Grass-G/jee-main-questions huggingface dataset:
Subject: Chemistry — Inorganic Chemistry — P-block
Difficulty: Easy / Moderate/ Tough

Question: Identify the incorrect statement

1) S_2 is paramagnetic like oxygen.
2) Rhombic and monoclinic sulphur have S_8 molecules.
3) S_8 ring has a crown shape.
4) The S-S-S bond angles in the S_8 and S_6 rings are the same.

Correct option: 4

Solution:
The statement "The $v = \frac{\omega}{k}$ bond angle in
the $= \frac{200\pi}{\left( \frac{5\pi}{4} \right)}$ and  = 160 m/s
$\Phi_{1} = \Phi_{2} = \Phi_{3} = \Phi_{4}\ $rings are the same." is
incorrect.
Both $C = \frac{\varepsilon_{0}A}{3}$ and
$C = \frac{\left( \frac{k\varepsilon_{0}A}{3} \right)\left( \frac{\varepsilon_{0}A}{2.4} \right)}{\frac{k\varepsilon_{0}A}{3} + \frac{\varepsilon_{0}A}{2.4}}$ molecule
have the same bond length but different bond angles.
The$\frac{\varepsilon_{0}A}{3} = \frac{\left( \frac{k\varepsilon_{0}A}{3} \right)\left( \frac{\varepsilon_{0}A}{2.4} \right)}{\frac{k\varepsilon_{0}A}{3} + \frac{\varepsilon_{0}A}{2.4}}$`<!- ->`3
k=2.4 k+3molecule has a puckered ring with crown conformation,
whereas $\frac{(A - C)}{D}$ molecule has a chair conformation as shown
below.
Subject: Chemistry — Coordination Compounds — Applications of crystal field theory
Difficulty: Tough

Question: Among the following ions, which one has the highest
paramagnetism?


Correct option: 2

Solution:
(1) Cr3+, 3d3
[IMAGE] n = 3
(2) Fe2+, 3d5 [IMAGE] n =
Subject: Chemistry — Inorganic Chemistry — P-block nitrogen
Difficulty: Moderate

Question: The number of hydrogen atom (s) attached to phosphorus atom in
hypophosphorus acid is:
gkbiks QkLQksjl vEy esa QkWLQksjl ijek.kq ls tqM+s gq, gkbMªkstu ijek.kq
¼Bksl½


Answer: 2

Solution:
Hypophosphorus acid gkbiksQkWLQksjl vEy
[IMAGE] Number of hydrogen atom(s) attached to phosphorus atom is 2 which are
called as reducing hydrogen.
QkWLQksjl ijek.kq ls tqM+s gkbMªkstu ijek.kq nks gSa] tks
fd vipf;r gkbMªkstu dgykrs gSaA
Subject: Chemistry — Inorganic Chemistry — D & F -Block Elements
Difficulty: Easy

Question: Which element gives maximum balanced oxide?

1) V
2) Cr
3) Mn
4) Fe

Correct option: 3

Solution:
Among the transition metals Mn forms maximum no. of oxides.
Subject: Chemistry —  — 

Question: Given that (in[IMAGE] ) at T = 298 K:
[IMAGE] are 228.8,120.3 & 129.8 respectively. Specific
conductance for 0.2 N [IMAGE] solution is
[IMAGE] then value of pH of the given solution of
[IMAGE] will be nearly: (Take log 2 = 0.3)



Solution:
Subject: Chemistry — Organic Chemistry — Aromatic compounds Final
Difficulty: Moderate

Question: Complete the following reactions


Correct option: 1

Solution:
Subject: Chemistry — Surface Chemistry — Purification and Preparation of Colloid
Difficulty: Moderate

Question: Bleeding is stopped by the application of ferric chloride. This
is because

1) the blood starts flowing in opposite direction
2) the blood reacts and forms a solid, which seals the blood vessel
3) the blood is coagulated and thus the blood vessel is sealed
4) the ferric chloride seals the blood vessel.

Correct option: 3

Solution:
Blood is a colloidal solution containing a -ve charge
colloidal particle (Albuminoid), bleeding can be stopped by use of alum
or FeCl3 solution. The addition of Al3+ or Fe3+ causes coagulation of
blood, so bleeding stops.
Subject: Chemistry — Physical Chemistry — Atomic Structure
Difficulty: Tough

Question: Calculate the energy required to excite one litre Hydrogen gas
at 1 atm and 298 K to first excited state of atomic hydrogen. The energy
for the dissociation of H-H bond is 436 kJ mol-1.


Answer: 98

Solution:
Determination of number of moles of hydrogen gas
$n = \frac{PV}{RT} = \frac{1 \times 1}{0.082 \times 298} = 0.0409$
The concerned reaction is
$H_{2} \longrightarrow 2H;\Delta H = kJ{mol}^{- 1}$
Energy required to bring 0.0409 moles of hydrogen gas to atomic state
$= 436 \times 0.0409 = 17.83kJ$
1 mole of H2 gas has 6.02 × 1023 molecules
0.0409 mole of H2 gas $= \frac{6.02 \times 10^{23}}{1} \times 0.0409$
Since 1 molecule of H2 gas has 2 hydrogen atoms
6.02 × 1023 × 0.0409 molecules of H2 gas
$= 2 \times 6.02 \times 10^{23} \times 0.0409\ $
$= 4.92 \times 10^{22}\ atoms$
Energy required to excite an electron from the ground state to the next
excited state
$= 13.6\left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right)eV = 13.6 \times \left( \frac{1}{1} - \frac{1}{4} \right) = 13.6 \times \frac{3}{4} = 10.2eV = 1.632 \times 10^{- 21}kJ$
Therefore energy required to excite 4.92 × 1022 electrons
$= 1.632 \times 10^{- 21} \times 4.92 \times 10^{22}kJ$
$= 8.03 \times 10 = 80.3kJ$
Therefore total energy required $= 17.83 + 80.3 = 98.17kJ$
Subject: Chemistry — IUPAC Nomenclature — IUPAC -Nomenclature of chain terminating Functional groups
Difficulty: Easy

Question: The correct IUPAC name of following compound is

1) 4-Aminomethyl-3-hydroxycyclohex-5-ene-1-carboxylic acid
2) 2-Aminomethyl-5-carboxycyclohex-3-en-1-ol
3) 4-Aminomethyl-5-hydroxycyclohex-2-ene-1-carboxylic acid
4) 3 -Hydroxy-4-aminomethylcyclohex-5-en-1-oic acid

Correct option: 3

Solution:
acid
Subject: Chemistry — Physical Chemistry — Chemical Equilibrium
Difficulty: Moderate

Question: If 10-4 dm3 of water is introduced into a 1.0 dm3 flask at 300
K, how many moles of water are in the vapour phase when equilibrium is
established ?
(Given: Vapour pressure of H2O at 300 K is 3170 Pa; R= 8.314 J K-1
mol-1)
 rks lkE;
LFkkfir gks tkus ij ty ds fdrus eksy ok"i izkoLFkk esa gksaxs 
( 300 K ij H2O dk ok"i nkc 3170 Pa

1) 5.56 × 10-3 mol
2) 1.53 × 10-2 mol
3) 4.46 × 10-2 mol
4) 1.27 × 10-3 mol

Correct option: 4

Solution:
PV = nRT
V = 1 dm3 = 10-3 m3
P = 3170 Pa
R = 8.314 J K-1 mol-1
T = 300 K
3170 × 10-3 = n × 8.314 × 300
n = $\frac{3170 \times 10^{- 3}}{8.314 \times 300}$ =1.27 × 10^-3 mol
Subject: Chemistry — Physical Chemistry — 
Difficulty: Easy

Question: Benzene and toluene form nearly ideal solutions. At 20ºC, the
vapour pressure of benzene is 75 torr and that of toluene is 22
torr. The parial vapour pressure of benzene of 20ºC for a solution
containing 78g of benzene and 46g of toluene in torr is.


Answer: 50

Solution:
[IMAGE] P_B = 50 torr
Subject: Chemistry — Chemical Kinetics — Integrated Rate law : Second Order & Pseudo first order
Difficulty: Moderate

Question: Compound A and B react to form C and D in a reaction that was
found to be second-order over all and second-order in A. The rate
constant -at 30ºC is 0.622 L mol-1 min-1. What is the half-life of A
when 4.10 × 10-2 M of A is mixed with excess B?


Answer: 39.2

Solution:
A + B → C + D
Second order with respect to A
then,
$t_{1/2} = \frac{1}{{Ka}_{0}} \Rightarrow t_{12} = \frac{1}{0.622 \times 4.10 \times 10^{- 2}} = 39.2min$
Subject: Chemistry — Organic Chemistry — Organic Chemistry
Difficulty: Moderate

Question: Which is the decreasing order of stability?

1) (i) < (ii) < (iii)
2) (i) > (ii) > (iii)
3) (iii) > (ii) > (i)
4) (i) > (iii) > (i)

Correct option: 3

Solution:
Subject: Chemistry — Physical Chemistry — Ionic Equilibrium
Difficulty: Moderate

Question: Calculate the change in pH of 0.1 M Na_2HA solution after
diluting it to ten times the original volume. (Given that for H_3A:
[IMAGE] = 10^-4, [IMAGE] = 10^-7,
[IMAGE] = 10^-11)


Answer: 0

Solution:
pH of solutions of amphiprotic species is independent of
concentration of solution.
Subject: Chemistry — Inorganic Chemistry — Chemical Bonding
Difficulty: Tough

Question: Consider all the atoms of C_2H_2 are on Z-axis. Find total no.
of orbitals which involve in overlapping along Z-axis


Answer: 6

Solution:
Total no. of orbitals = [4-sp orbitals from 2 carbon atoms +
two 1s-orbitals from 2 hydrogen atoms] = 6
Subject: Chemistry — Physical Chemistry — Electrochemistry
Difficulty: Easy

Question: The specific conductance of a solution is [IMAGE] and conductivity is [IMAGE] The cell constant would be



Solution:
Subject: Chemistry — Physical Chemistry — Ionic Equilibrium
Difficulty: Tough

Question: The minimum volume of the water needed to dissolve 1 g of BaSO4
(Ksp = 10-10) is about: Assume no reaction of cation/anion.
[Mol. mass (BaSO4) = 233 u]
1g BaSO4 (Ksp =10-10) dks ?kksyus ds fy, ty 
yxHkx........


Answer: 430

Solution:
Solubility of BaSO_4 =
$\sqrt{K_{5p}} = \sqrt{10^{- 10}} = 1 \times 10^{- 5}M$
$\therefore$ wt. of BaSO4 = 1 × 10-5 × 233 = 233 × 10-5 g/litre.
$\therefore$ Volume of water needed to dissolve 1 g BaSO4 is equal to
$\frac{1}{233 \times 10^{- 5}} \approx$ 430 litres
BaSO_4 =
$\sqrt{K_{5p}} = \sqrt{10^{- 10}} = 1 \times 10^{- 5}M$
$\therefore$ BaSO4 = 1 × 10-5 × 233 = 233 × 10-5 xzke/yhVj.
$\therefore$ 1gBaSO4 dks ?kksyus ds fy, ty
Subject: Chemistry — Physical Chemistry — Chemical equilibrium
Difficulty: Moderate

Question: The equilibrium constant at 298 K for a reaction
$A + B \rightleftharpoons C + D$ is 100. If the initial concentration of
all the four species were 1 M each, then equilibrium concentration of D
(in mol L-1) will be


Answer: 1.818

Solution:
$A\ \ \ \ \ \ \ \ \ + \ \ \ \ \ \ \ \ \ B\ \ \ \ \ \ \ \ \ \rightleftharpoons \ \ \ \ \ \ \ \ \ \ C\ \ \ \ \ \ \ \ \ + \ \ \ \ \ \ \ \ D$
t = 0 1 1 1 1
$t_{eq}$ $1 - x\ $ $1 - x$ $1 + x$ $1 + x$
$\Rightarrow \frac{(1 + x)^{2}}{(1 - x)^{2}} = 100 \Rightarrow \frac{1 + x}{1 - x} = 10$
$\Rightarrow 1 + x = 10 - 10x \Rightarrow 11x = 9$
$\Rightarrow x = \frac{9}{11}$
$\Rightarrow \lbrack D\rbrack = 1 + \frac{9}{11} \Rightarrow \lbrack D\rbrack = 1.818$
Subject: Chemistry — Atomic Structure — Quantum numbers & Electronic configuration
Difficulty: Moderate

Question: Given is the electronic configuration of element X:
dlh rRo X dk bysDVªkWfud K L M N
2 8 11 2
The number of electrons present with = 2 in an atom of element X is:
rRo X ds,d ijek.kq esa  = 2 okys bysDVªkWuks

1) 3
2) 6
3) 5
4) 4

Correct option: 1

Solution:
$X_{23}:1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{3}4s^{2}$
No. of electron with$\mathcal{l =}2$are$3\left( 3d^{3} \right)$
$\mathcal{l =}2$
Subject: Chemistry — Organic Chemistry — Hydrocarbon
Difficulty: Moderate

Question: Arrange the following alkanols A, B and C in order of their
reactivity towards acid catalysed dehydration:-,flM mRizsfjr futZyhdj.k

1) A > B > C
2) B > A > C
3) B > C > A
4) C > B > A

Correct option: 3

Solution:
n catalytic dehydration carbocation is formed as an
intermediate.
n catalytic fMgkbMªs"ku dkcksZds"ku,d:i esa curk gSA
Rate of dehydration of alcohol $\propto$ Stability of carbocation
"kjkc
Subject: Chemistry — Organic Chemistry — HALOALKANES & HALOARENES
Difficulty: Easy

Question: When phenol reacts with CHCl_3 and KOH, the product
obtained would be

1) Salicylaldehyde
2) p-hydroxyl benzaldehyde
3) Both (a) and (b)
4) Chloretone

Correct option: 3

Solution:
+ 2CHCl_3 + 6KOH →
Subject: Chemistry — Periodic table — Shielding Effect & Z_eff
Difficulty: Easy

Question: How many of the following have greaterZ_eff than Silicon atom:
Na, Mg, Al, P, Cl, S, N, O, C, F


Answer: 7

Solution:
N, O, F, P, C, S, Cl
Subject: Chemistry — Inorganic Chemistry — Periodic table
Difficulty: Easy

Question: The correct order of electronegativity on Pauling scale is

1) $F > Cl > O > S$
2) $Li > Na > K > Rb > Cs$
3) $Be < B < N < C$
4) Both (A) and (B)

Correct option: 2

Solution:
Electronegativity of elements generally increases across the
period (less increase) and decreases down the group (more decrease).
Subject: Chemistry — Organic Chemistry — ALKENE
Difficulty: Moderate

Question: A gas decolourised by ${KMnO}_{4}$ solution but givesno
precipitate with ammoniacal cuprous chloride is

1) Ethane
2) Methane
3) Ethene
4) Acetylene

Correct option: 3

Solution:
Subject: Chemistry — Chemical Kinetics — Rate of chemical reaction and Dependence of Rate : Basic
Difficulty: Tough

Question: If rate constant is numerically the same for the three reactions
of first, second and third order respectively. Assume all the reactions
of the kind A→ products. Which of the following is correct

1) if = 1 then r_1= r_2 = r_3
2) if < 1 thenr_1> r_2 > r_3
3) if > 1 then r_3 < r_2 < r_1
4) All

Correct option: 4

Solution:
Rate law for: I^storder II^ndorder III^rdorder
Rate = R_3 = 
Thanwe can say < 1 then r_1> r_2 >
r_3
 > 1 then r_3 < r_2 < r_1
Subject: Chemistry — Organic Chemistry — Aldehydes, Ketones & Carboxylic Acids
Difficulty: Easy

Question: Which one of the following reactions is a method for the
conversion of a ketone into a hydrocarbon?

1) Aldol condensation
2) Reimer-Thiemann reaction
3) Cannizzaro reaction
4) Wolf-Kirshner reduction

Correct option: 4

Solution:
Subject: Chemistry —  — 

Question: EAN of the central metal in the complexes K_2[Ni(CN)_4],

1) 36, 35, 86
2) 34, 35, 84
3) 34, 35, 86
4) 34, 36, 86


Solution:
Subject: Chemistry — Inorganic Chemistry — Chemical Bonding
Difficulty: Easy

Question: The correct bond angle order of CCl_4, SiCl_4, GeCl_4 and
SnCl_4 is

1) CCl_4> SiCl_4> GeCl_4> SnCl_4
2) CCl_4< SiCl_4< GeCl_4< SnCl_4
3) CCl_4 = SiCl_4 = GeCl_4 = SnCl_4
4) None of these

Correct option: 3

Solution:
CCl_4, SiCl_4, GeCl_4 and SnCl_4 all are sp^3 hybridised
symmetrical species there fore bond angle will be same.
Subject: Chemistry — Salt Analysis — IVth Group
Difficulty: Moderate

Question: An aqueous solution of colourless metal sulphate $M$, gives a
white precipitate with $NH4OH$. This was soluble in excess of $NH4OH$.
On passing $H2S$ gas through this solution a white precipitate is
formed. The metal $M$ in the salt is

1) Ca
2) Ba
3) Al
4) Zn

Correct option: 4

Solution:
${Zn}^{2 +} + 2{NH}_{4}OH \longrightarrow Zn(OH)_{2} \downarrow (\ white\ ) + 2{{NH}_{4}}^{+}$
$Zn(OH)_{2} + 4{NH}_{4}OH \longrightarrow Zn\left( {NH}_{3} \right)_{4} + 6H_{2}O$
Subject: Chemistry — Physical Chemistry — Chemical Equilibrium
Difficulty: Moderate

Question: 2 mole each of SO_3, CO, SO_2 and CO_2 is taken in a one lit.
Vessel. If KC for
${SO}_{3}(g) + CO(g) \rightleftharpoons {SO}_{2}(g) + {CO}_{2}(g)$. Is

1) Total no. of moles at equilibrium is less than 8
2) n(SO_3) + n(CO_2) = 4
3) [n(SO_2)/n(CO)] < 1
4) Both (B) and (C).

Correct option: 4

Solution:
${SO}_{3}(g) + CO(g) \rightleftharpoons {SO}_{2}(g) + {CO}_{2}(g)$
$9 = \left( \frac{2 + x}{2 - x} \right)^{2}$
2 2 2 2
2+x 2+x 2-x 2-x x = 1
n_eq = 3 + 3 + 1 + 1 = 8 n(SO_3) + n(CO_2) = 4
$\frac{n\left( {SO}_{2} \right)}{n(CO)} = \frac{1}{2} < 1$
Therefore, (D) option is correct.
Subject: Chemistry — Inorganic Chemistry — p block
Difficulty: Easy

Question: Carbon suboxide C_3O_2 has

1) Linear structure
2) Bent structure
3) Trigonal planar structure
4) Distorted tetrahedral structure

Correct option: 1

Solution:
Carbon suboxide has linear structure with
C - C bond length equal to [IMAGE] and
C - O bond length equal to [IMAGE]
Subject: Chemistry — Organic Chemistry — Organic compounds containing Nitrogen
Difficulty: Moderate

Question: In the series of reaction
[IMAGE] X and Y are respectively


Correct option: 3

Solution:
Subject: Chemistry — Organic Chemistry — Hydrocarbon
Difficulty: Moderate

Question: Kolbe's electrolysis of sodium maleate at anode gives.
Kolbe

1) Only acetylene dsoy acetylene
2) Only CO2 dsoy CO2
3) Acetylene + CO2
4) CO2 + H2

Correct option: 3

Solution:
Subject: Chemistry — Inorganic Chemistry — P-Block Element
Difficulty: Easy

Question: Increasing order of solubility is

1) CaCO_3, KHCO_3, NaHCO_3
2) NaHCO_3, KHCO_3, CaCO_3
3) KHCO_3, NaHCO_3, CaCO_3
4) CaCO_3, NaHCO_3, KHCO_3

Correct option: 4

Solution:
CaCO_3, > NaHCO_3 > KHCO_3 is the increasing order
of solubility.
Subject: Chemistry — Physical Chemistry — Redox Reaction
Difficulty: Moderate

Question: Which of the following substances acts as an oxidizing as well as
a reducing agent?

1) Na_2O
2) SnCl_2
3) Na_2O_2
4) NaNO_2

Correct option: 4

Solution:
NaNO_2, (Sodium nitrite) act both as oxidizing as well as
reducing agent because in it N atom is in +3 oxidation state
(intermediate oxidation state)
Oxidizing property
[IMAGE] Reducing property
Subject: Chemistry — Coordination Compounds — Werner's Theory
Difficulty: Tough

Question: 1 g of complex was passed through a cation
exchanger to produce HCl. The acid liberated was diluted to 1 liter.
What will be the molarity of acid solution [Molecular weight of complex
= 266.5] ?


Answer: 0.0075

Solution:
1 mole of complex will give two moles of Cl- ion, i.e. 2 mole
HCl
$\lbrack Cr{(H_{2}O)}_{5}Cl\rbrack{Cl}_{2} \cdot H_{2}O \rightleftharpoons {\lbrack Cr{(H_{2}O)}_{5}Cl\rbrack}^{+ 2} + 2{Cl}^{-}$
1g
mole of HCl$= \frac{2 \times 1}{266.5}$ (for HCl, V.f. = 1
[IMAGE] N = M)
$\therefore N = \frac{2/266.5}{1} = 0.0075$
Subject: Chemistry — Organic Chemistry — Chemistry in everyday life
Difficulty: Moderate

Question: Which of the following is not an antiseptic drug

1) Iodoform
2) Dettol
3) Gammexane
4) Genatian violet

Correct option: 3

Solution:
It is an insecticide.
Subject: Chemistry — Chemical Bonding — Applications of H-bond
Difficulty: Moderate

Question: Amongst NH3, PH3, AsH3 and SbH3 the one with highest boiling
point is

1) NH3 because of lower molecular weight
2) SbH3 because of higher molecular weight
3) PH3 because of H-bonding
4) AsH3 because of lower molecular weight

Correct option: 2

Solution:
As molecular weight increases, magnitude of Van der Waal's
force of attraction increases.
Therefore, boiling point increases. Though NH3 has strong H-bonding
but boiling point of SbH3 is highest due to highest molecular weight.
Boiling point NH3 = 238.5 K and SbH3 = 254.6 K.
Subject: Chemistry — Salt Analysis — Precipitation Reactions
Difficulty: Moderate

Question: When a mixture containing phosphate is heated with conc. $HNO3$
and ammonium molybdate solution, a canary yellow precipitate is formed.
The formula of the yellow precipitate is:
tc QkWLQsV eksfyCMsV foy;u

1) $(NH4)3PO4$
2) $(NH4)3PO4.12MoO4$
3) $(NH4)3PO4.12MoO3$
4) $(NH4)3PO4.(NH4)2MO4$

Correct option: 3

Solution:
$NaHPO4\ + \ 12(NH4)2MoO4\ + \ 23HNO3\ \rightarrow$
$(NH4)3PO4.12MoO3\ + \ 2MaNO3\ + \ 21NH4NO3\ + \ 12H2O$
(Canary yellowdsusjhihyk)
Subject: Chemistry — Organic Chemistry — Hydrocarbon
Difficulty: Moderate

Question: Which of the following undergoes decarboxylation most readily on
being heated ?
nsrk


Correct option: 4

Solution:
$\beta$-keto acid decarboxylate readily
$\beta$-dhVks vEy rsth
Subject: Chemistry — Organic Chemistry — Hydrocarbon
Difficulty: Moderate

Question: Hydration of [IMAGE] in presence of H2SO4 / HgSO4 gives
 H2SO4 / HgSO4


Correct option: 1

Solution:
Subject: Chemistry — Inorganic Chemistry — Metallurgy
Difficulty: Moderate

Question: Main function of roasting is

1) To remove volatile substances
2) Oxidation
3) Reduction
4) Slag formation

Correct option: 1

Solution:
To remove volatile substance.
S_8 + 8O_2 → 8SO_2 ↑; P_4 + 5O_2 → P_4 O_10 ↑
4 As + 3O_2 → 2AS_2 O_3 ↑
Subject: Chemistry — Inorganic Chemistry — P-block nitrogen
Difficulty: Moderate

Question: Which of the following is the wrong statement ?

1) ONCl and ONO^- are not isoelectronic.
2) O_3 molecule is bent
3) Ozone is violet-black in solid state
4) Ozone is diamagnetic gas.


Solution:
(1) ONCl = 8 + 7 + 17 = 32e^-
ONO- = 8 + 7 + 8 + 1 = 24e^- (correct)
(2) [IMAGE] Central atom O is sp^2 hybridised with
1 lone pair, so bent shape (correct)
(3) Ozone is violet-black in solid state.
(4) O_3 has no unpaired electrons, so diamagnetic (correct)
(1) ONCl = 8 + 7 + 17 = 32e^-
ONO- = 8 + 7 + 8 + 1 = 24e^- (lgh)
(2) vkWDlhtu ijek.kq sp2 ladfjr gS
rFkk;g,d,dkdh bySDVªkWu vr bldh eqM+h gqbZ vkd`fr
(bent) gksxh (lgh)
(3) Bksl voLFkk es vkstksu cSaxuh&dkyk gksrk
Subject: Chemistry — Physical Chemistry — Gaseous state
Difficulty: Moderate

Question: For a real gas (van der temperature and
low pressure, the pressure may be ($V_{m}$⇒ molar volume)

1) $a.b.V_{m}$
2) $\frac{V_{m}}{a.b}$
3) $\frac{a}{V_{m} \cdot b}$
4) $\frac{b}{a \cdot V_{m}}$

Correct option: 3

Solution:
A real gas behaves ideally at Boyle's temperature in low pressure
range.
$Z = \frac{{PV}_{M}}{RT} = 1$
$P = \frac{RT}{V_{M}} = \frac{R\frac{a}{Rb}}{V_{M}} = \frac{a}{V_{M} \cdot b}$
Subject: Chemistry — Physical Chemistry — Chemical Equilibrium
Difficulty: Moderate

Question: What is the equilibrium constant expression for the reaction
P4 (s) + 5O2 (g)[IMAGE] P4 O10 (s) ?

2) Kc =
3) Kc =

Correct option: 2

Solution:
P4(g) + 5O2 (g)[IMAGE] P4O10(g) K_c =
$\frac{\left\lbrack P_{4}O_{10}(s) \right\rbrack}{\left\lbrack P_{4}(s) \right\rbrack\left\lbrack O_{2}(g) \right\rbrack^{5}}$
Subject: Chemistry — Inorganic Chemistry — Environmental Chemistry
Difficulty: Easy

Question: The disease that occurs due to eating high Hg content fish.

1) Minimata disease
2) Beri-beri disorder
3) Kwashiorkar disese
4) Osteosclerosis

Correct option: 1

Solution:
Explanation: Minamata disease is a neurological syndrome caused
by severe mercury poisoning. Fishes that present in Hg contaminated
water bodies efficiently store Hg in the form of highly toxic organic
compound methylmercury in their bodies. Minimata disease is caused by
eating these fishes.
Subject: Chemistry — Equivalent Concept — Equivalent weight, n-factor and Normality for Oxidant and
Difficulty: Moderate

Question: In the ionic equation
$2K^{+}B{rO}_{3}^{-} + 12H^{+} + 10e^{-} \longrightarrow {Br}_{2} + 6H_{2}O + 2K^{+}$,
the equivalent weight of KBrO3 will be


Correct option: 1

Solution:
${BrO}_{2} \longrightarrow {Br}_{2}$
+5 0
$\therefore(V.f.){BrO}_{3}^{-} = 5$
Subject: Chemistry — Organic Chemistry — Biomolecule & Polymers
Difficulty: Moderate

Question: The presence or absence of hydroxy group on which carbon atom of
sugar differentiates RNA and DNA.

1) $1^{st}$
2) $\ 2^{nd\ }$
3) $3^{rd}$
4) $4_{\ }^{th\ }$

Correct option: 2

Solution:
RNA and DNA has ribose and deoxyribose sugars, which differs in
absence of hydroxy group at $2^{nd\ }$ carbon
Subject: Chemistry — Chemical Kinetics — Integrated rate law : Zero and First Order Reaction
Difficulty: Tough

Question: Two substances $A\ (t_{1/2} = 5min)$ and $B\ (t_{1/2} = 15min)$
are taken in such a way that initially The time after
which both the concentration will be equal is: (Assume that reaction is
first order)

2) ^ = lbrackfrac - fracrbrack t$
$2ln2 = lbrackfrac - fracrbrack t$
$2 = leftlbrack frac - frac rightrbrack t$
$Rightarrow 2 = frac times t$
$t = 15minute.$

Answer: 15

Solution:
$C_{t} = C_{0}e^{- Kt}$
According to question
$C_{A,t} = C_{B},t$
$\frac{C_{A}}{C_{B}} = \frac{e^{- K_{B}t}}{e^{- K_{A}t}} \Rightarrow \frac{C_{A}}{C_{B}} = e^{(K_{A} - K_{B})t}$
$4 = e^{\lbrack\frac{ln2}{5} - \frac{ln2}{15}\rbrack xt}$
$\Rightarrow ln4 = \lbrack\frac{ln2}{5} - \frac{ln2}{15}\rbrack t$
$ln(2)^{2} = \lbrack\frac{ln2}{5} - \frac{ln2}{15}\rbrack t$
$2ln2 = \lbrack\frac{ln2}{5} - \frac{ln2}{15}\rbrack t$
$2 = \left\lbrack \frac{1}{5} - \frac{1}{15} \right\rbrack t$
$\Rightarrow 2 = \frac{2}{15} \times t$
$t = 15minute.$
Subject: Chemistry — Organic Chemistry — Biomolecules
Difficulty: Moderate

Question: Starting with three different amino acid molecules, how many
different tripeptide molecules are formed


Answer: 6

Solution:
Six type of tripe tide molecules are formed.
Subject: Chemistry — Physical Chemistry — Thermodynamics
Difficulty: Moderate

Question: The volume of a gas decreases from [IMAGE] to
[IMAGE] when a sample of gas is compressed by an average
pressure of 0.6 atm. During this process 10 of heat is liberated. The
change in internal energy is

1) - 2.16 J
2) 12.156 J
3) 2.16 J
4) 101.3 J

Correct option: 3

Solution:
Here t Change in
[IMAGE] Pressure (P) = 0.6 atm and heat liberated (q)=10 J
But 1 liter-atm = 101.3 J
Hence [IMAGE]. We also know that heat is liberated,
therefore it would be negative. Thus change in
Subject: Chemistry — Organic Chemistry — Alcohols, Phenols and Ethers
Difficulty: Moderate

Question: Reaction of phenol with dil. HNO_3 gives

1) p and m-nitrophenols
2) o- and p-nitrophenols
3) Picric acid
4) o- and m-nitrophenols

Correct option: 2

Solution:
Subject: Chemistry — Organic Chemistry — Alkyl And Aryl Halide
Difficulty: Moderate

Question: Which one of the following reagents is not suitable for the
elimination reaction?

4) Nal

Correct option: 4

Solution:
The given conversion is carried out in the absence of 1.37
Bromopropane reacts with F_G$= \frac{GMm}{(R + h)^{2}}$ and yields
iodopropane.
$(4\pi\mu Bb)\Delta n$
Subject: Chemistry — Organic Chemistry — Hydrocarbon
Difficulty: Moderate

Question: 1-Bromo-3-chlorocyclobutane will react with two moles of Na in
ether producing?
1-czkseks&3&Dyksjks Na ds nks eksy


Correct option: 1

Solution: