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-01-Q12 | Which is not true for CO molecule? | [] | Bond order of CO is equal to N_2 | Bond enthalpy of CO is equal
to N_2 | CO is a strong ligand | CO is a good π acid | null | null | Bond enthalpy of CO is greater than N_2 because CO has electro
negativity difference which provide strength to bond length. | [] | Chemistry | single_correct | false | JEE Main | CH-01 Chemistry Paper 1.docx | |||
CH-10-Q8 | Which gives mono substituted product | [] | o -dinitrobenzene | m -dinitrobenzene | p -dinitrobenzene | Nitrobenzene | 2 | null | [IMAGE] gives only mono-substitution product as
[IMAGE] group is meta directing and only one m -position
is possible in m-dinitrobenzene. | [
"images/image39.png",
"images/image40.png"
] | Chemistry | Organic Chemistry | Organic Chemistry | Moderate | single_correct | true | JEE Main | CH-10 Chemistry Paper-1 30 Oct New Jee main.docx |
CH-05-Q10 | In which species positive charge is delocalised ? | [] | 2 | null | [
"images/image23.png"
] | Chemistry | Organic Chemistry | Some Basic Principles and Techniques | Moderate | single_correct | true | JEE Main | CH-05 Chemistry Paper 10 October.docx | |||||
CH-05-Q3 | The amount of arsenic pentasulphide (As_2S_5) that can be
obtained when 35.5arsenic acid (H_3AsO_4) is treated with excess H_2S
in the presence of conc. HCl (assuming 100 conversion) is | [] | 0.333 mol | 0.125 mol | 0.25 mol | 0.50 mol | 2 | null | H_3AsO_4 + H_2S + HCL [IMAGE] As_2S_5
Initial moles of H_3AsO_4 = $\frac{35.5}{142}\mathbf{=}\frac{1}{4}$
mol
Applying POAC on As atoms; moles of As_2S_5$= \frac{1}{8}$ mol | [
"images/image1.png"
] | Chemistry | Physical Chemistry | Some Basic Concepts of Chemistry | Easy | single_correct | true | JEE Main | CH-05 Chemistry Paper 10 October.docx |
CH-20-Q23 | The conductivity of a saturated solution of ${BaSO}_{4}$is
$3.06 \times 10^{- 6}{ohm}^{- 1}{cm}^{- 1}$ and its
equivalentconductance is $1.53{ohm}^{- 1}{cm}^{- 1}{equivalent}^{- 1}$.
The $K_{sp}$ of the $BaSO_{4}$ is $X \times 10^{- 6}$ value of X is. | [] | null | 4 | $\lambda m = \frac{1000K}{S} = \frac{1000 \times 3.06 \times 10^{- 6}}{S} = 1.53$
$S = 2 \times 10^{- 3}\frac{mol}{litre}$
$K_{sp({BaSO}_{4})} = S^{2} = {(2 \times 10^{- 3})}^{2} = 4 \times 10^{- 6}$ | [] | Chemistry | Physical Chemistry | Electrochemistry | Easy | numerical | false | JEE Main | CH-20 Chemistry Paper 12 Dec..docx | ||||
CH-20-Q3 | If the bond dissociation energies of XY, X_2 and Y_2 (all
diatomic molecules) are in the ratio of 1: 1: 0.5 and $\Delta_{f}H$ for
the formation of XY is $- 200kJ{mole}^{- 1}$. The bond dissociation
energy of X_2 will be | [] | $100\ k\ J\ {mole}^{- 1}$ | $800\ k\ J\ {mole}^{- 1}$ | $300\ k\ J\ {mole}^{- 1}$ | $400\ k\ J\ {mole}^{- 1}$ | 2 | null | $XY \longrightarrow X_{(g)} + Y_{(g)};\Delta H = + akJ/mole$....(i)
$X_{2} \longrightarrow 2X;\Delta H = + akJ/mole$...(ii)
$Y_{2} \longrightarrow 2Y;\Delta H = + 0.5akJ/mole$...(iii)
$\frac{1}{2} \times (ii) + \frac{1}{2} \times (iii) - (i),$ gives
$\frac{1}{2}X_{2} + \frac{1}{2}Y_{2} \longrightarrow XY$
$\Delta H = (... | [] | Chemistry | Physical Chemistry | Thermochemistry | Moderate | single_correct | false | JEE Main | CH-20 Chemistry Paper 12 Dec..docx |
CH-25-Q20 | Which of the following compounds on hydrolysis gives propyne ? | [] | CaC2 | Mg2C3 | Al4C3 | Cu2Cl2 | 2 | null | $2{Mg}^{+ 2}\left( \overset{¯}{C} \equiv C - C^{3 -} \right)\overset{H_{3}O^{+}}{\rightarrow}Mg(OH)_{2} + CH \equiv C - {CH}_{3}(\ Propyne)$ | [] | Chemistry | Organic Chemistry | Hydrocarbon | Moderate | single_correct | false | JEE Main | CH-25 Chemistry Paper 7 Jan.docx |
CH-13-Q8 | Mark the oxide which is amphoteric in character | [] | CO_2 | SiO_2 | SnO_2 | CaO | 3 | null | SnO_2 + 2NaOH [IMAGE] NA_2 SnO_3 +
H_2O
SnO_2 + 4HC[IMAGE] SnCl_4 + 2H_2O | [
"images/image18.png",
"images/image19.png"
] | Chemistry | Inorganic Chemistry | P-Block Element | Moderate | single_correct | true | JEE Main | CH-13 Chemistry Paper 4 Nov..docx |
CH-09-Q3 | In the following reaction, [IMAGE] | [
"images/image5.png"
] | Bromine is oxidized and carbonate is reduced | Bromine is reduced and water is oxidized | Bromine is neither reduced nor oxidized | Bromine is both reduced and oxidized | 1 | null | [IMAGE] In this reaction bromine is oxidized as
well as reduced. | [
"images/image6.png"
] | Chemistry | Physical Chemistry | Redox Reaction | Moderate | single_correct | true | JEE Main | CH-09 Chemistry Paper 26 October.docx |
CH-02-Q24 | How many of the following compounds give benzene on reaction with
PhMgBr?
(a) [IMAGE] (b) | [
"images/image156.png",
"images/image157.png",
"images/image158.png",
"images/image159.png",
"images/image160.png",
"images/image161.png",
"images/image162.png",
"images/image163.png",
"images/image164.png"
] | null | null | [
"images/image165.png"
] | Chemistry | single_correct | true | JEE Main | CH-02 Chemistry paper 2.docx | ||||||||
CH-21-Q17 | Amongst $Ni(CO)_{4},\left\lbrack Ni(CN)_{4} \right\rbrack^{2 -}$
and ${NiCl}_{4}^{2 -}$ | [] | $Ni(CO)4$and$NiCI42–$ are diamagnetic and
$\lbrack Ni(CN)4\rbrack 2–$ is paramagnetic. | $NiCI42–$and$\lbrack Ni(CN)4\rbrack 2–$ are diamagnetic and
$Ni(CO)4$ is paramagnetic. | $Ni(CO)4$and$\lbrack Ni(CN)4\rbrack 2–$ are diamagnetic and
$NiCI42$- is paramagnetic. | $Ni(CO)4$is diamagnetic and $NiCI42–$ and
$\lbrack Ni(CN)4\rbrack 2–$ are paramagnetic. | 3 | null | In complex, $\lbrack Ni(CO)4\rbrack$ nickel is in zero
oxidation state. The $CO$ is strong field ligand and, therefore compels
for the pairing of electrons. The hybridisation scheme is as shown in
figure.
$Niº(\lbrack Ar\rbrack\ 3d8\ 4s2)$
[IMAGE] Four pairs of electrons from four CO.
$\mathbf{sp}\mathbf{3}$hybrid orbi... | [
"images/image9.png",
"images/image10.png",
"images/image11.png",
"images/image12.png",
"images/image11.png",
"images/image13.png",
"images/image9.png",
"images/image10.png",
"images/image11.png",
"images/image12.png",
"images/image11.png",
"images/image13.png"
] | Chemistry | Coordination Compounds | Valence Bond Theory and Crystal Field Theory (Part II) | Tough | single_correct | true | JEE Main | CH-21 Chemistry Paper 19 Dec. Eng Hindi.docx |
CH-27-Q37 | 8 × 10-6 M AgNO3 solution is gradually added in 1 L of 10-4 M
KCl solution. Upto what volume of AgNO3 solution being added (in L),
precipitation of AgCl will not take place? (Ksp of AgCl = 2 × 10-10)
ikru uk gks ik;s (AgCl | [] | null | 1 | $2 \times 10^{- 10} = \left( \frac{8 \times 10^{- 8} \times V}{V + 1} \right)\left( \frac{10^{- 4}}{V + 1} \right)$
(V + 1)^2 = 4V
(V -1)^2 = | [] | Chemistry | Physical Chemistry | Ionic Equilibrium | Easy | numerical | false | JEE Main | CH-27 Chemistry Paper 2 15 Jan.docx | ||||
CH-09-Q5 | In the reaction [IMAGE] the
[IMAGE] acts as | [
"images/image7.png",
"images/image8.png"
] | Reducing agent | Oxidizing agent | Bleaching agent | None of the above | 2 | null | [
"images/image9.png"
] | Chemistry | Physical Chemistry | Redox Reaction | Moderate | single_correct | true | JEE Main | CH-09 Chemistry Paper 26 October.docx | |
CH-22-Q20 | Initially, the root mean square (rms) velocity of N_2 molecules
atcertain temperature is u. If this temperature is doubled and allthe
nitrogen molecules dissociate into nitrogen atoms, then the new rms
velocity will be | [] | 2u | 4u | 14u | null | null | The root mean square velocity for nitrogen initially is
calculated as, (d+200)
The dissociation of all the nitrogen molecules into nitrogen atoms
results in the new root mean square velocity, that
is,$t = \sqrt{\frac{2d}{2}}$ | [] | Chemistry | Physical Chemistry | States of matter | Moderate | single_correct | false | JEE Main | CH-22 Chemistry Paper 19 Dec. Eng.docx | |
CH-09-Q8 | Starting from propanoic acid, the following reactions were
carried out
[IMAGE] What is the compound Z | [
"images/image16.png"
] | 2 | null | [
"images/image21.png",
"images/image22.png",
"images/image23.png"
] | Chemistry | Organic Chemistry | Organic compounds containing Nitrogen | Tough | single_correct | true | JEE Main | CH-09 Chemistry Paper 26 October.docx | |||||
CH-08-Q3 | For the chemical equilibrium, CaCO_3 (s) $\rightleftharpoons$
CaO(s) +CO_2 (g), [IMAGE] can be determined for which
one of the following plots | [
"images/image3.png"
] | 1 | null | For the reaction,
CaCO_3 (g) = CaO(s) + CO_2 (g)
$K_{p} = P_{\infty},$ and $K_{c} = \left\lbrack {CO}_{2} \right\rbrack$
( $\because$ [CaCO_3 ] = 1 and for solids$\ \rbrack$
According to Arrhenius equation we have
$K = Ae^{- \Delta H^{*},/RT}$
Taking logarithm, we have
logK_p = log A $- \frac{\Delta H_{r}^{0}}{RT(2... | [
"images/image8.jpeg",
"images/image9.png",
"images/image9.png",
"images/image10.png"
] | Chemistry | Physical Chemistry | Chemical and Ionic Equilibrium | Tough | single_correct | true | JEE Main | CH-08 Chemistry Paper 23 October.docx | ||||
CH-24-Q23 | The equilibrium
${SO}_{2}(g) + \frac{1}{2}O_{2}(g) \rightleftharpoons {SO}_{3}(g)$ is
established in a container of 4L at a particular temperature. If the
number of moles of SO_2, O_2 and SO_3 at equilibrium are 2, 1 and 4
respectively then find the value of equilibrium constant. | [] | null | 4 | ${SO}_{2}(g) + \ \ \ \ \ \ \frac{1}{2}O_{2}(g)\ \ \ \rightleftharpoons \ \ \ \ {SO}_{3}(g)$
No. of mole 2 1 4
Conc. $\frac{2}{4}$ $\frac{1}{4}$ $\frac{4}{4}$
$K_{C} = \frac{\left\lbrack {SO}_{3} \right\rbrack}{\left\lbrack {SO}_{2} \right\rbrack\left\lbrack O_{2} \right\rbrack^{1/2}} = \frac{1}{(1/2)(1/4)^{1/2}} = \fra... | [] | Chemistry | Physical Chemistry | Chemical Equilibrium | Moderate | numerical | false | JEE Main | CH-24 Chemistry Paper 31 Dec..docx | ||||
CH-25-Q17 | Write the major product of following reactions.
Ph - Br + CH_3 - Br [IMAGE] (report mixture)
Ph - Br + CH_3 - Br | [
"images/image72.png",
"images/image71.png"
] | Ph - Ph - CH_3 - CH_3 + Ph - CH_3 | Ph - CH_3 + CH_3 - CH_3 + Ph - CH_3 | Ph - Ph + CH_3 - CH_3 + Ph - OH | None of these | 1 | null | Ph - Br + CH_3 - Br
[IMAGE] Ph - Ph + CH_3 - CH_3 + Ph - CH_3
These are wurtz reaction oqVZ~t | [
"images/image73.png",
"images/image71.png"
] | Chemistry | Organic Chemistry | Hydrocarbon | Moderate | single_correct | true | JEE Main | CH-25 Chemistry Paper 7 Jan.docx |
CH-28-Q4 | The enthalpies of neutralization of a weak base AOH and a strong
base BOH by HCI are -12250 mol respectively.
When one mole of HCI is added to a solution containing 1 mole of AOH and
1 mole of BOH, the enthalpy change was -12500 cal/ mol. In what ratio
the acid is distributed between AOH and BOH? | [] | 2: 1 | 2: 3 | 1: 2 | None of these | 1 | null | $- 12250x - 13000(1 - x) = - 12500$
$750x = 500 \Rightarrow x = 2/3$
So, required ratio is = $\frac{2}{1}$ | [] | Chemistry | Physical Chemistry | Thermodynamics | Easy | single_correct | false | JEE Main | CH-28 Chemistry Paper FST 16 Jan.docx |
CH-04-Q25 | are INTEGER ANSWER TYPE Questions.
(The answer of each of the questions is 2 digits integer)] | [] | null | 04 | [] | Chemistry | numerical | false | JEE Main | CH-04 Paper 02 Chemistry with Hindi 22-8-2020.docx | ||||||||
CH-24-Q21 | The pH of which salt solution is independent of its
concentration ?
A. (CH_3COO)C_5H_5NH
B. NaH_2PO_4
C. Na_2HPO_4
D. NH_4CN | [] | 1, 2, 3, 4 | 1, 4 | 2, 3 | 1, 2, 3 | 1 | null | pH of amphiprotic salt solutions and weak acid-weak base salt
solutions is independent of their concentration. | [] | Chemistry | Physical Chemistry | Ionic Equilibrium | Easy | single_correct | false | JEE Main | CH-24 Chemistry Paper 31 Dec..docx |
CH-18-Q8 | Which of the following pairs of compounds are enantiomers | [] | 1 | null | [IMAGE] non-super impossible mirror image stereoisomers. | [
"images/image27.png",
"images/image28.png",
"images/image29.png"
] | Chemistry | Stereoisomerism | Definition and Properties of Enantiomers, Diastereomers, Meso | Moderate | single_correct | true | JEE Main | CH-18 Chemistry Paper 27 Nov. Third.docx | ||||
CH-16-Q24 | A 0.2 g sample containing copper (II) was analyzed
eudiometrically, where copper (II)is reduced tocopper (I) by iodide
ions. $2{Cu}^{2 +} + 4I^{-} \longrightarrow 2CuI + I_{2}$
If 20 mL of 0.1 M Na2S2O3 solution is required for titration of the
liberated iodine, then the percentage of copper in the sample will be | [] | null | 63.5 | 2 moles of Cu2+ = 1 mole of I2
= 2 moles of hypo.
so moles of hypo used = 20 × 10-3 × 0.1 = 2 m moles = moles of copper
hence
$\% of\ copper = \frac{2 \times 10^{- 3} \times 63.5}{0.2} \times 10\% = 63.5\%$ | [] | Chemistry | Equivalent Concept | Eudiometric/Eudiometric Titration, Calculation of Available | Tough | numerical | false | JEE Main | CH-16 Chemistry Paper 25 Nov. Fisrt.docx | ||||
CH-01-Q18 | CrO_3 dissolves in aqueous NaOH to give | [] | Cr_2O_7 ^2 | CrO_4 ^2 | Cr(OH)_3 | Cr(OH)_2 | null | null | [IMAGE] gives the chromate ions in the basic medium | [
"images/image57.png",
"images/image58.png"
] | Chemistry | single_correct | true | JEE Main | CH-01 Chemistry Paper 1.docx | |||
CH-13-Q1 | The vapour pressure lowering caused by the addition of 100 g of
sucrose (molecular mass = 342 to 1000 g of water if the vapour
pressure of pure water at 25ºC is 23.8 mm Hg. | [] | 1.25 mm Hg | 0.125 mm Hg | 1.15 mm Hg | 00.12 mm Hg | 2 | null | Given molecular mass of sucrose = 342
Moles of sucrose [IMAGE] = 0.292 mole
Moles of water N [IMAGE] = 55.5 moles and vapour pressure
of pure water P^0 = 23.8 mm Hg
According to Raoult's law
[IMAGE] = 0.125 mm Hg. | [
"images/image1.png",
"images/image2.png",
"images/image3.png",
"images/image4.png"
] | Chemistry | Physical Chemistry | Moderate | single_correct | true | JEE Main | CH-13 Chemistry Paper 4 Nov..docx | |
CH-04-Q21 | The hydrated salt Na_2CO_3.xH_2O undergoes 63 loss in mass on
heating and becomes anhydrous.Calculate the value of x.
Na_2CO_3.xH_2O 63 x | [] | null | 10 | [IMAGE] Let 100 g of Na2CO3. xH2O be present
[IMAGE] mole of H2O formed 6678 + 1134 x = 1800 x
666 x = 6678
[IMAGE] gy- [IMAGE] Na2CO3. xH2O [IMAGE] 100 g
[IMAGE] H2O 6678 + 1134 x = 1800 x
666 x = 6678 | [
"images/image45.png",
"images/image46.png",
"images/image47.png",
"images/image48.png",
"images/image49.png",
"images/image50.png",
"images/image47.png",
"images/image46.png",
"images/image48.png",
"images/image49.png"
] | Chemistry | Physical Chemistry | Mole Concept | Moderate | numerical | true | JEE Main | CH-04 Paper 02 Chemistry with Hindi 22-8-2020.docx | ||||
CH-21-Q14 | The difference between the wave number of 1st line of Balmer
series and last line of paschen series for Li2+ ion is:
Li2+ fy, ckWej Js.kh dh 1st Js.kh dh vfUre
js[kk | [] | $\frac{R}{36}$ | $\frac{5R}{36}$ | 4R | $\frac{R}{4}$ | 4 | null | For 1st line of Balmer series
ckej Js.kh dh 1^st fy,
${\overline{v}}_{1} = R_{H}(3)^{2}\left\lbrack \frac{1}{(2)^{2}} - \frac{1}{(3)^{2}} \right\rbrack = 9R\left( \frac{5}{36} \right) = \frac{5}{4}R$
For last line of Pachen series
ik'pu Js.kh dh vfUre fy,
${\overline{v}}_{2} = R_{H}(3)^{2}\left\lbrack \frac{1}{(3)^{2}}... | [] | Chemistry | Atomic Structure | Spectrum | Moderate | single_correct | false | JEE Main | CH-21 Chemistry Paper 19 Dec. Eng Hindi.docx |
CH-26-Q30 | MX2 dissociates into M2+ and X- ions in an aqueous
solution, with a degree of dissociation $(\alpha)$ of 0.5 The ratio of
the observed depres sion of freezing point of the aqueous solution to
the value of the de pression of freezing point in the absence of ionic
diss ociation is
MX2,d 0.5 dh,d fo;kstu ek=k (degree of
d... | [] | null | 2 | ${MX}_{2}\ \ \ \ \ \ \ \ \rightleftharpoons \ \ \ \ M^{2 +} + 2X^{-}$
$m_{0}\left( \begin{matrix}
1 - \alpha)\ \ \ \ m_{0}\alpha\ \ \ \ 2m_{0}\alpha
\end{matrix} \right.\ $ $;m = m_{0}(1 + 2\alpha)$
$\therefore m = m_{0}(1 + 2 \times 0.5) = 2m_{0}$ (as given) k
$\frac{\left( - \Delta T_{f} \right)_{observed\ }}{\left( ... | [] | Chemistry | Physical Chemistry | Moderate | numerical | false | JEE Main | CH-26 Chemistry Paper 1 15 Jan.docx | |||||
CH-19-Q23 | Metallic magnesium has a hexagonal close-packed structure and a
density of 1.74 g/cm3. Assume magnesium atoms to be spheres of radius r.
74.1 of the space are occupied by atoms. Calculate the volume of each
atom and the atomic radius r. (Mg = 24.31) | [] | null | 1.60 | Packing efficiency of hcp
$= \frac{6\ \times \ vol.\ \ of\ 1\ atom}{Vol.\ \ of\ unit\ cell}$
Volume of unit cell $= \frac{Z \times M}{N_{A} \times d}$
$= \frac{6 \times 24.31}{6.023 \times 10^{23} \times 1.74}$
$= 13.9179 \times 10^{- 23}{cm}^{3}$
So, volume of 1 atom
$= \frac{Packing\ fraction \times Vol.\ of\ unit\ c... | [] | Chemistry | HCP & CCP structures | Tough | numerical | false | JEE Main | CH-19 Chemistry Paper 28 Nov. Fourth.docx | |||||
CH-13-Q9 | The metal which does not form ammonium nitrate by reaction with
dilute nitric acid is | [] | Al | Fe | Pb | Mg | 3 | null | Lead form nitric oxide with dil. HNO_3.
3Pb + 8 HNO_3 [IMAGE] 3Pb(NO_3)_2 + 2 NO +
4H_2O | [
"images/image20.png"
] | Chemistry | Inorganic Chemistry | P-Block Element | Moderate | single_correct | true | JEE Main | CH-13 Chemistry Paper 4 Nov..docx |
CH-10-Q15 | Which one of the following is least reactive in a nucleophilic
substitution reaction? | [] | 4 | null | The non reactivity of the chlorine atom in vinyl chloride may
be explained from the molecular orbital point of view as follows. If the
chlorine atom has sp^2 hybridization the C -Cl bond will be a σ bond
and the two lone pairs of electron would occupy the other two
[IMAGE] orbitals. This would leave a p -orbital
contai... | [
"images/image74.png",
"images/image75.jpeg"
] | Chemistry | Organic Chemistry | Organic Chemistry | Moderate | single_correct | true | JEE Main | CH-10 Chemistry Paper-1 30 Oct New Jee main.docx | ||||
CH-13-Q4 | If two substances A and B have [IMAGE] = 1: 2
and have mole fraction in solution 1: 2 then mole fraction of A in
vapours. | [
"images/image11.png"
] | 0.33 | 0.25 | 0.52 | 0.2 | 2 | null | Relationship between mole fraction of a component in the vapour
phase and total vapour pressure of an ideal solution. | [
"images/image12.png",
"images/image13.png"
] | Chemistry | Physical Chemistry | Moderate | single_correct | true | JEE Main | CH-13 Chemistry Paper 4 Nov..docx | |
CH-21-Q13 | Which of the following cannot evolve more than one gas (vapour)
if heated in dry test tube.
b.k ufydk | [] | ${NaNO}_{3}(s)$ | ${MgCO}_{3}(s)$ | ${FeSO}_{4}(s)$ | $\left( {NH}_{4} \right)_{2}{Cr}_{2}O_{7}(s)$ | 2 | null | $2{NaNO}_{3}(s)\frac{800^{\circ}C}{\Delta} \longrightarrow {Na}_{2}O(s) + N_{2}(g) + \frac{5}{2}O_{2}(g)$
${MgCO}_{3}(s)\overset{\Delta}{\longrightarrow}MgO(s) + {CO}_{2}(g)$
$2{FeSO}_{4}(s)\overset{\Delta}{\longrightarrow}{Fe}_{2}O_{3}(s) + {SO}_{2}(g) + {SO}_{3}(g)$
$\left( {NH}_{4} \right)_{2}{Cr}_{2}O_{7}(s)\overse... | [] | Chemistry | Salt Analysis | Dry test | Moderate | single_correct | false | JEE Main | CH-21 Chemistry Paper 19 Dec. Eng Hindi.docx |
CH-05-Q2 | A 1.5 m solution of acetic acid (I) is mixed with 3 m solution of
the acetic acid (II) to prepare 2m solution (m is molality of solution).
Select the correct statement(s) | [] | Mass ratio of solvents mixed$\left( \frac{I}{II} \right)$ is
$\frac{1}{2}$ | Mass ratio of solvents mixed$\left( \frac{I}{II} \right)$ is
$\frac{2}{1}$ | Mass ratio of solvents mixed$\left( \frac{I}{II} \right)$ is
$\frac{109}{59}$ | Mass ratio of solvents mixed$\left( \frac{I}{II} \right)$ is
$\frac{59}{109}$ | 3 | null | Let mass of solvent in 1.5 m solution = m_1 Kg and let mass of
solvent in 3 m solution = m_2 Kg.
Therefore molality of resulting solution
$= \frac{1.5m_{1} + 3m_{2}}{m_{1} + m_{2}} = 2$
$\therefore\frac{m_{1}}{m_{2}} = \frac{2}{1}$
Therefore required ratio $= \frac{109}{59}$ | [] | Chemistry | Physical Chemistry | Some Basic Concepts of Chemistry | Moderate | single_correct | false | JEE Main | CH-05 Chemistry Paper 10 October.docx |
CH-07-Q5 | At P is drawn select
correct graph. | [] | 3 | null | Boyle temperature gas behaves as ideal gas (Z = 1) in a range of
low pressure but at high pressure Z> | [] | Chemistry | Physical Chemistry | Gaseous state | Moderate | single_correct | false | JEE Main | CH-07 Chemistry Paper 17 October.docx | ||||
CH-14-Q23 | In a metal oxide, there is 20 oxygen by weight. It equivalent
weight is | [] | null | 32 | [
"images/image94.png",
"images/image95.png"
] | Chemistry | Physical Chemistry | Electrochemistry | Easy | numerical | true | JEE Main | CH-14 Chemistry Paper 7 Nov..docx | |||||
CH-04-Q11 | 15 mL of a gaseous hydrocarbon was exploded with 72 mL of oxygen.
The volume of gases on cooling was found to be 57 mL, 30 mL of which was
absorbed by aq. KOH and the rest was absorbed in a solution of alkalline
pyrogallol. Then the formula of hydrocarbon is:
15 mL 72 mL
57 mL 30 mL KOH | [] | C_2H_2 (2) C_2H_4 | 45: 8 | C_2H_6 | C_3H_6 | null | 2 | [] | Chemistry | Physical Chemistry | Gaseous State | Tough | numerical | false | JEE Main | CH-04 Paper 02 Chemistry with Hindi 22-8-2020.docx | |
CH-08-Q17 | Acedic strength of Boron trihalide is in order of | [] | 1 | null | Concentration of Lewis acid of boron tri halides is increased
in following order. | [
"images/image13.png"
] | Chemistry | Inorganic Chemistry | p block | Easy | single_correct | true | JEE Main | CH-08 Chemistry Paper 23 October.docx | ||||
CH-25-Q7 | Which acid will decarboxylate with greatest difficulty ? | [] | 1 | null | Bridge has carbon sp^2.
lsrq 'kh"kZ okys dkcZu sp^2 gksrs gSaA | [] | Chemistry | Organic Chemistry | Hydrocarbon | Moderate | single_correct | false | JEE Main | CH-25 Chemistry Paper 7 Jan.docx | ||||
CH-01-Q21 | How many number of mole of R-MgX consume in given structure? | [
"images/image60.png"
] | null | null | 08 | [] | Chemistry | single_correct | true | JEE Main | CH-01 Chemistry Paper 1.docx | |||||||
CH-26-Q3 | Equimolar solutions in the same solvent have | [] | same boiling point but different freezing point | same freezing point but different boiling point | same boiling and same freezing points | differnet boiling and freezing points
eku foyk;d | 3 | null | According to Raoult's law equimolal solutions of all the
substances in the same solvent will show equal elevation in boiling
points as well as equal depression in freezing point.
gy- vuqlkj lHkh inkFkZ | [] | Chemistry | Physical Chemistry | Easy | single_correct | false | JEE Main | CH-26 Chemistry Paper 1 15 Jan.docx | |
CH-20-Q4 | The molar heat capacity of water at constant pressure is
$75\ {J\ K}^{- 1}{mol}^{- 1}$. When 1.0 kJ of heat is supplied to 100 g
of water which is free to expand the increase in temperature of water is | [] | 6.6 K | 1.2 K | 2.4 K | 4.8 K | 3 | null | Heat capacity of water per gram$= \frac{75}{18} = 4.17J$
$Q = mst;1000 = 100 \times 4.17 \times T$
$T = \frac{1000}{100 \times 4.17} = 2.4K$ | [] | Chemistry | Physical Chemistry | Thermodynamics | Easy | single_correct | false | JEE Main | CH-20 Chemistry Paper 12 Dec..docx |
CH-15-Q21 | The volume of colloidal particle VC as compared to the volume of
a solute particle in a true solution VS could be. Find 10^Vc/ Vs^ | [] | null | 0.001 | For true solution the diameter range is 1 to 10 and for
colloidal solution diameter range is 10 to 10,000
$\frac{V_{0}}{V_{8}} = \frac{(4/3)\pi r_{c}^{3}}{(4/3)\pi r_{s}^{3}} = {(\frac{r_{c}}{r_{s}})}^{3}$
Ratio of diameters = (10/1)3 = 103
Vc/ Vs = 103 =0.001 | [] | Chemistry | Surface Chemistry | Coagulation, Protection And application of colloid | Moderate | numerical | false | JEE Main | CH-15 Chemistry Paper 11 Nov..docx | ||||
CH-21-Q2 | The Bouveault-Balance reduction involves
ckWosYV&csysal | [] | $C_{2}H_{5}OH/Na$ | ${\ LiAlH}_{4}$ | $C_{2}H_{5}{MgX}^{-}$ | $Zn/HCl$ | 1 | null | $\begin{matrix}
C_{3}H_{7}{COOC}_{2}H_{5} \\
\ Ethylbuty\ rate\
\end{matrix}\overset{NaC_{2}H_{5}OH}{\rightarrow}\begin{matrix}
C_{3}H_{7}{CH}_{2}OH \\
\ Butylalcohol\
\end{matrix}$ | [] | Chemistry | Alcohol, Phenol and Ethers | Uses of alcohol, Phenol and Ethers | Moderate | single_correct | false | JEE Main | CH-21 Chemistry Paper 19 Dec. Eng Hindi.docx |
CH-01-Q25 | How many molecules have two lone pairs on the central atom?
H_2O, SF_4, I_3¯, XeF_5¯, XeO_3, XeOF_4, PCl_3,
NCl_3, ClF_3, XeF_2, NO_2¯, CO_3^2 | [] | null | null | (i) H_2O
[IMAGE] (ii) ClF_3 | [
"images/image74.png",
"images/image75.png",
"images/image76.png"
] | Chemistry | single_correct | true | JEE Main | CH-01 Chemistry Paper 1.docx | |||||||
CH-27-Q28 | For the reaction equilibrium, N2O4 (g)[IMAGE] 2NO2
(g) the concentrations of N2O4 and NO2 at equilibrium are 4.8 × 10-2
and 1.2 × 10-2 mol L-1 respectively. The reaction for Kc =___ ×
10-3 | [
"images/image37.png",
"images/image38.png"
] | null | 3 | = 4.8 × 10-2 mol L-1, = 1.2 × 10-2 mol L-1
$K_{c} = \frac{\left. \ {\lbrack NO}_{2} \right\rbrack^{2}}{\left\lbrack N_{2}O_{4} \right\rbrack} = \frac{1.2 \times 10^{- 2} \times 1.2 \times 10^{- 3}}{4.8 \times 10^{- 2}}$
= 0.3 × 10-2 = 3 × 10-3 mol L-1 | [] | Chemistry | Physical Chemistry | Chemical Equilibrium | Moderate | numerical | true | JEE Main | CH-27 Chemistry Paper 2 15 Jan.docx | ||||
CH-26-Q38 | In a saturated solution of the sparingly soluble strong
electrolyte AgIO3 (Molecular mass = 283), the equilibrium which sets in
is:
AgIO3(s) [IMAGE] Ag+(aq) + IO3- (aq)
If the solubility product constant Ksp of AgIO3 at a given temperature
is 1.0 × 10-8, what is the mass of AgIO3 contained in 100 mL of its
saturated so... | [
"images/image56.png"
] | null | 2.83 | AgIO3(s)[IMAGE] Ag+(aq) + IO-3(aq) [s =
Ksp = s2
or = 1.0 × 10-4 × 283 g/L = 2.83 × 10-3 | [
"images/image57.png"
] | Chemistry | Physical Chemistry | Ionic Equilibrium | Moderate | numerical | true | JEE Main | CH-26 Chemistry Paper 1 15 Jan.docx | ||||
CH-12-Q5 | In the Bragg's equation for diffraction of X-rays, n
represents for | [] | Quantum number | An integer | Moles | 2 | null | Bragg's equation is $n\lambda = 2\theta\ $
Where n is an integer i.e. 1,2,3,4 etc. | [] | Chemistry | Physical Chemistry | Easy | single_correct | false | JEE Main | CH-12 Chemistry Paper 2 Nov..docx | ||
CH-22-Q7 | Identify the correct statement | [] | Iron corrodes in oxygen-free water. | Iron corrodes more rapidly in salt water because its
electrochemical potential is higher. | Corrosion of iron can be minimized by forming a contact with
another metal with a higher reduction potential. | Corrosion of iron can be minimized by forming an impermeable
barrier at its surface. | 4 | null | The statement "corrosion of iron can be minimized by forming
an impermeable barrier at its surface." is correct. | [] | Chemistry | Physical Chemistry | Electrochemistry | Easy | single_correct | false | JEE Main | CH-22 Chemistry Paper 19 Dec. Eng.docx |
CH-14-Q21 | If the [IMAGE] electrode is diluted to 100 times
then the change in e.m.f. is increase of _____________ mV | [
"images/image87.png"
] | In crease of 59 mV | Decrease of 59 mV | Increase of 29.5 mV | Decrease of 29.5 mV | null | 59 | [
"images/image88.png",
"images/image89.png"
] | Chemistry | Physical Chemistry | Electrochemistry | Moderate | numerical | true | JEE Main | CH-14 Chemistry Paper 7 Nov..docx | |
CH-18-Q3 | The IUPAC name of | [
"images/image3.png"
] | 5, 5-Diethyl-4, 4-dimethylpentane | 3-Ethyl-4,4-dimethylheptane | 1, 1-Diethyl-2, 2-dimethylpentane | 4, 4-Dimethyl-5, 5-diethylpentane | 2 | null | [
"images/image4.png"
] | Chemistry | IUPAC and Structural Isomerism | Nomenculture | Moderate | single_correct | true | JEE Main | CH-18 Chemistry Paper 27 Nov. Third.docx | |
CH-26-Q40 | The solubility of CaF2 (Ksp = 5.3 × 10-9) in 0.1 M solution of
NaF would be: Assume no reaction of cation/anion. n × 10-7. Find the
value of n.
NaF ds 0.1 M CaF2 (Ksp = 5.3 × 10-9).......
gksxh u ekusaA n × 10-7. n | [] | null | 5.3 | (a) NaF[IMAGE] Na+ + F-
0.1 0.1 0.1
CaF2 [IMAGE] Ca2+ + 2F-
x (2x + 0.1) 0.1
Ksp = x (0.1)2 = 5.3 × 10-9
So vr, x = 5.3 × 10-7 M. | [
"images/image58.png",
"images/image59.png"
] | Chemistry | Physical Chemistry | Ionic Equilibrium | Tough | numerical | true | JEE Main | CH-26 Chemistry Paper 1 15 Jan.docx | ||||
CH-06-Q3 | Select which type of overlapping is responsible for
$\pi$-character in Si-N bond of H_3SiNCO? | [] | 4 | null | $\pi$-character in Si-N bond is due to
$3d\ \pi\ \leftarrow \ 2P\ \pi$ back bonding. | [] | Chemistry | Inorganic Chemistry | Chemical Bonding | Easy | single_correct | false | JEE Main | CH-06 Chemistry Paper 14 October.docx | ||||
CH-14-Q6 | When a lead storage battery is discharged | [] | [IMAGE] is evolved | Lead sulphate is consumed | Lead is formed | Sulphuric acid is consumed | 4 | null | [IMAGE] Sulphuric acid is consumed on discharging. | [
"images/image31.png"
] | Chemistry | Physical Chemistry | Electrochemistry | Moderate | single_correct | true | JEE Main | CH-14 Chemistry Paper 7 Nov..docx |
CH-07-Q15 | Asthma patient use a mixture of..... for respiration | [] | $O_{2}$ and $N_{2}O$ | $O_{2}$ and $He$ | $O_{2}$ and $NH_{3}$ | $O_{2}$ and $CO$ | 2 | null | A mixture of $O_{2}$ and He is used for respiration as helium is
inert and light gas and diffuse rapidly. | [] | Chemistry | Organic Chemistry | Chemistry in everyday life | Moderate | single_correct | false | JEE Main | CH-07 Chemistry Paper 17 October.docx |
CH-07-Q10 | The structure given below is known as | [
"images/image5.png"
] | Penicillin F | Penicillin G | Penicillin K | Ampicillin | 2 | null | It is the known structure of penicillin G. | [] | Chemistry | Organic Chemistry | Chemistry in everyday life | Tough | single_correct | true | JEE Main | CH-07 Chemistry Paper 17 October.docx |
CH-19-Q21 | Among the triatomic molecules/ions, BeCl2, N3-, N2O, NO2+, O3,
SCl2, ICl2-, I3- and XeF2, the total number of linear
molecules(s)/ion(s) where the hybridization of the central atom does not
have contribution from the d-orbital(s) is | [] | null | 4 | [
"images/image22.png"
] | Chemistry | Chemical Bonding | Hybridization | Moderate | numerical | true | JEE Main | CH-19 Chemistry Paper 28 Nov. Fourth.docx | |||||
CH-16-Q12 | The average charge on each O atom and average bond order of I-O
bond in IO65- is | [] | -1 and 1.67 | - 5/6 and 1.67 | -5/6 and 1.33 | -5/6 and 1.167 | 4 | null | [
"images/image35.png"
] | Chemistry | Chemical Bonding | Resonance and Bond order Calculation | Easy | single_correct | true | JEE Main | CH-16 Chemistry Paper 25 Nov. Fisrt.docx | |
CH-03-Q20 | has four choices | [] | , | , | , | out of which ONLY ONE is correct]
Atomic masses: [H = 1, D = 2, Li = 7, C = 12, N = 14, O = 16, F =
19, Na = 23, Mg = 24, Al = 27, Si = 28, P = 31, S = 32, Cl = 35.5, K =
39, Ca = 40, Cr = 52, Mn = 55, Fe = 56, Cu = 63.5, Zn = 65, As = 75, Br
= 80, Ag = 108, I = 127, Ba = 137, Hg = 200, Pb = 207]
ijek. = 1, D = 2, Li =... | 4 | null | (2)
[IMAGE] A 100
(1) [IMAGE] for 3p & 4s electrons. Hence, total = 6+2 =8
(2) m = 0 for all s-electron and two p - electron and one d- electron.
Hence total = 13
(3) No. of unpaired electrons = | [
"images/image3.png",
"images/image4.png"
] | Chemistry | Physical Chemistry | Atomic Structure | Moderate | single_correct | true | JEE Main | CH-03 Paper 01 Chemistry with Hindi 21-8-2020.docx |
CH-24-Q14 | In above question, concentration of Triethyl ammonium ion
([C_6NH_16^+]) in resulting solution will be | [] | 100 K_b | 200 K_b | 10 K_b | K_b | 4 | null | $K_{b} = \frac{\left\lbrack {OH}^{-} \right\rbrack\left\lbrack C_{6}{NH}_{16}^{+} \right\rbrack}{\left\lbrack C_{6}{NH}_{15} \right\rbrack} \Rightarrow \frac{\frac{10^{- 2}}{2}\left\lbrack C_{\theta}{NH}_{16}^{+} \right\rbrack}{\frac{10^{- 2}}{2}} \Rightarrow \left\lbrack C_{6}{NH}_{16}^{+} \right\rbrack = K_{b}$ | [] | Chemistry | Physical Chemistry | Ionic Equilibrium | Moderate | single_correct | false | JEE Main | CH-24 Chemistry Paper 31 Dec..docx |
CH-19-Q12 | The shortest distance between Ist and Vth layer of HCP
arrangement is | [] | $8\sqrt{\frac{2}{3}}r$ | $4\sqrt{\frac{3}{2}}r$ | $16\frac{\sqrt{2}}{3}r$ | $8\sqrt{\frac{3}{2}}r$ | 1 | null | The shortest distance between Ist and Vth layer of HCP
arrangement is
$= 2C = 2 \times 4\sqrt{\frac{2}{3}}r = 8\sqrt{\frac{2}{3}}r$ | [] | Chemistry | HCP & CCP structures | Moderate | single_correct | false | JEE Main | CH-19 Chemistry Paper 28 Nov. Fourth.docx | |
CH-04-Q14 | The chemical formula of Phosphorous acid is | [] | H3PO4 (2) H3PO3 | H3PO2 | H2PO3 | 2 | null | [] | Chemistry | Physical Chemistry | Periodic Table | Easy | single_correct | false | JEE Main | CH-04 Paper 02 Chemistry with Hindi 22-8-2020.docx | ||
CH-02-Q25 | How many monochloro isomeric products (Including stereoisomers) is
obtained in given reaction? | [
"images/image166.png"
] | null | null | [IMAGE] Total monochloro products are 5. | [
"images/image167.png"
] | Chemistry | single_correct | true | JEE Main | CH-02 Chemistry paper 2.docx | |||||||
CH-15-Q11 | All the metal ions contain t2g6 eg0 configurations. Which of the
following complex will be paramagnetic? | [] | 1 | null | (1) [FeCl(CN)4(O2)]4-; O2 is O22-; CN- is strong field
ligand, so compels for the pairing of electrons.
Fe3+→ 3d5 (2) K4[Fe(CN)6]; CN- is strong field ligand so compels for the
pairing of electrons.
Fe2+→ 3d6 (3) [Co(NH3)6]Cl3; NH3 is strong field ligand and 3d6 configuration
has higher CFSE compelling for the pairin... | [
"images/image20.png",
"images/image21.png",
"images/image21.png",
"images/image21.png"
] | Chemistry | Coordination Compounds | Valence Bond Theory + Crystal Field Theory (Part-I) | Tough | single_correct | true | JEE Main | CH-15 Chemistry Paper 11 Nov..docx | ||||
CH-17-Q18 | Which of the following curves represents the Henry's law? | [] | 1 | null | where m = mass of gas absorbed by
given volume of the solvent.
P = pressure of gas | [
"images/image63.png"
] | Chemistry | Moderate | single_correct | true | JEE Main | CH-17 Chemistry Paper 26 Nov. Second.docx | ||||||
CH-13-Q5 | Two solutions A and B are separated by semi permeable membrane.
If liquid flows from A to B then | [] | A is less concentrated than B | A is more concentrated than B | Both have same concentration | None of these | 1 | null | Osmosis occurs from dilute solution to concentrate solution.
Therefore solution A is less concentrated than B. | [] | Chemistry | Physical Chemistry | Easy | single_correct | false | JEE Main | CH-13 Chemistry Paper 4 Nov..docx | |
CH-15-Q13 | The decomposition of N2O5 in chloroform was followed by
measuring the volume of O2 gas evolved: 2N2O5(CCI4) → 2N2O4(CCI4) +
O2(g). The maximum volume of O2 gas obtained was 100 cm3. In 500
minutes, 90 cm3 of O2 were evolved. The first order rate constant (in
min-1) for the disappearance of N2O5 is | [] | $\frac{2303}{500}$ | $\frac{2303}{500}\log\frac{100}{90}$ | $\frac{2303}{500}\log\frac{90}{100}$ | $\frac{100}{10 \times 500}$ | 1 | null | $kt = ln(\frac{C_{0}}{C_{1}})$
$2N_{2}O_{5} \longrightarrow 2N_{2}O_{4} + O_{2}$
$\begin{matrix}
t = 0 & 200{cm}^{3}\ \ \ \ \ \ \ \ \ \ \ 0\ \ \ \ \ \ \ \ \ \ \ \ 0
\end{matrix}$
$\begin{matrix}
t = t & 20{cm}^{3} & 180{cm}^{3} & 90{cm}^{3}
\end{matrix}$
$\begin{matrix}
t = \infty & 0 & \ \ \ \ \ \ \ \ 200{cm}^{3} & 10... | [] | Chemistry | Chemical Kinetics | Method to monitor the progress of Reactions (Titration Method | Moderate | single_correct | false | JEE Main | CH-15 Chemistry Paper 11 Nov..docx |
CH-17-Q13 | $B(OH)_{3} + NaOH \rightleftharpoons Na\lbrack B(OH)_{4}\rbrack$
How can this reaction be made to proceed in forward direction? | [] | Addition of cis - 1, 2-diol | Addition of boron | Addition of trans-1, 2-diol | Addition of Na2HPO4 | 1 | null | [
"images/image45.png"
] | Chemistry | P-block | Oxides, Hydroxides, Oxyacids, Borax | Easy | single_correct | true | JEE Main | CH-17 Chemistry Paper 26 Nov. Second.docx | |
CH-28-Q32 | The Gibbs energy for the decomposition of ${Al}_{2}O_{3}$ at
500°C is as follows:
$\frac{2}{3}{Al}_{2}O_{3} \rightarrow \frac{4}{3}Al + O_{2};\Delta_{r}G = + 966{kJmol}^{- 1}$
The potential difference needed for electrolytic reduction of
${Al}_{2}O_{3}$ at 500° is at least | [] | null | 2.5 | $\frac{2}{3}{Al}_{2}O_{3} \longrightarrow \frac{4}{3}Al + O_{2}$
$\Delta_{r}G = + 966kJ{mol}^{- 1} = 966 \times 10^{3}J{mol}^{- 1}$
$\Delta G = - nFE_{cell\ }$
$966 \times 10^{3} = - 4 \times 96500 \times E_{cell\ }$
$E_{cell\ } = 2.5V$ | [] | Chemistry | Physical Chemistry | Electrochemistry | Moderate | numerical | false | JEE Main | CH-28 Chemistry Paper FST 16 Jan.docx | ||||
CH-15-Q6 | Which of the following will exhibit maximum ionic conductivity? | [] | 1 | null | Conductivity a number of ions in the solution.
(1) [IMAGE] 4: 1 electrolyte.
It contains maximum number of ions i.e. | [
"images/image4.png"
] | Chemistry | Coordination Compounds | Werner's Theory | Moderate | single_correct | true | JEE Main | CH-15 Chemistry Paper 11 Nov..docx | ||||
CH-14-Q10 | [IMAGE] in basic medium is reduced to | [
"images/image41.png"
] | 2 | null | [IMAGE] is first reduced to manganate and then
to insoluble manganese dioxide. Colour change first from purple to green
and finally becomes colorless. | [
"images/image41.png",
"images/image46.png"
] | Chemistry | Inorganic Chemistry | D & F -Block Elements | Moderate | single_correct | true | JEE Main | CH-14 Chemistry Paper 7 Nov..docx | ||||
CH-07-Q25 | Two vessels whose volumes are in the ratio 2: 1 contain
nitrogen and oxygen at 2500 mm and 1000 mm pressures respectively when
they are connected together what will be the pressure of the resulting
mixture (in meters) ? | [] | null | 2 | $\frac{V_{N_{2}}}{V_{O_{2}}} = \frac{2}{1}$
$P_{N_{2}} = 2500mm,P_{O_{2}} = 1000mm$
$P_{now} = \frac{P_{N_{2}} \cdot V_{N_{2}} + P_{0_{2}} \cdot V_{0_{2}}}{V_{N_{2}} + V_{0_{2}}} = \frac{2500 \cdot V_{N_{2}} + 1000 \cdot \frac{V_{N_{2}}}{2}}{V_{N_{2}} + \frac{V_{N_{2}}}{2}}$
$= \frac{3000{VN}_{2} \cdot 2}{3VN_{2}} = 20... | [] | Chemistry | Physical Chemistry | Gaseous state | Easy | numerical | false | JEE Main | CH-07 Chemistry Paper 17 October.docx | ||||
CH-16-Q7 | Which of the following is correct order of stability | [] | Tl3+> Bi3+ | PbO2> PbO | BiI5< BiF5
(D) Sn2+ = Ge2+ | 3 | null | BiI5 does not exists because of I- being very strong reducing
agent. So it reduces Bi5+ to Bi3+ and forms BiI | [] | Chemistry | Periodic table | Oxidation states & Inert pair effect | Easy | single_correct | false | JEE Main | CH-16 Chemistry Paper 25 Nov. Fisrt.docx | |
CH-12-Q25 | of Fe in Alloy Invar is | [] | null | 64 | of Fe in Alloy Invar is 64 | [] | Chemistry | Inorganic Chemistry | Metallurgy | Easy | numerical | false | JEE Main | CH-12 Chemistry Paper 2 Nov..docx | ||||
CH-26-Q4 | Regular use of which of the following fertilizers increases the
acidity of soil? | [] | Superphosphate of lime | Ammonium sulphate | Potassium nitrate | Urea
fdl mojZd e`nk | 2 | null | (NH4)2SO4 + 2H2O → (2H+ + SO42-) + 2NH4OH
Strong acid Weakbase
(NH_4)_2 SO_4 on hydrolysis produces strong acid H_2SO_4, which
increases the acidity of the soil.
(NH4)2SO4 + 2H2O → (2H+ + SO42-) + 2NH4OH
çcy vEy nqcZy {kkj
(NH4)2 SO4 ds tyvi?kV~u ij çcy vEy H_2SO_4, curk | [] | Chemistry | Inorganic Chemistry | P-block nitrogen | Moderate | single_correct | false | JEE Main | CH-26 Chemistry Paper 1 15 Jan.docx |
CH-25-Q25 | The number of $\pi$-bonds present in pent-1-en-4-yne is:
isUV-1-bZu-4-vkbZu v.kqvksa $\pi$cU/kksa | [] | null | 3 | Pent-4-ene-1-yne
${CH}_{2} = CH - {CH}_{2} - C = CH$
No. of $\pi$ -bonds: $1(C = C) + 2(C \equiv C) = 3$
${CH}_{2} = CH - {CH}_{2} - C = CH$
$\pi$ - la $1(C = C) + 2(C \equiv C) = 3$ | [] | Chemistry | Organic Chemistry | Hydrocarbon | Easy | numerical | false | JEE Main | CH-25 Chemistry Paper 7 Jan.docx | ||||
CH-07-Q16 | Electrolysis of fused sodium hydride liberate hydrogen at the | [] | Anode | Cathode | Cathode and anode both | None of these | 1 | null | $Na \rightleftharpoons {Na}^{+} + H^{-}$
At anode: $H^{-} \rightarrow H + e^{-}$
$H + H \rightarrow H_{2}$ | [] | Chemistry | Inorganic Chemistry | Hydrogen And Its Compound | Moderate | single_correct | false | JEE Main | CH-07 Chemistry Paper 17 October.docx |
CH-09-Q11 | Primary nitro compounds when crystalline solids which on
treatment with NaOH gives react with HNO_2 forms | [] | Red solution | Blue solution | White precipitate | Yellow colouration | 1 | null | [
"images/image36.png"
] | Chemistry | Organic Chemistry | Organic compounds containing Nitrogen | Moderate | single_correct | true | JEE Main | CH-09 Chemistry Paper 26 October.docx | |
CH-06-Q21 | [IMAGE] No. of monochloro structure isomers in products is | [
"images/image33.png"
] | null | 4 | Alkyl group of this molecule has four types of replaceable
hydrogen atoms | [] | Chemistry | Organic Chemistry | Hydrocarbons | Easy | numerical | true | JEE Main | CH-06 Chemistry Paper 14 October.docx | ||||
CH-18-Q24 | The number of water molecule (s) directly bonded to the metal
center in CuSO4. 5H2O is | [] | null | 4 | [
"images/image40.png"
] | Chemistry | Chemical Bonding | H- Bonding | Moderate | numerical | true | JEE Main | CH-18 Chemistry Paper 27 Nov. Third.docx | |||||
CH-08-Q25 | No. of atoms in one molecule of sulphur is | [] | null | 8 | [
"images/image31.png",
"images/image32.jpeg"
] | Chemistry | Inorganic Chemistry | p-block elements | Easy | numerical | true | JEE Main | CH-08 Chemistry Paper 23 October.docx | |||||
CH-17-Q21 | How many positional isomers are possible for
dimethylcyclohexane? | [] | 3 | 4 | 5 | 6 | 2 | null | [
"images/image68.png"
] | Chemistry | Sub | Number of Structural Isomers | Moderate | single_correct | true | JEE Main | CH-17 Chemistry Paper 26 Nov. Second.docx | |
CH-20-Q1 | For a reaction, $X(g) \rightarrow Y(g) + Z(g)$ the half life
period is 10 min. In What period of time would the concentration of X be
reduced to 10 of original concentration. | [] | 20 min | 33 min | 15min | 25 min | 2 | null | $X(g) \rightarrow Y(g) + Z(g)$
The reaction is a first order reaction hence,
$K = \frac{0.693}{t_{1/2}} = \frac{2.303}{t}log\frac{a}{a - x} = \frac{0.693}{10\min}$
$= \frac{2.303}{t}log\frac{a}{a/10} = \frac{0.693}{10} = \frac{2.303}{t}log10$
$\therefore t = \frac{2.303 \times 10}{.693} = 33min$ | [] | Chemistry | Physical Chemistry | Chemical Kinetics | Moderate | single_correct | false | JEE Main | CH-20 Chemistry Paper 12 Dec..docx |
CH-12-Q6 | A solid is made of two elements X and Z. The atoms Z are in
CCP arrangement while the atom X occupy all the tetrahedral sites.
What is the formula of the compound | [] | XZ | XZ_2 | X_2Z | X_2Z_3 | 3 | null | Tetrahedral sites one double comparable to octahedral sites
then ratio of X and Z respectively 2: 1 since formula of the
compound X_2 Z. | [] | Chemistry | Physical Chemistry | Moderate | single_correct | false | JEE Main | CH-12 Chemistry Paper 2 Nov..docx | |
CH-20-Q18 | The correct order of increasing C-O bond length of
$CO,{CO}_{3}^{2 -},{CO}_{2}$ is | [] | ${CO}_{3}^{2 -} < {CO}_{2} < CO$ | ${CO}_{2} < {CO}_{3}^{2 -} < CO$ | $CO < {CO}_{3}^{2 -} < {CO}_{2}$ | $CO < {CO}_{2} < {CO}_{3}^{2 -}$ | 4 | null | B.O. in CO i.e.$:\overset{-}{C} = \overset{+}{O}:$ is 3, that
of$O = C = O$is 2 while that of${CO}_{3}^{2 -}$ ion is 1. | [] | Chemistry | Inorganic Chemistry | Chemical Bonding | Easy | single_correct | false | JEE Main | CH-20 Chemistry Paper 12 Dec..docx |
CH-12-Q22 | The number of octahedral voids in a unit cell of a cubical
closest packed structure is | [] | null | 4 | Number of atoms in the cubic close packed structure = | [] | Chemistry | Physical Chemistry | Moderate | numerical | false | JEE Main | CH-12 Chemistry Paper 2 Nov..docx | |||||
CH-17-Q14 | Volume of HCP unit cell is | [] | 24r3 | 8r3 | 16r3 | 24r3 | 1 | null | Volume of hexagon:
[IMAGE] So [IMAGE] and Area of hexagonal
surface[IMAGE] volume of hexagon = area of base × height | [
"images/image46.png",
"images/image47.png",
"images/image48.png",
"images/image49.png",
"images/image50.png",
"images/image51.png"
] | Chemistry | HCP & CCP structures | Tough | single_correct | true | JEE Main | CH-17 Chemistry Paper 26 Nov. Second.docx | |
CH-02-Q23 | Among the following inter halogen compounds, the total number of
compounds that exist in the gaseous state at room temperature is | [
"images/image151.png"
] | null | null | [IMAGE] - Gases
[IMAGE] - liquids | [
"images/image152.png",
"images/image153.png",
"images/image154.png",
"images/image155.png"
] | Chemistry | single_correct | true | JEE Main | CH-02 Chemistry paper 2.docx | |||||||
CH-13-Q10 | Nitrogen is liberated by the thermal decomposition of only | [] | NH_4 NO_2 | NaN_3 | (NH_4)_2 Cr_2 O_7 | All the these | 4 | null | NH_4 NO_2 [IMAGE] N_2 + 2H_2O
2NaN_3[IMAGE] 2Na + 3N_2
(NH_4)_2 Cr_2 O_7[IMAGE] N_2 + Cr_2
O_3 + 4H_2O | [
"images/image21.png",
"images/image22.png",
"images/image23.png"
] | Chemistry | Inorganic Chemistry | P-Block Element | Moderate | single_correct | true | JEE Main | CH-13 Chemistry Paper 4 Nov..docx |
CH-14-Q15 | distinguished by testing with | [
"images/image66.png",
"images/image67.png"
] | Phenyl hydrazine | Hydroxylamine | Fehling solution | Sodium bisulphite | 3 | null | [
"images/image68.png",
"images/image69.png"
] | Chemistry | Organic Chemistry | Aldehydes, Ketones & Carboxylic Acids | Easy | single_correct | true | JEE Main | CH-14 Chemistry Paper 7 Nov..docx | |
CH-17-Q17 | Given standard electrode potentials:
Fe3+ + 3e¯ → Fe; E° = -0.036 volt
Fe2+ + 2e¯ → Fe; E° = -0.440 volt
The standard electrode potential E° for Fe3+ + e¯ Fe2+ | [] | -0.476 volt | -0.404 volt | -0.440 volt | -0.772 volt | 4 | null | [IMAGE] E° = 0.772 Volt | [
"images/image56.png",
"images/image57.png",
"images/image58.png"
] | Chemistry | Electrochemistry | Concept of △G | Moderate | single_correct | true | JEE Main | CH-17 Chemistry Paper 26 Nov. Second.docx |
CH-01-Q4 | In which of the following pairs both are monobasic, proton donor
acid when dissolved in water? | [] | H_3PO_2, H_3BO_3 | H_3PO_2, HClO_4 | HClO_3, H_3BO_3 | H_3PO_2, H_3PO_3 | null | null | [IMAGE] Both are monobasic and proton donor acid.
But H_3BO_3 is not proton donor acid.
H_3BO_3 + H_2O → [B(OH)_4]^¯ + H^+
H_3BO_3 is only Lewis acid because it accept OH¯ from water. | [
"images/image12.png"
] | Chemistry | single_correct | true | JEE Main | CH-01 Chemistry Paper 1.docx | |||
CH-02-Q10 | The first noble gas compound formed was........... and reason
behind this discovery was......... | [] | [IMAGE] ionization enthalpy IE of
[IMAGE] of Xe | [IMAGE] ionization enthalpy (IE) of
[IMAGE] of Xe | [IMAGE] ionization enthalpy (IE) of
[IMAGE] of Xe | [IMAGE] ionization enthalpy (IE) of
[IMAGE] of Xe | null | null | (4) It is theoretical. | [] | Chemistry | single_correct | false | JEE Main | CH-02 Chemistry paper 2.docx | |||
CH-17-Q6 | Which Fisher projection represents the given wedge dash
structure | [
"images/image10.png"
] | 1 | null | First decide the (R/S) configuration in wedge-dash and then
draw Fischer projection for same configuration. | [] | Chemistry | Isomerism | R/S & D/L Naming. | Moderate | single_correct | true | JEE Main | CH-17 Chemistry Paper 26 Nov. Second.docx | ||||
CH-23-Q25 | A gaseous mixture of three gases A, B and C has a pressure of 10
atm. The total number of moles of all the gases is 10. If the partial
pressure of A and B are 3.0 and 1.0 atm respectively and if C has mol.
wt. of 2.0, what is the weight of C in g present in the mixture ? | [] | null | 12 | Pressure of Total mixture = 10 atm
$P_{A} + P_{B} + P_{c} = 10$
$3 + 1 + P_{c} = 10 \Rightarrow P_{c} = 6\ atm\ $
Total moles of mixture = 10
$N_{A} + n_{B} + n_{C} = 10$
$\frac{P_{A}}{P_{B}} = \frac{n_{A}}{n_{B}} = \frac{3}{1} \Rightarrow \frac{P_{B}}{P_{C}} = \frac{n_{B}}{n_{C}} = \frac{1}{6}$
Let
$n_{A} = K \Rightar... | [] | Chemistry | Physical Chemistry | Ideal gases | Moderate | numerical | false | JEE Main | CH-23 Chemistry Paper 31 Dec. FST.docx | ||||
CH-23-Q9 | Predict the hybridisation and geometry of the following
complexes. | [] | 2 | null | (2) In the paramagnetic octahedral complex, [Fe(CN)6]3-, the iron
ion is in +3 oxidation state and has the electronic configuration
represented as shown below.
Fe3+,[Ar]3d5 (CN)6]3- [IMAGE] d2sp3 hybrid orbital & octahedral.
(3) In the paramagnetic and tetrahedral complex the
manganese is in +2 oxidation state and the ... | [
"images/image28.png",
"images/image29.png"
] | Chemistry | Inorganic Chemistry | Coordination compounds final | Moderate | single_correct | true | JEE Main | CH-23 Chemistry Paper 31 Dec. FST.docx | ||||
CH-07-Q24 | Two flask A & B have capacity of 1 litre and 2 litre
respectively. Each of them contain 1 mole of a gas. The temperature of
the flask are so adjusted that average speed of molecules in "A" is
twice that in "B" & pressure in flask "A" is x times of that in "B".
Then value of x is | [] | null | 8 | $V_{A} = 1\ lit,V_{B} = 2\ lit$
${(u_{avg})}_{A} = 2 \cdot {(u_{avg})}_{B}$
$\sqrt{\frac{8RT_{A}}{\pi M}} = 2 \cdot \sqrt{\frac{8RT_{B}}{\pi M}}$
$\frac{T_{A}}{T_{B}} = \frac{4}{1}$
$P_{A} = \frac{n_{A} \cdot R \cdot T_{p}}{V_{A}} \cdot P_{B} = \frac{n_{B} \cdot R \cdot T_{B}}{V_{B}}$
$\frac{P_{A}}{P_{B}} = \frac{T_{A}... | [] | Chemistry | Physical Chemistry | Gaseous state | Tough | numerical | false | JEE Main | CH-07 Chemistry Paper 17 October.docx | ||||
CH-15-Q2 | The donor sites of (EDTA)4- are ? | [] | O atoms only | N atoms only | Two N atoms and four O atoms | Three N atoms and three O atoms | 3 | null | It attaches to the central metal ion through four carboxylate
group oxygen atoms and the two amine nitrogen. | [
"images/image1.png"
] | Chemistry | Coordination Compounds | General introduction of complex salts and definitions to be | Moderate | single_correct | true | JEE Main | CH-15 Chemistry Paper 11 Nov..docx |
CH-15-Q15 | The substance undergoes first order decomposition. The
decomposition follows two parallel first order reactions as:
[IMAGE] K1 = 1.26 × 10-4 sec-1 and K2 = 3.8 × 10-5 sec-1
The percentage distribution of B and C | [
"images/image22.png"
] | 80 B and 20 C | 76.83 B and 23.17 C | 90 B and 10 C | 60 B and 40 C | 2 | null | $\%\ of\ B = \frac{k_{1} \times 100}{k_{1} + k_{2}} = \frac{1.26 \times 10^{- 4} \times 100}{12.6 \times 10^{- 5} + 3.8 \times 10^{- 5}} = 76.83\%$
$\%\ of\ C = \frac{k_{2} \times 100}{k_{1} + k_{2}} = \frac{3.8 \times 10^{- 5} \times 100}{12.6 \times 10^{- 5} + 3.8 \times 10^{- 5}} = 23.17\%$ | [] | Chemistry | Chemical Kinetics | Complication in 1^stOrder Reaction | Tough | single_correct | true | JEE Main | CH-15 Chemistry Paper 11 Nov..docx |
CH-26-Q24 | (CH_3)_3CMgCl on reaction with D_2O produces
(CH_3)_3CMgCl, D_2O | [] | (CH_3)_3CD | (CH_3)_3OD | (CD_3)_3CD | (CD_3)_3OD | 1 | null | [
"images/image41.png"
] | Chemistry | Organic Chemistry | Hydrocarbon | Moderate | single_correct | true | JEE Main | CH-26 Chemistry Paper 1 15 Jan.docx |
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