is $98 \% \mathrm{H}_2 \mathrm{SO}_4$ by mass and has a density of $1.80 \mathrm{~g} \mathrm{~mL}^{-1}$ Volume of acid required to make one litre of $0.1 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4$ solution is:

Options:
A $16.65 \mathrm{~mL}$
B $22.20 \mathrm{~mL}$
C $5.55 \mathrm{~mL}$
D $11.10 \mathrm{~mL}$

The correct answer is:
$5.55 \mathrm{~mL}$

Solution:
Molarity of $\mathrm{H}_2 \mathrm{SO}_4$ solution
$=\frac{98 \times 1000}{98 \times 100} \times 1.84=18.4$
Suppose $\mathrm{V} \mathrm{ml}$ of this $\mathrm{H}_2 \mathrm{SO}_4$ is used to prepare 1 . of $0.1 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4$
$\begin{aligned}
& \therefore \mathrm{V} \times 18.02=1000 \times 0.1 \\
& \mathrm{~V}=\frac{1000 \times 0.1}{18.02}=5.55 \mathrm{ml}
\end{aligned}$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
Concentrated sodium hydroxide can separate a mixture of:
A. $\mathrm{Al}^{3+}$ and $\mathrm{Cr}^{3+}$
B. $\mathrm{Zn}^{2+}$ and $\mathrm{Pb}^{2+}$
C. $\mathrm{Cr}^{3+}$ and $\mathrm{Fe}^{3+}$
D. $\mathrm{Al}^{3+}$ and $\mathrm{Zn}^{3+}$

Options:
A $\mathrm{Zn}^{2+}$ and $\mathrm{Pb}^{2+}$
B $\mathrm{Cr}^{3+}$ and $\mathrm{Fe}^{3+}$
C $\mathrm{Al}^{3+}$ and $\mathrm{Zn}^{3+}$
D $\mathrm{Al}^{3+}$ and $\mathrm{Cr}^{3+}$

The correct answer is:
$\mathrm{Cr}^{3+}$ and $\mathrm{Fe}^{3+}$

Solution:
When the $\mathrm{Cr}^{3+}$ reacts with the conc. $\mathrm{NaOH}$, then following reaction, takes place,
$\mathrm{Cr}^{3+}+3 \mathrm{OH}^{-} \rightarrow \mathrm{Cr}(\mathrm{OH})_3$
Which on further reactions gives $\mathrm{Cr}(\mathrm{OH})_4^{-}$(green coloured solution)
$\mathrm{Cr}(\mathrm{OH})_3+\mathrm{OH}^{-} \rightarrow \mathrm{Cr}(\mathrm{OH})_4^{-}$(green coloured solution)
Whereas $\mathrm{Fe}^{3+}$ reaction with the conc. $\mathrm{NaOH}$ gives the $\mathrm{Fe}(\mathrm{OH})_3$ precipitate
$\mathrm{Fe}^{3+}+3 \mathrm{OH}^{-} \rightarrow \mathrm{Fe}(\mathrm{OH})_3$ (precipitate)
So, as the $\mathrm{Cr}(\mathrm{OH})^{4-}$ is in solution and $\mathrm{Fe}(\mathrm{OH})_3$ (precipitate)
formed is not in solution so, it can be separated out from the solution.

Topic: Chemistry, Practical Chemistry, JEE Main
Concentric metallic hollow spheres of radii R and 4R hold charges Q1 and Q2 respectively. Given that surface charge densities of the concentric spheres are equal, the potential difference VR-V(4R) is:

Options:
A Q24πε0R
B 3Q24πε0R
C 3Q116πε0R
D 3Q14πε0R

The correct answer is:
3Q116πε0R

Solution:
As given the surface charge density for both the spheres is same, hence

σ1=σ2∴Q1A1=Q2A2⇒Q14πR2=Q24π4R2∴Q2=16Q1

From above diagram, the electric potential for both the spheres is

VInner=KQ1R+KQ24R=KRQ1+16Q14=5KQ1R.

In the same way for outer sphere,

VOuter=KQ14R+KQ24R=K4R16Q1+Q1=17KQ14R.

Now the potential difference is,

∆V=Vinner-Vouter

=5KQ1R-17KQ14R=3Q116πε0R.

Topic: Physics, Electrostatics, JEE Main, JEE Main 2020 (03 Sep Shift 2)
Concrete mixture is made by mixing cement, stone and sand in a rotating cylindrical drum. If the drum rotates too fast, the ingredients remain stuck to the wall of the drum and proper mixing of ingredients does not take place. The maximum rotational speed of the drum in revolutions per minute (rpm) to ensure proper mixing is close to :
(Take the radius of the drum to be 1.25 m and its axle to be horizontal) :

Options:
A 27.0
B 0.4
C 1.3
D 8.0

The correct answer is:
27.0

Solution:
To break the circular motion max velocity at top point
v max =Rg
⇒ w max = v maxR=gR=101.25
w max rpm=602π101.25=27

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2016 (10 Apr Online)
Condensation polymerization among the following polymer is:

Options:
A Teflon
B Polystyrene
C PVC
D Dacron

The correct answer is:
Dacron

Solution:
Dacron is condensation polymer of Glycol and Terphthalic acid.

Topic: Chemistry, Polymers, JEE Main