datavorous/entrance-exam-dataset huggingface dataset:
Combustion of glucose \(\left(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6\right)\) produces \(\mathrm{CO}_2\) and water. The amount of oxygen (in g) required for the complete combustion of \(900 \mathrm{~g}\) of glucose is :
[Molar mass of glucose in \(\mathrm{g} \mathrm{mol}^{-1}=180\) ]

Options:
A 480
B 800
C 960
D 32

The correct answer is:
960

Solution:
$\begin{aligned} & \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_{6(\mathrm{~s})}+6 \mathrm{O}_{2(\mathrm{~g})} \longrightarrow 6 \mathrm{CO}_{2(\mathrm{~g})}+6 \mathrm{H}_2 \mathrm{O}_{(\theta)} \\ & \frac{900}{180} \\ & =5 \mathrm{~mol} \quad 30 \mathrm{~mol}\end{aligned}$
Mass of $\mathrm{O}_2$ required $=30 \times 32=960 \mathrm{gm}$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2024 (08 Apr Shift 1)
Combustion of glucose takes place according to the equation
C 6 H 12 O 6  + 6  O 2  → 6  CO 2  + 6  H 2 O; ΔH = −72 kcal/mol. How much energy will be released by the combustion of 1.6 g of glucose (Molecular mass of glucose = 180 g/mol )?

Options:
A 0.064 kcal
B 0.64 kcal
C 6.4 kcal
D 64 kcal

The correct answer is:
0.64 kcal

Solution:
Combustion of glucose takes place according to the equation
C 6 H 12 O 6  +  6O 2  →  6CO 2  +  6H 2 O; ΔH = -72 kcal
ÎHâperâ1.6âgâ=â 72Ã1.6 180 â=â0.64âkcal

Topic: Chemistry, Thermodynamics (C), JEE Main
Combustion of Hydrogen in a fuel cell at 300âK is represented as 2â H 2 g â+â O 2 g â2â H 2 O g . If ÎH and ÎG are â241 .80 kJ mol â1 and â228.60 kJ mol â1 respectively for H2O then the value of ÎS for the above process is:

Options:
A +44.JK-1mol-1
B -88 JK-1mol-1
C +88 JK-1mol-1
D -44.JK-1mol-1

The correct answer is:
-88 JK-1mol-1

Solution:
ΔS= ΔH−ΔG T = −241.8− −228.6 300
=-44 JK-1mole-1

Topic: Chemistry, Thermodynamics (C), JEE Main
Combustion of Hydrogen in a fuel cell at 300âK is represented as 2 H 2 g + O 2 g â2 H 2 O g .
If âH and âG are -241.60 kJ mol-1 and -228.40 kJ mol-1 are respectively for H2O then the value of âS for the above process is

Options:
A 4.4ââJâ K -1
B â88ââJâ K -1
C +88 Jâ K -1
D −44 J  K -1

The correct answer is:
−88  J  K -1

Solution:
âS=âH - âGT=-241.6--228.4300
For 1 mole of H2O=-44 Jâ K -1
Hence, for 2 moles =â88 Jâ K -1

Topic: Chemistry, Thermodynamics (C), JEE Main
Commelina produces:

Options:
A Cleistogamous flowers
B Chasmogamous flowers
C Both cleistogamous and chasmogamous types of flowers
D Either cleistogamous or chasmogamous flowers at a time

The correct answer is:
Both cleistogamous and chasmogamous types of flowers

Solution:
To avoid cross pollination and favor self-pollination in some plants, their flower remains permanently closed. Such flowers are called closed marriage flower or cleistogamous flowers. This behavior is seen commonly in grass family. Some plants produce flowers which exposes their reproductive organ to favor cross pollination, these flowers are called open marriage flowers or chasmogamous flowers. Commelina produces both cleistogamous and chasmogamous types of flowers

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
Comment upon the habitats and external features of animals belonging to class-Amphibia and Reptilia.

Options:

None

Solution:
Amphibians
(i) They can dwell in aquatic as well as terrestrial habitats.
They are ectothermic or (cold blooded).
(ii) They are tetrapods having (4 limbs) which facilitate movement on land.
(iii) Their limbs have evolved from the pectoral and pelvic fins.
(iv) Skin is thin, covered by mucus and remains mostly moist. It also serves as an accessory source of oxygen.
(v) They breathe through gills and lung gills usually appear in the larval stage, replaced by lungs in the adults stage.
(vi) Their heart is three chambered with two atria and one ventricle.
(vii) Females are oviparous and fertilisation is mostly external.
(viii) Larva is a tadpole, which metamorphose into adult e.g., Rana frog, Nectureus (mud puppy), Salamandera (salamander).
Reptiles
(i) They are mostly terrestrial animals and their body is covered by dry, and cornified skin, epidermal scales or scutes.
(ii) In reptiles the mode of locomotion is creeping and crawling.
(iii) Lungs are well developed and present in all stages of life.
(iv) Claws are present in toes.
(v) Appendages are well adapted for land movement.
(vi) Heart possesses a partially divided ventricle and 2 atria.
(vii) They lay amniotic eggs which are inclubated on land.
(viii) They are poikilothermic or cold blooded animals.

Topic: Biology, Animal Kingdom
Common cold is not cured by antibiotics because it is

Options:
A caused by a virus
B caused by a Gram positive bacterium
C caused by a Gram negative bacterium
D not an infectious disease

The correct answer is:
caused by a virus

Solution:
Common cold is not cured by antibiotics because it is caused by a virus, called rhinovirus.

Topic: Biology, Human Health and Diseases, NEET, NEET 2011 (Mains)
Common name of Benzene - 1, 2 - diol is -

Options:
A catechol
B o-cresol
C quinol
D resorcinol

The correct answer is:
catechol

Solution:
IUPAC name : Benzene-1,2-diol Common name: catechol

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (04 Apr Shift 2)
Commonly in a mature fertilised ovule n, 2n and 3n condition is respectively found in:

Options:
A Antipodals, synergids and integuments
B Antipodals, zygote and PEN
C Egg, nucellus and endosperm
D Endosperm, nucellus and egg

The correct answer is:
Antipodals, zygote and PEN

Solution:
In a fertillised ovule, zygote formed after fertilization is 2n and PEN formed after triple fusion is 3n, while the remaining antipodal or synergid is n.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
Companion cells are closely associated with

Options:
A sieve elements
B vessel elements
C trichomes
D guard cells

The correct answer is:
sieve elements

Solution:
Companion cells are characteristic elements of phloem tissue associated with the sieve tubes (sieve elements) in the angiosperms. The siever tubes and companion cells are related ontogenically as they develop from the same mother cell. The companion cells and sieve tubes maintain close cytoplasmic connections through plasmodesmata.

Topic: Biology, Anatomy of Flowering Plants, NEET, NEET 2012 (Screening)
Compare heat of hydrogenation of the following

Options:
A I>II>III
B III>II>I
C II>III>I
D II>I>III

The correct answer is:
III>II>I

Solution:
For isomeric alkene
Heat of hydrogenation â1Stabilities of alkene
More the number of α - hydrogen atoms, more is the stability of alkene and less is the heat of hydrogenation value.

Topic: Chemistry, General Organic Chemistry, JEE Main
Compare the atomic and ionic sizes of actinoids with that of the lanthanoids

Options:
A The atomic and ionic sizes of lanthanoids and actinoids increase with increase in the atomic number due to lanthanoid and actinoid contraction.
B The atomic and ionic sizes of lanthanoids and actinoids decrease with increase in the atomic number due to lanthanoid and actinoid contraction.
C The atomic and ionic sizes of lanthanoids and actinoids remain constant with increase in the atomic number due to lanthanoid and actinoid contraction.
D Cannot be decided without further information

The correct answer is:
The atomic and ionic sizes of lanthanoids and actinoids decrease with increase in the atomic number due to lanthanoid and actinoid contraction.

Solution:
Correct Option is : (B)
The atomic and ionic sizes of lanthanoids and actinoids decrease with increase in the atomic number due to lanthanoid and actinoid contraction.

Topic: Chemistry, d and f Block Elements, NEET
Compare the energies of following sets of quantum numbers for multielectron system.
(A) \(\mathrm{n}=4,1=1\)
(B) \(\mathrm{n}=4, \mathrm{l}=2\)
(C) \(\mathrm{n}=3,1=1\)
(D) \(\mathrm{n}=3,1=2\)
(E) \(\mathrm{n}=4,1=0\)
Choose the correct answer from the options given below :

Options:
A \(\text { (B) } > \text { (A) } > \text { (C) } > \text { (E) } > \text { (D) }\)
B \(\text { (E) } < (\text { C }) < \text { (D) } < (\text { A }) < (\text { B })\)
C \(\text { (E) } > (\text { C }) > (\text { A }) > (\text { D }) > (B)\)
D \(\text { (C) } < \text { (E) } < \text { (D) } < \text { (A) } < \text { (B) }\)

The correct answer is:
\(\text { (C) } < \text { (E) } < \text { (D) } < \text { (A) } < \text { (B) }\)

Solution:
Energy level can be determined by comparing ( $\mathrm{n}+\ell$ ) values
(A) $\mathrm{n}=4, \ell=1 \Rightarrow(\mathrm{n}+\ell)=5$
(B) $\mathrm{n}=4, \ell=2 \Rightarrow(\mathrm{n}+\ell)=6$
(C) $\mathrm{n}=3, \ell=1 \Rightarrow(\mathrm{n}+\ell)=4$
(D) $\mathrm{n}=3, \ell=2 \Rightarrow(\mathrm{n}+\ell)=5$
(E) $\mathrm{n}=4, \ell=0 \Rightarrow(\mathrm{n}+\ell)=4$
For same value of $(n+\ell)$, orbital having higher value of $n$, will have more energy.
(B) $>(\mathrm{A})>(\mathrm{D})>(\mathrm{E})>(\mathrm{C})$

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2024 (09 Apr Shift 1)
Compare the stability of following carbocations.

Options:
A III > II > I
B II > III > I
C III > I > II
D II > I > III

The correct answer is:
III > I > II

Solution:
Aromatic compound is the more stable compound

Anti Aromatic compound is the least stable compound

1 is the non aromatic compound

Topic: Chemistry, General Organic Chemistry, JEE Main
Compare the statement I and II and choose the correct option.
Statement I In the flowering plants due to higher accumulation of auxins dormancy of lateral buds occurs.
Statement II In Maryland Mammoth (a tobacco variety) flowering occurred at different time at different latitude due to gibberellin concentration.

Options:
A Statement I is true, but II is false
B Statement I is false, but II is true
C Both statements are true
D Both statements are false

The correct answer is:
Statement I is true, but II is false

Solution:
Auxin is responsible for apical dominance in apical buds, which inhibits the lateral buds to grow.
Plants require a certain day length to flower. This relative length of day and night needed for flowering is called as photoperiod. The effect of photoperiod or daily duration of light hours and dark periods on the growth and development of plants, especially flowering, is called photoperiodism.
This phenomenon was first discovered by Garner and Allard (1920, 1922) on Maryland Mammoth variety of tobacco. They observed that the Maryland Mammoth could be made to flower only by reducing the light hours with artificial darkening. It could be made to remain vegetative in winter by providing extra light.

Topic: Biology, Plant - Growth and Development, JIPMER, JIPMER 2016
Compared to blood our lymph has :

Options:
A More RBCs and less WBCs
B No plasma
C Plasma without proteins
D More WBCs and no RBCs

The correct answer is:
More WBCs and no RBCs

Solution:
No Solution Available

Topic: Biology, Body Fluids and Circulation, NEET, NEET 2009 (Mains)
Compared with the alkaline earth metals, the alkali metals exhibit

Options:
A Greater hardness
B Smaller ionic radii
C Lower ionisation energies
D Highest boiling points

The correct answer is:
Lower ionisation energies

Solution:
Because of the larger size and smaller nuclear charge, alkali metals have low ionisation potential relative to alkaline earth metals

Topic: Chemistry, s Block Elements, JEE Main
Complete combustion of 1.80 g of an oxygen containing compound CxHyOz gave 2.64 g of CO2 and 1.08 g of H2O. The percentage of oxygen in the organic compound is :

Options:
A 53.33
B 50.33
C 51.63
D 63.53

The correct answer is:
53.33

Solution:
nCO2=2.6444=0.06

nc=0.06

weight of carbon =0.06×12=0.72 gm

nH2O=1.0818=0.06

nH=0.06×2=0.12

weight of H=0.12 gm

= 1.8 - (0.72 + 0.12)

= 0.96 gram

% weight of oxygen =0.961.8×100

The percentage of oxygen in the organic compound = 53.3%.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2021 (25 Feb Shift 1)
Complete combustion of 3 g of ethane gives x×1022molecules of water. The value of x is ______ (Round off to the Nearest Integer).
Use :NA=6.023×1023; Atomic masses in u: C: 12.0 ; O: 16.0 ; H: 1.0

Options:

The correct answer is:
18

Solution:
C2H6+ 72 O2→2CO2 + 3H2O

MWC2H6=30g

30g of C2H6 gives rise to 3 moles of H2O

3g of C2H6 gives rise to 0.3 moles of H2O

No. of molecules =0.3×6.023×1023
=18.069×1022

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2021 (18 Mar Shift 1)
Complete hydrolysis of cellulose gives

Options:
A
D-ribose

B
D-glucose

C
L-glucose

D
D-fructose

The correct answer is:
D-glucose

Solution:

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2003
Complete hydrolysis of cellulose gives

Options:
A
D-ribose

B
D-glucose

C
L-glucose

D
D-fructose

The correct answer is:
D-glucose

Solution:

Topic: Chemistry, Biomolecules, JEE Main
Complete hydrolysis of starch gives:

Options:
A Galactose and fructose in equimolar amounts
B Glucose and galactose in equimolar amounts
C Glucose and fructose in equimolar amounts
D Glucose only

The correct answer is:
Glucose only

Solution:
On complete hydrolysis of starch, glucose is formed. Amylase is an enzyme that catalyzes the hydrolysis of starch into sugars.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2015 (10 Apr Online)
Complete reaction of $2.0 \mathrm{~g}$ of calcium (at. wt. = 40 ) with excess $\mathrm{HCL}$ produces $1.125 \mathrm{~L}$ of $\mathrm{H}_{2}$ gas. Complete reaction of the same quantity of another metal "M" with excess $\mathrm{HCL}$ produces $1.85 \mathrm{~L}$ of $\mathrm{H}_{2}$ gas under indentical conditions. The equivalent weight of "M" is closest to

Options:
A 23
B 9
C 7
D 12

The correct answer is:
12

Solution:
$1.125 \mathrm{~L}$ of $\mathrm{H}_{2}$ produced by $0.1$ equivalent of metal $1.85 \mathrm{~L}$ a of $\mathrm{H}_{2}$ will be produced by $=\frac{0.1 \times 1.85}{1.125}$ equivalents $\therefore$ No of gram equivalent of metal
$\begin{array}{l}
=\frac{2}{\text { Equivalent weight }}=\frac{2}{x} \\
\therefore \frac{0.1}{1.125} \times 1.85=\frac{2}{x} \\
x=12.16
\end{array}$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
Complete reduction of benzene-diazonium chloride with $\mathrm{Zn} / \mathrm{HCl}$ gives:

Options:
A
Aniline

B
Phenylhydrazine

C
Azobenzene

D
Hydrazobenzene

The correct answer is:
Aniline

Solution:
On reduction with $\mathrm{Zn}$ and $\mathrm{HCl}, \mathrm{C}_6 \mathrm{H}_5 \mathrm{~N}_2 \mathrm{Cl}$ forms aniline as the main product.

Topic: Chemistry, Amines, JEE Main, JEE Main 2014 (11 Apr Online)
Complete the following paragraph by selecting the correct option for the blanks A, B, and C from the options given below.

The symbiotic association formed by fungi with roots of higher plants is known as .........A...........It is mainly formed by members from the genus........B............The fungal symbiont in these associations is responsible for absorbing .........C..........from the soil and passing it to the plant. It also has many other benefits like resistance to root borne pathogens, tolerance to salinity and drought, etc. ...........D......... seeds cannot germinate and establish without the presence of ..........A........

Options:
A A: Lichen; B: Rhizobium; C: Phosphorus; D: Cycas
B A: Mycorrhiza; B: Glomus; C: Potassium; D: Cycas
C A: Mycorrhiza; B: Glomus; C: Phosphorus; D: Pinus
D A: Lichen; B: Rhizobium; C: potassium; D: Pinus

The correct answer is:
A: Mycorrhiza; B: Glomus; C: Phosphorus; D: Pinus

Solution:
Mycorrhiza is the symbiotic association formed by fungi with the roots of higher plants. Many members from the genus Glomus form this association. The fungal symbiont in these associations is responsible for absorbing phosphorus from the soil and passing it to the plant. Some plants have an obligate association with the mycorrhizae. For example, Pinus seeds cannot germinate and establish without the presence of mycorrhizae.

Topic: Biology, Microbes in Human Welfare, NEET
Complete the following paragraph by selecting the correct options for the blanks A, B and C.

..........A......... is the structural and functional unit between the embryo and the maternal body. It is connected to the embryo by ...........B............. It also acts as an ..........C......... tissue and secretes many hormones essential for maintaining pregnancy.

Options:
A A: placenta, B: chorionic villi, C: endocrine
B A: placenta, B: umbilical cord, C: endocrine
C A: placenta, B: umbilical cord, C: exocrine
D A: placenta, B: chorionic villi, C: exocrine

The correct answer is:
A: placenta, B: umbilical cord, C: endocrine

Solution:
Placenta is the structural and functional unit between the embryo and the matenrnal body. It is responsible for transportation of essential nutrients and gases to the foetus and also the removal of foetal waste matter. It is connected to the embryo by an umbilical cord. It also acts an endocrine tissue and secretes many hormones like HCG, progesterone, HPL, etc. which are essential for maintaining pregnancy.

Topic: Biology, Human Reproduction, NEET
Complete the following paragraph by selecting the correct options for the blanks A, B and C.

Saliva is mainly produced by three pairs of salivary glands, the ....A.... (cheek), the ....B.... (lower jaw) and the ......C...... (below the tongue). These glands situated just outside the buccal cavity secrete salivary juice into the buccal cavity.

Options:
A A: Sublingual, B: Parotid, C: Sub-Maxillary/Sub-Mandibular
B A: Sub-Maxillary/Sub-Mandibular, B:Sublingual, C: Parotid
C A: Parotid, B :Sub-Maxillary/Sub-Mandibular, C: Sublingual
D A: Lingual, B: Parotod, C: Maxillary

The correct answer is:
A: Parotid, B :Sub-Maxillary/Sub-Mandibular, C: Sublingual

Solution:
The three pairs of salivary glands are:
1. Parotids (cheek): The largest salivary glands which have Stenson's duct. They produce salivary amylase or ptyalin.
2. Sub-maxillary/sub-mandibular (lower jaw) glands are the salivary glands with Wharton's duct.
3. Sublingual (below the tongue) glands: The smallest salivary glands with ducts of Rivinus.

Topic: Biology, Digestion and Absorption, NEET
Complete the missing links
CH 3 CHBrCH 3 → alc.KOH X → HBr, Peroxiden Y → CH 3 ONa Z

Options:
A

X	Y	Z

CH3CH=CH2	CH3CH(Br)CH2Br	CH3CH(OH)CH3

B

X	Y	Z

CH3CH=CH2	CH3CH2CH2Br	CH3CH2OCH2CH3

C

X	Y	Z

CH3CH=CH2	CH3CH(Br)CH3	CH3CH2OCH2CH3

D

X	Y	Z

CH3CH=CH2	CH3CH2CH2Br	CH3CH2CH2OCH3

The correct answer is:

X	Y	Z

CH3CH=CH2	CH3CH2CH2Br	CH3CH2CH2OCH3

Solution:
CH3CHBrCH3→ alc.KOH CH3CH=CH2(X)→ HBr,peroxide CH3CH2CH2Br(Y)→ CH3ONa CH3CH2CH2OCH3(Z)

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Complex A has a composition of H12O6Cl3Cr. If the complex on treatment with conc. H2SO4 loses 13.5% of its original mass, the correct molecular formulas of A is:
[Given : atomic mass of Cr=52 amu and CI=35 amu]

Options:
A CrH2O6Cl3
B CrH2O3Cl3·3H2O
C CrH2O5ClCl2·H2O
D CrH2O4Cl2Cl·2H2O

The correct answer is:
CrH2O4Cl2Cl·2H2O

Solution:
Conc. H2SO4 acts as dehydrating agent.

Molar mass of given complex =266.5 g/mol.

On treating with conc. H2SO4 the mass lost by the complex =13.5100266.5≈36g

=2 moles of H2O

Formula of the complex =CrH2O4Cl2 Cl·2H2O

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2020 (03 Sep Shift 2)
Complex \( \left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right] \mathrm{Cl} \) has number of possible isomers. (en = ethylenediamine)

Options:
A \( 2 \)
B \( 1 \)
C \( 3 \)
D \( 4 \)

The correct answer is:
\( 3 \)

Solution:
Coen2Cl2Cl
Possible isomers
(i) Geometrical isomers
(ii) In trans form plane of symmetry present, so trans form is optically inactive but cis is optically active.
Total number of stereoisomer = 2 + 1 = 3

Topic: Chemistry, Coordination Compounds, JEE Main
Complex X Of composition CrH2O6Cln Has a spin only magnetic moment of 3.83 B.M. It reacts with AgNO3 And shows geometrical isomerism. The IUPAC nomenclature of X Is:

Options:
A Hexaaqua chromium(III) chloride
B Tetraaquadichlorido chromium(IV) chloride dihydrate
C Dichloridotetraaqua chromium(IV) chloride dihydrate
D Tetraaquadichlorido chromium(III) chloride dihydrate

The correct answer is:
Tetraaquadichlorido chromium(III) chloride dihydrate

Solution:
CrH2O6Clnμcomplexspin=3.83 B.M.
From data of magnetic moment oxidation number of Cr should be +3
Hence, complex is CrH2O6Cl3.

The above complex formula can have following different types of complexes-
1. CrH2O6Cl3 As chlorine atoms are outside of co-ordination sphere hence it can react with Silver nitrate to form silver chloride, but can not show Geometrical Isomerism because all 6 ligands are same i.e. aqua

2. CrH2O5ClCl2.H2O

This complex can react with silver nitrate as chlorine atoms are outside of co-ordination sphere, but can not show Geometrical Isomerism.

3. CrH2O4Cl2Cl.2H2O

This complex can react with silver nitrate as one chlorine atom is outside of co-ordination sphere and also show Geometrical isomerism. The IUPAC name of this complex is Tetraaquadichloridochromium(III) chloride dihydrate.

4. CrH2O3Cl33H2O

This complex can not react with silver nitrate because not chlorine atom is outside of co-ordination sphere, but this complex can show Geometrical isomerism(Facial and meridional ).

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2020 (09 Jan Shift 1)
Complex X of composition CrH2O6Cln has a spin only magnetic moment of 3.83 BM . It reacts with AgNO3 and shows geometrical isomerism. The IUPAC nomenclature of X is

Options:
A Tetraaquadichlorido chromium (IV) chloride dihydrate
B Tetraaquadichlorido chromium (III) chloride dihydrate
C Hexaaqua chromium (III) chloride
D Dichloridotetraaqua chromium (IV) chloride dihydrate

The correct answer is:
Tetraaquadichlorido chromium (III) chloride dihydrate

Solution:
This compound can be identified and named as follows
CrH2O6Cln
As Î¼complexspin=3.8 B.M. =n(n+2)
On solving n=3
Oxidation No. of Cr=+3
As compound also shows Geometrical isomerism so it will be [CrH2O)4Cl2Cl.2H2O
It is named as â Tetraaquadichloridochromium(III)chloride dihydrate

Topic: Chemistry, Coordination Compounds, JEE Main
Complexes ML5 of metals Ni and Fe have ideal square pyramidal and trigonal bipyramidal geometries, respectively. The sum of the 90o,120o and 180o L-M-L angles in the two complexes is_______

Options:

The correct answer is:
20

Solution:
∠120°=3;∠90°=6;∠180°=1⟹ Total =10
∠90°=8;∠180°=2⟹ Total =10

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2020 (08 Jan Shift 2)
Components of buffer solution are $0.1 \mathrm{M} \mathrm{HCN}$ and. $0.2 \mathrm{M} \mathrm{NaCN}$. What is the $p H$ of the solution

Options:
A 9.61
B 6.15
C 2.0
D 4.2

The correct answer is:
9.61

Solution:
$\begin{aligned} & p H=p K_a+\log \left[\frac{\text { salt }}{\text { acid }}\right] \\ & 9.30+\log \left[\frac{0.2}{0.1}\right]=9.30+0.3010=9.6\end{aligned}$

Topic: Chemistry, Ionic Equilibrium, JEE Main
Compound (A)  → 2 mol of  HIO 4 mol of glyoxalic acid. The compound 'A' is

Options:
A
B
C
D

The correct answer is:

Solution:
Here the compound 'A' is
Glyoxalic acid is (OHC-COOH) . Proceed reverse

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Compound A, C8H9Br , gives a white precipitate when warmed with alcoholic AgNO3 . Oxidation of A gives an acid B, C8H6O4. B easily forms anhydride on heating. Identity the compound A.

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
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