question_id string | question string | question_images list | option_1 string | option_2 string | option_3 string | option_4 string | correct_option int64 | numerical_answer string | solution string | solution_images list | subject string | topic string | subtopic string | difficulty string | question_type string | has_image bool | exam string | source_paper string |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
CH-28-Q3 | A solution of Na2CO3 is added drop by drop to one litre of a
solution containing 10-4 mole of Ba2+ and 10-5 mole of Ag+. If Ksp for
BaCO3 is 8.1 × 10-9 and Ksp for Ag2CO3 is 6.9 × 10-12, then which is
not true ? Assume no hydrolysis of CO32- ion. | [] | No precipitate of BaCO3 will appear until
$\left\lbrack {CO}_{3}^{2 -} \right\rbrack$ reaches 8.1 × 10-5 mol
per litre. | A precipitate of Ag2CO3 will appear when
$\left\lbrack {CO}_{3}^{2 -} \right\rbrack$ reaches 6.9 × 10-5 mol
per litre. | No precipitate of Ag2CO3 will appear until
$\left\lbrack {CO}_{3}^{2 -} \right\rbrack$ reaches 6.9 × 10-2 mole
per litre. | BaCO3 will be precipitated first. | 2 | null | For precipitation of Ag2CO3
$\left\lbrack {CO}_{3}^{2 -} \right\rbrack = \frac{K_{SP}}{\left\lbrack {Ag}^{+} \right\rbrack^{2}} = \frac{6.9 \times 10^{- 12}}{\left\lbrack 10^{- 5} \right\rbrack^{2}} = 6.9 \times 10^{- 2}$
and for precipitation of BaCO | [] | Chemistry | Physical Chemistry | Ionic equilibrium | Moderate | single_correct | false | JEE Main | CH-28 Chemistry Paper FST 16 Jan.docx |
CH-21-Q3 | Best method of preparing alkyl chloride is | [] | $ROH + SOCl_{2} \rightarrow$ | $ROH + PCl_{2} \rightarrow$ | $ROH + PCl_{3} \rightarrow$ | $ROH + HCl\overset{anhy\ ZnCl_{2}}{\rightarrow}$ | 1 | null | The chlorination of alcohol by $SOCl_{2}$ (thionyl chloride) is
the best method for the preparation of alkyl halides as in this method
all the other product are gaseous and thus halides are obtained on quite
pure state
gy $SOCl_{2}$ ¼ fFk;ksukby DyksjkbM½ n~okjk vYdksgy bl lHkh
gSykbM~l dkQh "kqn~/k voLFkk esa izkIr gk... | [] | Chemistry | Halogen | Preparation of Halogen containing compounds | Moderate | single_correct | false | JEE Main | CH-21 Chemistry Paper 19 Dec. Eng Hindi.docx |
CH-27-Q18 | The products A and B in the reaction
are given by the set | [
"images/image15.png"
] | CH3--CH2--CH(OH)--CH3 and CH3--CH2--CHCH2 | CH3--CHCH--CH3 and CH3--CH2--CH2--CH2OH | CH3--CHCH--CH3 and CH3--CH2--CHCH2 | CH3--CH2--CH2--CH2OH and CH3--CH2--CH CH2 | 3 | null | [IMAGE] alc. KOH gives elimination reaction hence the product A and B are
alc. KOH mUewyu | [
"images/image15.png",
"images/image16.png",
"images/image17.png"
] | Chemistry | Organic Chemistry | Hydrocarbon | Moderate | single_correct | true | JEE Main | CH-27 Chemistry Paper 2 15 Jan.docx |
CH-21-Q24 | $1\ g$of complex $\lbrack Cr(H2O)5Cl\rbrack Cl2.H2O$ was passed
through a cation exchanger to produce $HCl$. The acid liberated was
diluted to 1 litre. What will be the molarity of acid solution
?
ls ladqy $\lbrack Cr(H2O)5Cl\rbrack Cl2.H2O$ | [] | null | 0.0075 | 1 mole of complex will give two moles of $Cl–$ ion, i.e. 2 mole
$HCl$
ladqy ds 1 2 eksy $Cl–$ vFkkZr~ $HCl$ ds 2
eksy
$\left\lbrack Cr\left( H_{2}O \right)_{5}Cl \right\rbrack{Cl}_{2} \cdot H_{2}O \rightleftharpoons \left\lbrack Cr\left( H_{2}O \right)_{5}Cl \right\rbrack^{+ 2} + 2{Cl}^{-}$
1 g
mole of HCl$= \frac{2 \t... | [] | Chemistry | Coordination Compounds | Werner's Theory | Moderate | numerical | false | JEE Main | CH-21 Chemistry Paper 19 Dec. Eng Hindi.docx | ||||
CH-07-Q14 | Which one of the following is known as broad spectrum
antibiotics? | [] | Streptomycin | Ampicillin | Chloramphenicol | Penicillin G | 3 | null | Chloramphenicol is broad spectrum antibiotic used in the treatment
of typhoid, dysentery,
acute fever. | [] | Chemistry | Organic Chemistry | Chemistry in everyday life | Easy | single_correct | false | JEE Main | CH-07 Chemistry Paper 17 October.docx |
CH-09-Q22 | In XeO_3 and XeF_6 the oxidation state of Xe is | [] | null | 6 | The oxidation state of Xe in both XeO_3 and XeF_6 is +6
[IMAGE] x = +6 x = +6 | [
"images/image44.png",
"images/image45.png",
"images/image46.png",
"images/image47.png"
] | Chemistry | Physical Chemistry | Redox Reaction | Easy | numerical | true | JEE Main | CH-09 Chemistry Paper 26 October.docx | ||||
CH-15-Q25 | Find out the percentage of the reactant molecules crossing over
the activation energy barrier at 325 K, given that DH325 = 0.12 kcal,
Ea(b) = + 0.02 kcal. | [] | null | 80.65 | Given that, $\Delta H = 0.12 \times 10^{3}cal$
$E_{a(b)} = 0.02 \times 10^{3}cal$ $(E_{a}Never\ negative)$
$\because\Delta H = E_{a(f)} - E_{a(b)}$
$\therefore E_{a(f)} = 0.12 \times 10^{3} + 0.02 \times 10^{3}cal = 0.14 \times 10^{3}cal$ of molecule crossing over the barrier
$$= 100 \times e^{- E_{a}(f)/RT} = 100 \tim... | [] | Chemistry | Chemical Kinetics | Temperature dependence of rate | Tough | numerical | false | JEE Main | CH-15 Chemistry Paper 11 Nov..docx | ||||
CH-24-Q28 | Calculate in a 0.2 M solution of dichloroacetic acid
(K_a = 5 × 10^-2) that also contains 0.05 M H_2SO_4. Report your
answer after multiplying it by 400. | [] | Expecting α < < 1, 5×10^-2 = $\frac{0.2\alpha \times 0.1}{0.2}$
$\therefore$ a = 0.5 (not negligible)
So, solve quadratic | null | 60 | $\begin{matrix}
\ HA\ \ \ \ \ \ \ \ \ \ \ \ \rightleftharpoons \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ A^{-}\ \ \ \ + \ \ \ \ H^{+}
\end{matrix}$
t = 0 0.2 0.1
t = eq 0.2(1 - α) 0.2 α 0.1 + 0.2 α
$K_{a} = \frac{\left\lbrack H^{+} \right\rbrack\left\lbrack A^{-} \right\rbrack}{\lbrack HA\rbrack} = \frac{0.2\alpha(0.1 + 0.2\alpha)... | [] | Chemistry | Physical Chemistry | Ionic Equilibrium | Tough | numerical | false | JEE Main | CH-24 Chemistry Paper 31 Dec..docx | |||
CH-01-Q8 | The element having second highest ionization enthalpy in 3d-series. | [] | Cr | Mn | Fe | Ti | null | null | After removal one electron of Cr, it achieve stable half filled
configuration (3d5) which is more stable than other electronic
configuration, hence its second ionisation energy value will be high
other than. | [] | Chemistry | single_correct | false | JEE Main | CH-01 Chemistry Paper 1.docx | |||
CH-20-Q10 | The major product obtained when $Br_{2}/Fe$ is treated with | [
"images/image6.png"
] | 1 | null | The phenyl ring having H-N< group is activated while another
one is deactivated due to
[IMAGE], so electrophilic aromatic bromination0
will occur at para position with respect to H - N < group inactivated
ring. | [
"images/image11.png"
] | Chemistry | Organic Chemistry | REACTION MECHANISM | Tough | single_correct | true | JEE Main | CH-20 Chemistry Paper 12 Dec..docx | ||||
CH-13-Q24 | An aqueous solution of a weak monobasic acid containing 0.1 g
in 21.7g of water freezes at 272.813 K. If the value of K_f for water
is 1.86 k/m, what is the molecular mass of the monobasic acid. | [] | null | 60 | [
"images/image42.png"
] | Chemistry | Physical Chemistry | Moderate | numerical | true | JEE Main | CH-13 Chemistry Paper 4 Nov..docx | ||||||
CH-16-Q16 | How many millilitres of 0.1N H2SO4 solution will be required for
complete reaction with a solution containing 0.125 g of pure Na2CO3 | [] | 23.6 mL | 25.6 mL | 26.3 mL | 32.6 mL | 1 | null | $M.\ eq.\ ofH_{2}{SO}_{4} = m.\ eq.\ of{Na}_{2}{CO}_{3}$
$0.1 \times \frac{V}{1000} = \frac{0.125}{106} \times 2\ $
$V = 23.6mL$ | [] | Chemistry | Equivalent Concept | Equivalent Concept for Acid Base Titration and Precipitation | Easy | single_correct | false | JEE Main | CH-16 Chemistry Paper 25 Nov. Fisrt.docx |
CH-05-Q8 | In which of the following resonance is not possible ? | [] | 4 | null | Delocalization is possible in the cases where atoms with lone
pair & vacant p or d-orbital are adjacent. | [] | Chemistry | Organic Chemistry | Some Basic Principles and Techniques | Moderate | single_correct | false | JEE Main | CH-05 Chemistry Paper 10 October.docx | ||||
CH-27-Q22 | Chlorination of toluene in the presence of light and heat
followed by treatment with aqueous NaOH gives
tyh; NaOH ds lkFk mipkj ds ckn izdk"k | [] | o-cresol | p-cresol | 2, 4-dihydroxy toluene | Benzyl alcohol | 4 | null | [
"images/image29.png"
] | Chemistry | Organic Chemistry | Hydrocarbon | Moderate | single_correct | true | JEE Main | CH-27 Chemistry Paper 2 15 Jan.docx | |
CH-19-Q3 | Which is correct about the cyclic silicate | [] | The value of n is 12 | each Si atom is bonded with three oxygen atoms | each oxygen atom is bonded with two Si atoms | all the above are correct. | 1 | null | [IMAGE] General formula of cyclic silicates is [Sin O3n]2n- ¼ | [
"images/image5.png",
"images/image6.png",
"images/image7.png",
"images/image8.png"
] | Chemistry | Chemical Bonding | Multicentered species | Moderate | single_correct | true | JEE Main | CH-19 Chemistry Paper 28 Nov. Fourth.docx |
CH-12-Q13 | After partial roasting, the sulphide of copper is reduced by | [] | Reduction by carbon | Electrolysis | Self-reduction | Cyanide process | 3 | null | Self reduction: - Reduction of oxide ore of a metal by its own
sulphide
2Cu_2O + Cu_2 S → 6Cu + SO_2 | [] | Chemistry | Inorganic Chemistry | Metallurgy | Easy | single_correct | false | JEE Main | CH-12 Chemistry Paper 2 Nov..docx |
CH-11-Q19 | An engine operating between 150ºC and 25ºC takes 500 J heat
from a higher temperature reservoir if there are no frictional losses,
then work done by engine is | [] | 147.7 J | 157.75 J | 165.85 J | 169.95 J | 1 | null | $T_{2} = 150 + 273 = 423K$
$T_{1} = 25 + 273 = 298K$
$Q = 500K$
$\frac{W}{Q} = \frac{T_{2} - T_{1}}{T_{2}};W = 500\left( \frac{423 - 298}{423} \right) = 147.7J$ | [] | Chemistry | Physical Chemistry | Thermodynamics | Moderate | single_correct | false | JEE Main | CH-11 Chemistry Paper-2 1Nov New Jee main.docx |
CH-01-Q10 | In an atom an electron is moving with a speed equal to 600 m/s with
an accuracy upto 0.001 with what certainity its position can be
determined. | [] | 9.75 × 10^-3 m | 5.2 × 10^-4 m | 2 × 10^-3 m | 0.8 ×
10^-3 m | null | null | v = 600 m/sec | [
"images/image20.png",
"images/image21.png",
"images/image22.png"
] | Chemistry | single_correct | true | JEE Main | CH-01 Chemistry Paper 1.docx | |||
CH-23-Q8 | Which method of purification is represented by the equations ? | [
"images/image27.png"
] | Cupellation | Poling | Van Arkel | Zone refining | 3 | null | Purification of Ti and Zr are performed by Van Arkel method as
given in the question. | [] | Chemistry | Inorganic Chemistry | Metallurgy | Moderate | single_correct | true | JEE Main | CH-23 Chemistry Paper 31 Dec. FST.docx |
CH-23-Q16 | Which of the following reactions lead to chemical inertness | [] | Lead with dilute H2SO4 | Lead with conc. HCl | Aluminium with oxygen | All of above reactions | 4 | null | (1)
$Pb + H_{2}{SO}_{4} \longrightarrow {PbSO}_{4}\ (layer) + H_{2} \uparrow$
(2)
$Pb + Conc.HCl \longrightarrow {PbCl}_{2}(\ Coating) + H_{2} \uparrow$
(3) $2Al + 3/2O_{2} \longrightarrow {Al}_{2}O_{3}(s)$ | [] | Chemistry | Inorganic Chemistry | P-block | Moderate | single_correct | false | JEE Main | CH-23 Chemistry Paper 31 Dec. FST.docx |
CH-13-Q17 | The reaction of C_2H_5OH with H_2SO_4 does not
give | [] | Ethylene | Diethyl ether | Acetylene | Ethyl hydrogen | 3 | null | [
"images/image28.png"
] | Chemistry | Organic Chemistry | Alcohols, Phenols and Ethers | Moderate | single_correct | true | JEE Main | CH-13 Chemistry Paper 4 Nov..docx | |
CH-02-Q15 | What is the final product of the following reaction? | [
"images/image94.png"
] | null | null | [
"images/image99.png",
"images/image96.png"
] | Chemistry | single_correct | true | JEE Main | CH-02 Chemistry paper 2.docx | ||||||||
CH-16-Q1 | 20 ml of a gaseous hydrocarbon requires 100 ml of O_2 for
complete combustion and produces 60 ml of CO_2 gas. Determine the
molecular formula of the hydrocarbon. | [] | 3 | null | Let the molecular formula of the hydrocarbon is
[IMAGE] From ques. Vol. of CO_2 formed = 60 ml
20 x=60
x=3
vol. of O_2 used = 100 ml
Hence the hydrocarbon is [IMAGE] | [
"images/image5.png",
"images/image6.png",
"images/image7.png",
"images/image3.png"
] | Chemistry | Eudiometry | Eudiometry | Tough | single_correct | true | JEE Main | CH-16 Chemistry Paper 25 Nov. Fisrt.docx | ||||
CH-20-Q14 | Which set of quantum numbers are not possible from the following | [] | $n = 3,l = 2,m = 0,s = - \frac{1}{2}$ | $n = 3,l = 2,m = - 2,s = - \frac{1}{2}$ | $n = 3,l = 3,m = - 3,s = - \frac{1}{2}$ | $n = 3,l = 0,m = 0,s = - \frac{1}{2}$ | 3 | null | If $n = 3$ then, $l = 0$ to $n = - 1$&$m = - l\ to\ + l$ | [] | Chemistry | Inorganic Chemistry | Quantum number | Easy | single_correct | false | JEE Main | CH-20 Chemistry Paper 12 Dec..docx |
CH-02-Q9 | [IMAGE] of the solution in the anode compartment of
the following cell at [IMAGE] is x when
[IMAGE] Find out x | [
"images/image53.png",
"images/image54.png",
"images/image55.png",
"images/image56.png"
] | 1 | 2 | 3 | 4 | null | null | [
"images/image57.png",
"images/image58.png",
"images/image59.png",
"images/image60.png",
"images/image61.png",
"images/image62.png"
] | Chemistry | single_correct | true | JEE Main | CH-02 Chemistry paper 2.docx | ||||
CH-12-Q15 | Which compound does not form iodine form with alkali and iodine? | [] | Acetone | Ethanol | Diethyl ketone | Isopropyl alcohol | 3 | null | [IMAGE] Yellow ppt | [
"images/image2.png",
"images/image3.png"
] | Chemistry | Organic Chemistry | HALOALKANES & HALOARENES | Moderate | single_correct | true | JEE Main | CH-12 Chemistry Paper 2 Nov..docx |
CH-01-Q9 | For a reaction: 2A(g) [IMAGE] 5B(g), if the rate
constant for disappearance of is 4 × 10-2 M sec-1 then calculate
rate of formation of B after 10 seconds from start of the reaction. If
initially 10 M of A is taken. | [
"images/image17.png"
] | 4 × 10-2 Msec-1 | 0 | 0.1 Msec-1 | 2 × 10-1 Msec-1 | null | null | 2A (g) → 5B (g) | [
"images/image18.png",
"images/image19.png"
] | Chemistry | single_correct | true | JEE Main | CH-01 Chemistry Paper 1.docx | |||
CH-28-Q21 | Rate of the reaction is fastest when Z is | [] | Cl | ${OCOCH}_{3}$ | ${OC}_{2}H_{5}$ | ${NH}_{2}$ | 1 | null | Among the given option Cl- is the best leaving group hence the
rate of reaction will be fastest in case of RCOCl. | [] | Chemistry | Organic Chemistry | Organic Reaction mechanisms-I | Moderate | single_correct | false | JEE Main | CH-28 Chemistry Paper FST 16 Jan.docx |
CH-11-Q3 | Which of the following is the most powerful oxidizing agent | [] | F_2 | Cl2 | Br_2 | I_2 | 1 | null | Fluorine is a most powerful oxidizing agent because it consist
of E^0 = + 2.5 volt. | [] | Chemistry | Physical Chemistry | Redox Reaction | Easy | single_correct | false | JEE Main | CH-11 Chemistry Paper-2 1Nov New Jee main.docx |
CH-18-Q18 | A gas obeys the equation of state P(V - b) = RT (The parameter
b is a constant). The slope for an isochore will be | [] | Negative | Zero | R/(V - b) | 3 | null | $P(V - b) = RT;P = \frac{RT}{(V - b)}$
$P = (\frac{R}{(V - b)})T + 0$ | [] | Chemistry | Real Gas | Vander waal equation and virial equation of state | Easy | single_correct | false | JEE Main | CH-18 Chemistry Paper 27 Nov. Third.docx | |
CH-06-Q20 | Given that in H-atom, the transition energy for n = 1 to n = 2
is 10.2 eV, the energy for the same transition in Be^3+ is | [] | 20.4 eV | 30.6 eV | 40.8 eV | None of these | 4 | null | $\Delta$E = z^2 x 10.2 ev
=16 x 10.2 ev
=162.3 ev | [] | Chemistry | Physical Chemistry | Atomic Structure | Moderate | single_correct | false | JEE Main | CH-06 Chemistry Paper 14 October.docx |
CH-24-Q9 | (a)$\ {CuSO}_{4}.5H2O(s) \rightleftharpoons {CuSO}_{4} \cdot 3H_{2}O(s) + 2H_{2}O(g)$
K_P = 4 × 10^-4 atm^2
(b)$\ {Na}_{2}{SO}_{4} \cdot 10H_{2}O(s) \rightleftharpoons {Na}_{2}{SO}_{4.5}H_{2}O(s) + 5H_{2}O(g)$
K_P = 2.43 × 10^-8 atm^5
(c)
${Na}_{2}S_{2}O_{3} \cdot 5H_{2}O(s) \rightleftharpoons {Na}_{2}S_{2}O_{3} \cdot ... | [] | c > b > a V.P.
c > b > a R.H. | c < b < a V.P.
c > b > a R.H. | a > c > b V.P.
a > c > b R.H. | a > c > b V.P.
a < c < b R.H. | 1 | null | [IMAGE] = 2 × 10^-2
[IMAGE] = 243 × 10-10 = [IMAGE] = 3 × 10^-2
[IMAGE] = 6.4 × 10^-5 = [IMAGE] = 4 × 10^-2
Order of V.P c > b > a
and same as c > b > a | [
"images/image13.png",
"images/image14.png",
"images/image15.png",
"images/image16.png",
"images/image17.png",
"images/image18.png",
"images/image19.png",
"images/image20.png"
] | Chemistry | Physical Chemistry | Chemical Equilibrium | Moderate | single_correct | true | JEE Main | CH-24 Chemistry Paper 31 Dec..docx |
CH-25-Q26 | Consider the following statements: A hydrocarbon of molecular
formula C5H10 is a
1. monosubstituted alkene
2. disubstituted alkene
3. trisubstituted alkene
How many statement(s) is (are) correct? | [] | null | 3 | [] | Chemistry | Organic Chemistry | Hydrocarbon | Moderate | numerical | false | JEE Main | CH-25 Chemistry Paper 7 Jan.docx | |||||
CH-26-Q19 | When ethyl chloride and n-propyl chloride undergoes wurtz
reaction which is not obtained | [] | n-butane | n-pentane | n-hexane | isobutene
tks izkIr ugha gksrk | 4 | null | [IMAGE] (a) [IMAGE] (b) [IMAGE] (c) [IMAGE] Isobutane is not obtained. | [
"images/image19.png",
"images/image20.png",
"images/image21.png",
"images/image22.png"
] | Chemistry | Organic Chemistry | Hydrocarbon | Moderate | single_correct | true | JEE Main | CH-26 Chemistry Paper 1 15 Jan.docx |
CH-06-Q11 | ${CH}_{3}C \equiv {CCH}_{3}\overset{Na,{NH}_{3}(I)}{\longrightarrow}Q\frac{dil\ Cold\ }{{KMO}_{4}} \longrightarrow R$
If Q is an unsaturated hydrocarbon then products R is/are | [
"images/image11.png",
"images/image12.png",
"images/image13.png",
"images/image14.png",
"images/image15.png"
] | i and ii | ii and iii | Only v | Only iv | 2 | null | ${CH}_{3}C = {CCH}_{3}\overset{Na,{NH}_{3}(l)}{\rightarrow}$ | [
"images/image16.png"
] | Chemistry | Organic Chemistry | Hydrocarbons | Moderate | single_correct | true | JEE Main | CH-06 Chemistry Paper 14 October.docx |
CH-14-Q9 | When calomel reacts with [IMAGE] we get | [
"images/image35.png"
] | 1 | null | [
"images/image40.png"
] | Chemistry | Inorganic Chemistry | D & F -Block Elements | Moderate | single_correct | true | JEE Main | CH-14 Chemistry Paper 7 Nov..docx | |||||
CH-02-Q13 | A reaction follows the given concentration reaction (M) v/s time
graph. The instantaneous rate for this reaction at 20 seconds will be | [
"images/image82.png"
] | null | null | [
"images/image87.png",
"images/image88.png"
] | Chemistry | single_correct | true | JEE Main | CH-02 Chemistry paper 2.docx | ||||||||
CH-16-Q4 | A compound possess 8 sulphur by mass. The least molecular mass
is | [] | 300
(B) 400
(C) 155
(D) 355 | 2 | null | 8 sulphur by mass means - 8 g sulphur is present in 100 g
solid.
[IMAGE] 32 g sulphur (1 mole atom) will be present in
Q compound must be having at least one atom of
[IMAGE] min. mol. mass = 400 g. | [
"images/image15.png",
"images/image16.png",
"images/image15.png",
"images/image17.png"
] | Chemistry | Mole concept | Empirical Formula, % Composition of a given compound by mass, | Moderate | single_correct | true | JEE Main | CH-16 Chemistry Paper 25 Nov. Fisrt.docx | |||
CH-17-Q15 | KCl can be used in salt bridge as electrolyte in which of the
following cells? | [] | Zn | ZnCl2 || AgNO3 | Ag | Pb | Pb(NO3)2 || Cu(NO3)2 | Cu | Cu | CuSO4 || AuCl3 | Au | Fe | FeSO4 || Pb(NO3)2 | Pb | 3 | null | KCl can make precipitate with AgNO3, Pb(NO3)2 so can't be used
along these electrolyte. | [] | Chemistry | Electrochemistry | Electrochemical series & its Applications | Easy | single_correct | false | JEE Main | CH-17 Chemistry Paper 26 Nov. Second.docx |
CH-28-Q35 | The number of geometric isomers that can exist for square planar
[Pt (Cl) (py)
$\left. \ \left( {NH}_{3} \right)\left( {NH}_{2}OH \right) \right\rbrack^{+}$
is (py = pyridine ) | [] | null | 3 | The complex is of the type
M = metal
a, b, c, d = Monodentate ligands. | [
"images/image15.png"
] | Chemistry | Inorganic Chemistry | Coordination Compounds | Moderate | numerical | true | JEE Main | CH-28 Chemistry Paper FST 16 Jan.docx | ||||
CH-20-Q12 | 2-Methylbutane on reacting with bromine in the presence of
sunlight gives mainly | [] | 1 -bromo-2-methylbutane | 2 -bromo-2-methylbutane | 2 -bromo-3-methylbutane | 1 -bromo-3-methylbutane | 2 | null | [
"images/image12.png"
] | Chemistry | Organic Chemistry | ALKANE | Easy | single_correct | true | JEE Main | CH-20 Chemistry Paper 12 Dec..docx | |
CH-13-Q16 | When Phenol is heated with phthalic anhydride in concentrated
sulphuric acid and the hot reaction mixture is poured into a dilute
solution of sodium hydroxide, the product formed is | [] | Alizarin | Methyl orange | Fluorescein | Phenolphthalein | 4 | null | [
"images/image27.png"
] | Chemistry | Organic Chemistry | Alcohols, Phenols and Ethers | Moderate | single_correct | true | JEE Main | CH-13 Chemistry Paper 4 Nov..docx | |
CH-22-Q5 | Assertion: Among the carbon allotropes, diamond is an insulator,
whereas, graphite is a good conductor of electricity. Reason:
Hybridization of carbon in diamond and graphite aresp^3 and sp^2,
respectively. | [] | Both assertion and reason are correct, and the reason is the
correct explanation for the assertion. | Both assertion and reason are correct, but the reason is not the
correct explanation for the assertion. | Assertion is incorrect statement, but the reason is correct. | Both assertion and reason are incorrect. | 1 | null | Diamond has carbon atoms that are bonded via a covalent bond to
four other carbon atoms in a tetrahedral manner.
The$a = \frac{a_{0}}{\sqrt{2}} = \frac{bt}{m}$
hybridization of carbon atom in diamond is
$\frac{a_{0}}{\sqrt{2}} = \frac{bt}{m}$. In
$= \frac{10^{- 2}t}{0.1} = \frac{t}{10}$
diamond, carbon utilizes its unp... | [] | Chemistry | Inorganic Chemistry | Chemical Bonding | Moderate | single_correct | false | JEE Main | CH-22 Chemistry Paper 19 Dec. Eng.docx |
CH-07-Q21 | 20 volume $H_{2}O$ solution has a strength of about in gm/ml | [] | null | 6.71 | $\because 22.4\text{ litre }O_{2}$at N.T.P. obtained by 68
gmof$H_{2}O_{2}$
$\therefore 1\text{ litre }O_{2}$at N.T.P. obtained by
$\frac{68}{22.4} \times 20gm\text{ of }H_{2}O_{2} = 60.71gm\text{ of }H_{2}O_{2}$
$\therefore 1000mlO_{2}$ at N.T.P. obtained by $= 60.71\ gm\ of\ H_{2}O$
$\therefore 100mlO_{2}$ N.T.P. obt... | [] | Chemistry | Inorganic Chemistry | Hydrogen And Its Compound | Tough | numerical | false | JEE Main | CH-07 Chemistry Paper 17 October.docx | ||||
CH-18-Q22 | Which of the following groups (attached with benzene ring)
show + M effect? | [
"images/image36.png",
"images/image37.png"
] | null | 6 | [IMAGE] have + M group. | [
"images/image38.png",
"images/image39.png"
] | Chemistry | Organic Chemistry | Mesomeric Effect | Moderate | numerical | true | JEE Main | CH-18 Chemistry Paper 27 Nov. Third.docx | ||||
CH-27-Q8 | In case of nitrogen, NCl3 is possible but no NCl5 while in case
of phosphorus, PCl3 as well as PCl5 are possible. It is due to | [] | Availability of vacant d-orbital in P but not in N. | Lower electronegativity of P then N. | Lower tendency of H bond formation in P than N. | Occurrence of P in solid while N in gaseous state at room
temperature.
ukbVªkstu dh fLFkfr esa] NCl3 lEHko fdUrq NCl5 ugha] tcfd QkWLQksjl
ds fLFkfr esa PCl3 rFkk PCl5 nksuks lEHko gSA;g bl dkj.k gksrk | 1 | null | In phosphorous the vacant 3d-orbitals are available. So it can
increase its covalence beyond three. | [] | Chemistry | Inorganic Chemistry | P-block Halogen | Moderate | single_correct | false | JEE Main | CH-27 Chemistry Paper 2 15 Jan.docx |
CH-28-Q13 | Among the following compounds, the most acidic is | [] | p-nitrophenol | p-hydroxybenzoic acid | o-hydroxybenzoic acid | p-toluic acid | 1 | null | [IMAGE] Due to intramolecular hydrogen bonding in conjugate base of
o-Hydroxybenzoic acid, it is strongest acid. | [
"images/image9.png",
"images/image10.png"
] | Chemistry | Organic Chemistry | Organic Chemistry | Moderate | single_correct | true | JEE Main | CH-28 Chemistry Paper FST 16 Jan.docx |
CH-04-Q7 | Fluoxymesterone, C_20H_29FO_3, is an anabolic steroid. A solution
is prepared by dissolving 10.0 mg of the steroid in 500 mL of water. How
many moles of Fluoxymesterone are present in 1 mL of solutions.
C_20H_29FO_3 500 mL
10.0 mg 1 mL | [] | 1.16 × 10-10 | 1.19 × 10-17 (3) 5.95 × 10-8 | 2.38 ×
10-11 | null | 3 | Number of moles in 500 mL [IMAGE] 500 mL [IMAGE] Number of moles in 1 mL [IMAGE] =5.95 × 10-8.
1 mL [IMAGE] 5.95 × 10-8. | [
"images/image13.png",
"images/image13.png",
"images/image14.png",
"images/image14.png"
] | Chemistry | Physical Chemistry | Mole Concept | Tough | numerical | true | JEE Main | CH-04 Paper 02 Chemistry with Hindi 22-8-2020.docx | |
CH-21-Q5 | Ethanol is prepared industrially by | [] | Hydration of ethylene | Fermentation of sugars | Both the above | None of these,FkukWy vkS|ksfxd:i | 3 | null | Hydration of ethylene,Fkyhu
${CH}_{2} = {CH}_{2} + \overset{+}{HH}\overset{-}{{SO}_{4}} \rightarrow {CH}_{3} - {CH}_{2} - {HSO}_{4}$
${CH}_{3} - {CH}_{2}{HSO}_{4}\frac{H_{2}O}{\ Boil\ } \rightarrow {CH}_{3} - {CH}_{2} - OH + H_{2}{SO}_{4}$
Fermentation of sugars
"kdZjk dk fd.ou
$C_{12}H_{22}O_{11} + H_{2}O\overset{Int... | [] | Chemistry | Alcohol, Phenol and Ethers | Preparation of alcohol, Phenol and Ethers | Moderate | single_correct | false | JEE Main | CH-21 Chemistry Paper 19 Dec. Eng Hindi.docx |
CH-14-Q17 | [IMAGE] product in the reaction is | [
"images/image75.png"
] | None of these | 1 | null | [
"images/image79.png"
] | Chemistry | Organic Chemistry | Aldehydes, Ketones & Carboxylic Acids | Moderate | single_correct | true | JEE Main | CH-14 Chemistry Paper 7 Nov..docx | ||||
CH-06-Q14 | [IMAGE] Compound (A) is | [
"images/image21.png"
] | 4 | null | [
"images/image26.png"
] | Chemistry | Organic Chemistry | Hydrocarbons | Tough | single_correct | true | JEE Main | CH-06 Chemistry Paper 14 October.docx | |||||
CH-07-Q3 | An unknown gaseous hydrocarbon and oxygen gas are mixed in volume
ratio 1: 7 and exploded. The resulting mixture upon cooling occupied a
volume of V mL, half of which got absorbed in aq. KOH (absorbs $CO_{2}$)
and the remaining half was absorbed in alkalline pyrogallol solution
(absorbs $O_{2}$). Assuming all volumes t... | [] | The molecular formula of hydrocarbon can be $C_{2}H_{6}$ | The resulting mixture obtained after cooling contains 50 $O_{2}$
and 50 CO by mole. | The molecular formula of hydrocarbon can be $C_{3}H_{4}$ | Molecular mass of hydrocarbon is definitely less than 28 u | 3 | null | $CxHy + \left( x + \frac{y}{4} \right)O_{2} \longrightarrow {xCO}_{2} + \frac{y}{2}H_{2}O\mathcal{(l)}$
$i$ $a$ $\text{7a}$ 0
$f$ 0 $7a - \left( x + \frac{y}{4} \right)a$ xa
$\text{Given: }7a - \left( x + \frac{y}{4} \right)a = xa$
$\begin{aligned}
\therefore & 8x + y = 28
\end{aligned}$
Only $C_{3}H_{4}$ hydrocarbon s... | [] | Chemistry | Physical Chemistry | Gaseous state | Moderate | single_correct | false | JEE Main | CH-07 Chemistry Paper 17 October.docx |
CH-02-Q18 | The decreasing order of the rate of nitration of the following
compounds is:
(i) Benzene [IMAGE] (ii) Hexadeutero benzen [IMAGE] (iii) Nitrobenzene
(iv) Chlorobenzene | [
"images/image113.png",
"images/image113.png"
] | i > ii > iii > iv | i > ii > iv > iii | i = ii > iv > iii | i = ii > iii > iv | null | null | Rate of Nitration [IMAGE] Rate of E.S.R | [
"images/image114.png",
"images/image115.png",
"images/image116.png"
] | Chemistry | single_correct | true | JEE Main | CH-02 Chemistry paper 2.docx | |||
CH-04-Q20 | has four choices | [] | , | , | , | out of which ONLY ONE is correct] | null | 3 | [] | Chemistry | numerical | false | JEE Main | CH-04 Paper 02 Chemistry with Hindi 22-8-2020.docx | ||||
CH-25-Q1 | C_6 H_12 (P) has two types of alkenes that can be reduced to
one type of C_6H_14 (Q). Q is:
C_6 H_12 (P) nks rjg dh,Ydhu tks fd,d rjg,Ydsu C_6H_14
(Q) | [] | 2 | null | [
"images/image5.png"
] | Chemistry | Organic Chemistry | Hydrocarbon | Moderate | single_correct | true | JEE Main | CH-25 Chemistry Paper 7 Jan.docx | |||||
CH-28-Q20 | Hydrogen peroxide in its reaction with KIO4 and NH2OH
respectively, is acting as a | [] | reducing agent, oxidising agent | reducing agent, reducing agent | oxidising agent, oxidising agent | oxidising agent, reducing agent | 1 | null | ${KIO}_{4} + H_{2}O_{2} \rightarrow {KIO}_{3} + H_{2}O + O_{2}$
$H_{2}O_{2}$ acts as a reductant
$2{NH}_{2}OH + H_{2}O_{2} \rightarrow N_{2} + 4H_{2}O$
$H_{2}O_{2}$ acts as an oxidant. | [] | Chemistry | Inorganic Chemistry | p-Block (Nitrogen and Oxygen family) | Moderate | single_correct | false | JEE Main | CH-28 Chemistry Paper FST 16 Jan.docx |
CH-19-Q24 | In the scheme given below, the total number of intermolecular
aldol condensation products formed from is | [
"images/image23.png"
] | null | 1 | [
"images/image24.png"
] | Chemistry | Carbonyl Compound (Aldehyde and Ketone) | Aldol condensation | Moderate | numerical | true | JEE Main | CH-19 Chemistry Paper 28 Nov. Fourth.docx | |||||
CH-21-Q12 | A hydrogen sample is prepared in a particular excited state.
Photons of energy 2.55 eV get absorbed into the sample to take some of
the electrons to a further excited state B. Find orbit numbers of the
states A and B. Given the allowed energies of hydrogen atom:
$E_{1} = - 13.6eV,E_{2} = - 3.4eV,E_{3} = - 1.5eV,E_{4} =... | [] | A =2, B = 4 | A =4, B = 2 | A =2, B = 2 | A =4, B = 4 | 1 | null | Clearly,
$E_{4} - E_{2} = ( - 0.85) - ( - 3.4) = 2.55eV \Rightarrow A = 2,B = 4$
gy Li"V:i
$E_{4} - E_{2} = ( - 0.85) - ( - 3.4) = 2.55eV \Rightarrow A = 2,B = 4$ | [] | Chemistry | Atomic Structure | Bohr Model | Moderate | single_correct | false | JEE Main | CH-21 Chemistry Paper 19 Dec. Eng Hindi.docx |
CH-18-Q1 | The general formula CnH2nO2 could be for open chain | [] | diketones | carboxylic acids | diols | dialdehydes. | 2 | null | Diketones: CnH2n -2O2, Carboxylic acid: CnH2nO2,
Diols: CnH2n+2O2, Dialdehydes: CnH2n-2O2 | [] | Chemistry | IUPAC and Structural Isomerism | Chain Isomerism | Easy | single_correct | false | JEE Main | CH-18 Chemistry Paper 27 Nov. Third.docx |
CH-26-Q35 | A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some
of the CO2 is converted into CO on the addition of graphite. If the
total pressure at equilibrium is 0.8 atm, the value of K is:
dqN CO2, CO esa ifjofrZr gks tkrh gSA ij dqy nkc 0.8 atm gks
rks K | [] | null | 1.8 | CO2(g) + C(s)[IMAGE] 2CO (g)
0.5 atm
0.5-p 2p
Total pressure ¼dqy nkc½ = 0.5 - P + 2P = 0.8
P = 0.3
$K_{p} = \frac{P_{CO}^{2}}{P_{{CO}_{2}}} = \frac{(2P)^{2}}{(0.5 - P)} = \frac{(0.6)^{2}}{(0.5 - 0.3)}$
Kp = 1.8 | [
"images/image54.png"
] | Chemistry | Physical Chemistry | Chemical Equilibrium | Moderate | numerical | true | JEE Main | CH-26 Chemistry Paper 1 15 Jan.docx | ||||
CH-18-Q16 | For an electron, with n = 3 has only one radial node. The
orbital angular momentum of the electron will be | [] | 0 | $\sqrt{6}\frac{h}{2\pi}$ | $\sqrt{2}\frac{h}{2\pi}$ | $3(\frac{h}{2\pi})$ | 3 | null | Number of radial nodes
$= n\mathcal{- l -}1 = 1,n = 3.\therefore\mathcal{l} = 1$
Orbital angular momentum
$= \sqrt{\mathcal{l}(\mathcal{l} + 1)}\frac{h}{2\pi} = \sqrt{2}\frac{h}{2\pi}$ | [] | Chemistry | Atomic Structure | Quantum numbers & Electronic configuration | Moderate | single_correct | false | JEE Main | CH-18 Chemistry Paper 27 Nov. Third.docx |
CH-21-Q9 | The minimum voltage required to electrolyse of Al2O3 in the
Hall-Heroult process is
Given:$\Delta G_{f}^{\circ}\left( {Al}_{2}O_{3} \right) = - 1520kJ{mol}^{- 1}$; $\Delta G_{f}^{\circ}\left( {CO}_{2} \right) = - 394kJ{mol}^{- 1}$
If net reaction in Hall-Heroult process is:$3C + 2{Al}_{2}O_{3} \longrightarrow 4Al + 3{C... | [] | 16 | 14 | 17 | 15 | 1 | null | Net reaction in Hall-Heroult process is:
$3C + 2{Al}_{2}O_{3} \longrightarrow 4Al + 3{CO}_{2}$
or $4{AI}^{3 +}12e^{-} \longrightarrow 4Al$ number of electrons (n) = 12
$\Delta G^{\circ} = 3\Delta G_{f}^{\circ}\left( {CO}_{2} \right) - 2\Delta G_{f}^{\circ}\left( {AI}_{2}O_{3} \right)$
$$= 3 \times 394 - 2( - 1520) = 18... | [] | Chemistry | Inorganic Chemistry | Metallurgy | Tough | single_correct | false | JEE Main | CH-21 Chemistry Paper 19 Dec. Eng Hindi.docx |
CH-17-Q19 | Peptisation is | [] | Conversion of a colloidal into precipitate form | Conversion of precipitate into colloidal sol | Conversion of metal into colloidal sol by passage of electric
current | Conversion of colloidal sol into macromolecules | 2 | null | Process by which precipitate is converted into colloid is known
as peptisation. | [] | Chemistry | Surface Chemistry | Purification and Preparation of Colloid | Easy | single_correct | false | JEE Main | CH-17 Chemistry Paper 26 Nov. Second.docx |
CH-07-Q18 | When zeolite Hydrated sodium aluminium silicate) is treated with
hard water the sodium ions are exchanged with | [] | $OH^{-}$ ions | ${SO}_{4}^{2 -}$ ions | $Ca^{2 +}$ ions | $H^{+}$ ions | 3 | null | $\underset{\text{ Zeolita }}{Na_{2}Al_{2}}Si_{2}O_{8} \cdot xH_{2}O + Ca^{+ 2} \rightarrow {CaAl}_{2}{Si}_{2}O_{8} \cdot {xH}_{2}O + 2{Na}^{+}$ | [] | Chemistry | Inorganic Chemistry | Hydrogen And Its Compound | Moderate | single_correct | false | JEE Main | CH-07 Chemistry Paper 17 October.docx |
CH-25-Q24 | What will be the least molecular mass of an alkane which is
optically active?,Ydsu | [] | null | 100 | 12 × 7 + 16 × 1 = 100 | [
"images/image99.png"
] | Chemistry | Organic Chemistry | Hydrocarbon | Moderate | numerical | true | JEE Main | CH-25 Chemistry Paper 7 Jan.docx | ||||
CH-06-Q2 | Molecule AX4 have all bond angles equal and molecule is non-polar
also then which of the following conclusion is incorrect? | [] | Molecule may be tetrahedral. | Molecule may be square planar. | Central atom must have at least six valence electrons. | Central atom has either zero lone pair or two lone pairs | 3 | null | For AX_4 type molecule
If $\mu$ = 0, it is non-polar. It must be tetrahedral (or) square
planar.
Ex: CH4, SiH_4$\rightarrow \ $ tetrahedral (Zero lone pairs)
XeF_4, ICl_4$\rightarrow \ $ square planar (Two lone pairs) | [] | Chemistry | Inorganic Chemistry | Chemical Bonding | Moderate | single_correct | false | JEE Main | CH-06 Chemistry Paper 14 October.docx |
CH-19-Q25 | The numberof aldol reaction (s) that occurs in the given
transformation is | [
"images/image25.png"
] | null | 3 | [
"images/image26.png",
"images/image27.png",
"images/image28.png"
] | Chemistry | Carbonyl Compound (Aldehyde and Ketone) | Aldol condensation | Tough | numerical | true | JEE Main | CH-19 Chemistry Paper 28 Nov. Fourth.docx | |||||
CH-26-Q27 | A 0.004 M solution of Na2SO4 is isotonic with 0.010 M solution
of glucose at same temperature. The apparent percentage dissociation of
Na2SO4 is | [] | null | 75 | $\left( \pi_{obs\ } \right)_{{Na}_{2}{SO}_{4}} = \pi_{glucose\ }$
or $\frac{10}{4} = \frac{1 + 2\alpha}{1}$ or $10 = 4 + 8\alpha$
$\alpha = \frac{10 - 4}{8} = 0.75$ $\%\ of\ \alpha = 75\%$ | [] | Chemistry | Physical Chemistry | Moderate | numerical | false | JEE Main | CH-26 Chemistry Paper 1 15 Jan.docx | |||||
CH-23-Q2 | Decreasing order of boiling point of I to IV follow
Methylformate Ethylformate Iso-propylformate n-propylformate
I II III IV | [] | I > II > III > IV | III > IV > II > I | IV > III > II > I | I > II > IV > III | 3 | null | Boiling point µ molecular weight. | [] | Chemistry | Organic Chemistry | Chemistry in everyday life & POC | Easy | single_correct | false | JEE Main | CH-23 Chemistry Paper 31 Dec. FST.docx |
CH-09-Q12 | Aniline on treatment with excess of bromine water gives | [] | Aniline bromide | 0 - bromoaniline | p - bromoaniline | 2, 4, 6 - tribromoaniline | 4 | null | [
"images/image37.png"
] | Chemistry | Organic Chemistry | Organic compounds containing Nitrogen | Easy | single_correct | true | JEE Main | CH-09 Chemistry Paper 26 October.docx | |
CH-25-Q4 | [
"images/image12.png"
] | The compound is (CH3CH2)2CuLi. | The compound is CH3(CH2)5CH2Br. | The compound is (CH3CH2)2CuLi. | The compound is CH3(CH2)5CH2Br. | 1 | null | [IMAGE] CH3 - CH2 - CH2 - CH2 - CH2 - CH2 - CH2 - CH2 - CH3
n-nonane (n - uksusu) | [
"images/image13.png",
"images/image14.png"
] | Chemistry | Organic Chemistry | Hydrocarbon | Tough | single_correct | true | JEE Main | CH-25 Chemistry Paper 7 Jan.docx | |
CH-28-Q29 | $A^{-}(g) \rightarrow A^{2 +}(g)$ $\Delta H = 1100\ KJ/mol$
$A(g) \rightarrow A^{2 +}(g)$ $\Delta H = 1200\ KJ/mol$
Electron gain enthalphy of A is $P \times 10^{2}KJ/mol$. What is the
value of P ? | [] | $A(g) \rightarrow A^{2 +}(g) + 2e^{-}$ $\Delta H = 1200\ KJ/mol$.... | $A^{-}(g) \rightarrow A^{2 +}(g) + 3e^{-}$ $\Delta H = 1100\ KJ/mol$.... | Eq. | null | 1 | $A(g) + e^{-} \rightarrow A^{-}(g)$
$\Delta H = + E \cdot G \cdot E$....(1)
$A(g) \rightarrow A^{2 +}(g) + 2e^{-}$ $\Delta H = 1200\ KJ/mol$....(2)
$A^{-}(g) \rightarrow A^{2 +}(g) + 3e^{-}$ $\Delta H = 1100\ KJ/mol$....(3)
Eq. (3) = (2) - (1)
$- E \cdot G \cdot E + 1200 = 1100$
$- E \cdot G \cdot E = - 100\ KJ/mol$
$P... | [] | Chemistry | Inorganic Chemistry | Periodic table | Moderate | numerical | false | JEE Main | CH-28 Chemistry Paper FST 16 Jan.docx | |
CH-28-Q10 | The IUPAC name of neopentane is | [] | 2,2-dimethylpropane | 2-methylpropane | 2,2 -dimethylbutane | 2 -methylbutane | 1 | null | [
"images/image6.png"
] | Chemistry | Organic Chemistry | IUPAC Nomenclature | Moderate | single_correct | true | JEE Main | CH-28 Chemistry Paper FST 16 Jan.docx | |
CH-17-Q8 | The enantiomeric excess and observed rotation of a mixture
containing 6 gm of (+)-2-butanol and 4 (gm) (-)-2-butanol are
respectively (If the specific rotation of enantiomerically pure
(+)-2-butanol is +13.5 unit). | [] | 80, +2.7 unit | 20, -27 unit | 20, +2.7 unit | 80, -27 unit | 3 | null | [
"images/image30.png",
"images/image31.png",
"images/image32.png",
"images/image33.png"
] | Chemistry | Isomerism | Specific rotation, optical purity, enantiomeric excess and | Tough | single_correct | true | JEE Main | CH-17 Chemistry Paper 26 Nov. Second.docx | |
CH-23-Q3 | Acetylene may be prepared by electrolysis of | [] | potassium oxalate | potassium acetate | potassium maleate | potassium succinate | 3 | null | [IMAGE] + 2CO2 + 2NaOH + 2H2O | [
"images/image1.png",
"images/image2.png",
"images/image3.png"
] | Chemistry | Organic Chemistry | Preparation of hydrocarbon (Handout) | Moderate | single_correct | true | JEE Main | CH-23 Chemistry Paper 31 Dec. FST.docx |
CH-14-Q24 | The equivalent weight of [IMAGE] in acidic medium | [
"images/image96.png"
] | null | 49 | [
"images/image97.png",
"images/image98.png",
"images/image99.png"
] | Chemistry | Inorganic Chemistry | D & F -Block Elements | Moderate | numerical | true | JEE Main | CH-14 Chemistry Paper 7 Nov..docx | |||||
CH-10-Q24 | The molecular formula of diphenyl
methane[IMAGE], is [IMAGE] How many structural isomers are possible when one of the hydrogens is
replaced by a chlorine atom? | [
"images/image120.jpeg",
"images/image121.png"
] | null | 4 | Only four structural isomers are possible for diphenyl methane. | [
"images/image122.jpeg",
"images/image123.jpeg",
"images/image124.jpeg",
"images/image125.jpeg"
] | Chemistry | Organic Chemistry | Organic Chemistry | Moderate | numerical | true | JEE Main | CH-10 Chemistry Paper-1 30 Oct New Jee main.docx | ||||
CH-09-Q15 | Besides carbondioxide and nitrous oxide, the other greenhouse
gas is | [] | methane | carbon monoxide | both a and b | none of these | 3 | null | Explanation: Greenhouse gases such as carbondioxide, nitrogen
oxide, methane, carbon monoxide, water vapours and CFCs
(chlorofluorocarbons) causes greenhouse effect on plants. | [] | Chemistry | Inorganic Chemistry | Environmental Chemistry | Moderate | single_correct | false | JEE Main | CH-09 Chemistry Paper 26 October.docx |
CH-23-Q10 | Inversion of sucrose (C12H22O11) is first-order reaction and is
studied by measuring angle of rotation at different instant of time
[IMAGE] If $\left( r_{\infty} - r_{0} \right) = a$ and
$\left( r_{\infty} - r_{t} \right) = (a - x)$ (where $r_{0}$, $r_{t}$
and $r_{\infty}$are the angle of rotation at the start, at the ... | [
"images/image36.png"
] | $r_{0} = 2r_{t} - r_{\infty}$ | $r_{0} = r_{t} - r_{\infty}$ | $r_{0} = r_{t} - 2r_{\infty}$ | $r_{0} = r_{t} + r_{\infty}$ | 1 | null | Given,
$\left( r_{\infty} - r_{0} \right) = a,\left( r_{\infty} - r_{t} \right) = (a - x)$
At 50 Inversion
$\frac{a}{2} = (a - x)$
$\frac{\left( r_{\infty} - r_{0} \right)}{2} = \left( r_{\infty} - r_{t} \right)$
$\left( r_{\infty} - r_{0} \right) = 2r_{\infty} - 2r_{t}$
$r_{0} = 2r_{t} - r_{\infty}$ | [] | Chemistry | Physical Chemistry | Chemical Kinetics | Moderate | single_correct | true | JEE Main | CH-23 Chemistry Paper 31 Dec. FST.docx |
CH-01-Q22 | What is the number of Chiral Carbon in the compound C. | [
"images/image61.png"
] | null | null | [IMAGE] Compound (C) has one chiral C-atom. | [
"images/image62.png",
"images/image63.png"
] | Chemistry | single_correct | true | JEE Main | CH-01 Chemistry Paper 1.docx | |||||||
CH-20-Q22 | $MnO_{4}^{2 -}$(1 mole) in neutral aqueous medium
isdisproportionateto X mole of $MnO_{4}^{-}$ and Y mole of $MnO_{2}$.
3X+6Y = | [] | null | 4 | $MnO_{4}^{2 -}$in neutral aqueous medium isdisproportionate to
$\frac{2}{3}$ mole of $MnO_{4}^{-}$ and $\frac{1}{3}$ mole of $MnO_{2}$. | [] | Chemistry | Physical Chemistry | Redox reaction | Tough | numerical | false | JEE Main | CH-20 Chemistry Paper 12 Dec..docx | ||||
CH-08-Q4 | 3.2 moles of hydrogen iodide were heated in a sealed bulb at 444
ºC till the equilibrium state was reached. Its degree of dissociation at
this temperature was found to be 22. The numbers of moles of hydrogen
iodide present at equilibrium are | [] | 2.496 | 1.87 | 2 | 4 | 1 | null | $\frac{22}{100}$ × 3.2 = 0.704
[IMAGE] at equil. moles of HI = 3.2 - 0.704 = 2.496 | [
"images/image11.png"
] | Chemistry | Physical Chemistry | Chemical and Ionic Equilibrium | Moderate | single_correct | true | JEE Main | CH-08 Chemistry Paper 23 October.docx |
CH-08-Q9 | Sucrose is | [] | A reducing sugar | Not a reducing sugar | Partial reducing sugar | Mixed sugar | 2 | null | Sucrose is not a reducing sugar. | [] | Chemistry | Organic Chemistry | Biomolecules | Easy | single_correct | false | JEE Main | CH-08 Chemistry Paper 23 October.docx |
CH-25-Q21 | The major product obtained in the reaction | [
"images/image82.png",
"images/image83.png"
] | 3 | null | For photochemical bromination reactivity of hydrogen atom is 3º
H > 2º H > 1º H.
czksehuhdj.k ds fy, gkbMªkstu ijek.kq dh fØ
Øe 3º H > 2º H > 1º H gSA | [
"images/image88.png"
] | Chemistry | Organic Chemistry | Hydrocarbon | Moderate | single_correct | true | JEE Main | CH-25 Chemistry Paper 7 Jan.docx | ||||
CH-24-Q6 | Equilibrium constant for the reactions,
${NO}_{2} + {SO}_{2} \rightleftharpoons {SO}_{3} + NO$ is
$2{SO}_{3} \rightleftharpoons 2{SO}_{2} + O_{2}$ is
[IMAGE] then correct reaction is | [
"images/image5.png",
"images/image6.png",
"images/image7.png"
] | 2 | null | (i) $2NO + O_{2} \rightleftharpoons 2{NO}_{2}$
(ii) ${NO}_{2} + {SO}_{2} \rightleftharpoons {SO}_{3} + NO$
(iii) $2{SO}_{3} \rightleftharpoons 2{SO}_{2} + O_{2}$
Now, - 2 (ii) = (i) + (iii) | [
"images/image12.png"
] | Chemistry | Physical Chemistry | Chemical Equilibrium | Moderate | single_correct | true | JEE Main | CH-24 Chemistry Paper 31 Dec..docx | ||||
CH-23-Q5 | , Product (X) in this reaction is | [
"images/image6.png",
"images/image7.png",
"images/image8.png"
] | 1 | null | [
"images/image13.png",
"images/image14.png",
"images/image15.png",
"images/image16.png",
"images/image17.png",
"images/image18.png",
"images/image19.png"
] | Chemistry | Organic Chemistry | Carbonyl compound_Final | Tough | single_correct | true | JEE Main | CH-23 Chemistry Paper 31 Dec. FST.docx | |||||
CH-14-Q13 | An alkene of molecular formula [IMAGE] on
ozonolysis gives 2, 2 dimethyl propanal & 2 butanon, then the alkene is | [
"images/image60.png"
] | 2,2,4 -trimethyl -3 -hexene | 2,2,6 -trimethyl- 3 -hexene | 2,3,4 -trimethyl- 2 -hexene | 2,2,4 -trimethyl- 2 -hexene | 1 | null | [IMAGE] On the basis of product formation, it would be alkene. | [
"images/image61.png",
"images/image62.png",
"images/image63.png",
"images/image64.png"
] | Chemistry | Organic Chemistry | Aldehydes, Ketones & Carboxylic Acids | Moderate | single_correct | true | JEE Main | CH-14 Chemistry Paper 7 Nov..docx |
CH-26-Q18 | The Corey-House alkane synthesis is carried out by treating an
alkyl halide with | [] | Lithium metal | Copper metal | Lithium metal followed by reaction with cuprous iodide and then
treating the product with an alkyl halide | Cuprous iodide followed by reaction with alkyl halide
dksjh gkmLk,Ydsu la"ys'k.k ds lkFk,d,Ydkby gsykbM | 3 | null | [
"images/image16.png",
"images/image17.png",
"images/image18.png"
] | Chemistry | Organic Chemistry | Hydrocarbon | Moderate | single_correct | true | JEE Main | CH-26 Chemistry Paper 1 15 Jan.docx | |
CH-02-Q16 | In which reaction correct product is given ? | [] | null | null | [IMAGE] This is Wolf Kischner Reduction | [
"images/image104.png",
"images/image102.png"
] | Chemistry | single_correct | true | JEE Main | CH-02 Chemistry paper 2.docx | |||||||
CH-24-Q15 | Consider an aqueous solution, 0.1 M each in HOCN, HCOOH,
(COOH)_2 and H_3PO_4. For HOCN, we can write: K_a (HOCN) =
$\frac{\left\lbrack H^{+}I{OCN}^{-} \right\rbrack}{\lbrack HOCN\rbrack}$. in this equation is | [] | H^+ ions released by HOCN | Sum of H^+ ions released by all monoprotic acids | Sum of H^+ ions released only the first dissociation of all the
acids. | Overall H^+ ion concentration in the solution. | 4 | null | For HOCN^-, K_a =
$\frac{\left\lbrack H^{+} \right\rbrack\left\lbrack {OCN}^{-} \right\rbrack}{\lbrack HOCN}$
= total H^+ concentration of solution. | [] | Chemistry | Physical Chemistry | Ionic Equilibrium | Easy | single_correct | false | JEE Main | CH-24 Chemistry Paper 31 Dec..docx |
CH-18-Q14 | Consider the following statements:
(a) Electron density in the XY plane in orbital is zero
(b) Electron density in the XY plane in orbital is zero.
(c) 2s orbital has one nodal surface
(d) for 2pz orbital, XY is the nodal plane.
Which of these are incorrect statements | [] | a & c | b & c | Only b | a, b | 4 | null | (a) Electron density in the XY plane in orbital is not zero
(b) Electron density in the XY plane in orbital is not zero.
(c) 2s orbital has one nodal surface
(d) For 2pz orbital, XY is the nodal plane. | [] | Chemistry | Atomic Structure | Quantum mechanical model of atom, Shrodinger wave equation | Moderate | single_correct | false | JEE Main | CH-18 Chemistry Paper 27 Nov. Third.docx |
CH-20-Q24 | Given standard electrode potentials
The standard electrode
potential$E^{0}$for$Fe^{+ + +} + e^{-} \rightarrow Fe^{+ +}$ is | [] | null | 0.77 | $\Delta G^{\circ} = - nE^{0}F$
$Fe^{2 +} + 2e^{-} \rightarrow Fe$....(i)
$\Delta G^{0} = - 2 \times F \times ( - 0.440V) = 0.880F$
$Fe^{3 +} + 3e^{-} \rightarrow Fe$....(ii)
$\Delta G^{\circ} = - 3 \times F \times ( - 0.036) = 0.108F$
On subtracting equation (i) from (ii)
${Fe}^{3 +} + e^{-} \rightarrow {Fe}^{2 +}$
$\D... | [] | Chemistry | Physical Chemistry | Electrochemistry | Moderate | numerical | false | JEE Main | CH-20 Chemistry Paper 12 Dec..docx | ||||
CH-16-Q5 | When 100g of ethylene polymerises entirely to polyethene, the
weight of polyethene formed as per the equation [IMAGE] is | [
"images/image18.png"
] | 100g | 100ng | 2 | null | By applying POAC for C atoms
moles of ethylene × 2 = mole of polythene × n × 2
[IMAGE] wt. of polyethene = 100 g | [
"images/image19.png"
] | Chemistry | Mole concept | Stoichiometry, Equation based calculations (Elementary level | Moderate | single_correct | true | JEE Main | CH-16 Chemistry Paper 25 Nov. Fisrt.docx | ||
CH-08-Q19 | Lead pipes are corroded quickly by | [] | Dil. H_2SO_4 | Conc. H_2SO_4 | Acetic acid | Water | 3 | null | Organic acids dissolve lead in presence of oxygen | [
"images/image20.png"
] | Chemistry | Inorganic Chemistry | p block | Moderate | single_correct | true | JEE Main | CH-08 Chemistry Paper 23 October.docx |
CH-13-Q14 | Phenolphthalein is obtained by heating phthalic anhydride with
conc. H_2 SO_4 and | [] | Benzyl alcohol | Benzene | Phenol | Benzoic acid | 3 | null | [
"images/image25.png"
] | Chemistry | Organic Chemistry | Alcohols, Phenols and Ethers | Moderate | single_correct | true | JEE Main | CH-13 Chemistry Paper 4 Nov..docx | |
CH-26-Q26 | To 500 cm3 of water, 3.0 × 10-3 kg of acetic acid is added. If
23 of acetic acid is dissociated, what will be the depression in
freezing point? kf and density of water are 1.86 K kg mole-1 and 0.997
g cm-3 respectively
500 cm3 ty esa 3.0 × 10-3 kg,flfVd vEy Mkyk A;fn 23,flfVd
vEy rks fgekad esa fdruk voueu gksxk kf o t... | [] | null | 0.228 | Weight of water $= 500 \times 0.997 = 498.5g$
No. of moles of acetic acid
$= \frac{\ Wt.\ of\ {CH}_{3}COOH\ (ingm)\ }{\ Mol.wt.of\ {CH}_{3}COOH} = \frac{3 \times 10^{- 3} \times 10^{3}}{60} = 0.05$
Since 498.5 g of water has 0.05 moles of CH3COOH
1000 g of water has $= \frac{0.05 \times 1000}{498.5} = 0.1$
Determinatio... | [] | Chemistry | Physical Chemistry | Tough | numerical | false | JEE Main | CH-26 Chemistry Paper 1 15 Jan.docx | |||||
CH-06-Q15 | In which reactions major product is propane ?
(a) Propene | [] | b and d | a and d | b and c | 4 | null | (a) Propene
$\frac{(1){LiAlH}_{4}}{\ ether\ }\overset{H_{2}O}{\longrightarrow}$ No
reaction
(b) Glyceraldehyde $\frac{HI,\Delta}{Phosphorous}$ Propane
(c) Propanone $\overset{Zn.Hg,\ HCl}{\rightarrow}$ Propane
(d) Ethyl propanoate
$\frac{\ DIBAL - H\ }{\ low\ temp\ }\overset{H_{2}O}{\rightarrow}$
Propanal + Ethanol | [] | Chemistry | Organic Chemistry | Hydrocarbons | Moderate | single_correct | false | JEE Main | CH-06 Chemistry Paper 14 October.docx | |
CH-22-Q21 | The wavelength of a moving body of mass 0.1 mg is 3.31 × 10-29
m. The kinetic energy of the body in J would be | [] | null | 2 | $\lambda = \frac{h}{mv}$
$mv = \frac{h}{\lambda}$
$m^{2}v^{2} = \frac{h^{2}}{m\lambda^{2}}$
$\frac{1}{2}{mv}^{2} = \frac{h^{2}}{2m\lambda^{2}}$
$KE = \frac{h^{2}}{2m\lambda^{2}} = \frac{\left( 6.6 \times 10^{- 34} \right)^{2}}{2 \times 10^{- 4} \times \left( 3.31 \times 10^{- 29} \right)^{2}} = 2 \times 10^{- 6}$ | [] | Chemistry | Physical Chemistry | Atomic Structure | Moderate | numerical | false | JEE Main | CH-22 Chemistry Paper 19 Dec. Eng.docx | ||||
CH-26-Q37 | M(OH)x (producing Mx+ and OH- ions) has Ksp 4 × 10-12 and
solubility 10-4 M. The value of x is:
M(OH)x (tks Mx+ o OH- vk;u cukrk gS) 4 × 10-12 Ksp rFkk
bl 10-4 M gSA x | [] | null | 2 | Ksp = 4 × 10-12 = = =
xx. (10-4)1+x
$\therefore$ x = 2 | [] | Chemistry | Physical Chemistry | Ionic Equilibrium | Moderate | numerical | false | JEE Main | CH-26 Chemistry Paper 1 15 Jan.docx | ||||
CH-08-Q1 | Partial pressures of A, B, C and D on the basis of
gaseous system A + 2B = C = + 3D are A = C = 0.30
and D = 0.50 atm. The numerical value of equilibrium constant is | [] | 11.25 | 18.75 | 5 | 3.75 | 2 | null | A + 2 B = C + 3D | [
"images/image1.png"
] | Chemistry | Physical Chemistry | Chemical and Ionic Equilibrium | Easy | single_correct | true | JEE Main | CH-08 Chemistry Paper 23 October.docx |
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