Grass-G/jee-main-questions huggingface dataset:
Subject: Chemistry — Inorganic Chemistry — Metallurgy
Difficulty: Tough

Question: 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:$3C + 2{Al}_{2}O_{3} \longrightarrow 4Al + 3{CO}_{2}$
(Report your answer as voltage × 10)
gkWy&gSjkYV izØe eas Al2O3 ds oS|qr vi?kVu foHko
¼oksYVst½ dk eku gksxkA
 ∆Gºƒ (Al2O3) = -1520 kJ mol-1; ∆Gºƒ (CO2) = -394 kJ
mol-1
¼vius mÙkj

1) 16
2) 14
3) 17
4) 15

Correct option: 1

Solution:
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
$\Delta G^{\circ} = 3\Delta G_{f}^{\circ}\left( {CO}_{2} \right) - 2\Delta G_{f}^{\circ}\left( {AI}_{2}O_{3} \right)$
$$= 3 \times 394 - 2( - 1520) = 1858kJ$$
$\Delta G^{\circ} = - {nFE}_{cell}^{\circ}$
$- E_{cell\ }^{\circ} = \frac{\Delta G^{\circ}}{nF} = \frac{1858 \times 1000}{12 \times 96500} = 1.60V$
$1.60 \times 10 = 16$
$3C + 2{Al}_{2}O_{3} \longrightarrow 4Al + 3{CO}_{2}$;k $4{AI}^{3 +}12e^{-} \longrightarrow 4Al$ (n) = 12
$\Delta G^{\circ} = 3\Delta G_{f}^{\circ}\left( {CO}_{2} \right) - 2\Delta G_{f}^{\circ}\left( {AI}_{2}O_{3} \right)$
$$= 3 \times 394 - 2( - 1520) = 1858kJ$$
$\Delta G^{\circ} = - {nFE}_{cell}^{\circ}$
$- E_{cell\ }^{\circ} = \frac{\Delta G^{\circ}}{nF} = \frac{1858 \times 1000}{12 \times 96500} = 1.60V$
$1.60 \times 10 = 16$
Subject: Chemistry — Surface Chemistry — Purification and Preparation of Colloid
Difficulty: Easy

Question: Peptisation is

1) Conversion of a colloidal into precipitate form
2) Conversion of precipitate into colloidal sol
3) Conversion of metal into colloidal sol by passage of electric
current
4) Conversion of colloidal sol into macromolecules

Correct option: 2

Solution:
Process by which precipitate is converted into colloid is known
as peptisation.
Subject: Chemistry — Inorganic Chemistry — Hydrogen And Its Compound
Difficulty: Moderate

Question: When zeolite Hydrated sodium aluminium silicate) is treated with
hard water the sodium ions are exchanged with

1) $OH^{-}$ ions
2) ${SO}_{4}^{2 -}$ ions
3) $Ca^{2 +}$ ions
4) $H^{+}$ ions

Correct option: 3

Solution:
$\underset{\text{ Zeolita }}{Na_{2}Al_{2}}Si_{2}O_{8} \cdot xH_{2}O + Ca^{+ 2} \rightarrow {CaAl}_{2}{Si}_{2}O_{8} \cdot {xH}_{2}O + 2{Na}^{+}$
Subject: Chemistry — Organic Chemistry — Hydrocarbon
Difficulty: Moderate

Question: What will be the least molecular mass of an alkane which is
optically active?,Ydsu


Answer: 100

Solution:
12 × 7 + 16 × 1 = 100
Subject: Chemistry — Inorganic Chemistry — Chemical Bonding
Difficulty: Moderate

Question: Molecule AX4 have all bond angles equal and molecule is non-polar
also then which of the following conclusion is incorrect?

1) Molecule may be tetrahedral.
2) Molecule may be square planar.
3) Central atom must have at least six valence electrons.
4) Central atom has either zero lone pair or two lone pairs

Correct option: 3

Solution:
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)
Subject: Chemistry — Carbonyl Compound (Aldehyde and Ketone) — Aldol condensation
Difficulty: Tough

Question: The numberof aldol reaction (s) that occurs in the given
transformation is


Answer: 3

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

Question: 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


Answer: 75

Solution:
$\left( \pi_{obs\ } \right)_{{Na}_{2}{SO}_{4}} = \pi_{glucose\ }$
or $\frac{10}{4} = \frac{1 + 2\alpha}{1}$ or $10 = 4 + 8\alpha$
$\alpha = \frac{10 - 4}{8} = 0.75$ $\%\ of\ \alpha = 75\%$
Subject: Chemistry — Organic Chemistry — Chemistry in everyday life & POC
Difficulty: Easy

Question: Decreasing order of boiling point of I to IV follow
Methylformate Ethylformate Iso-propylformate n-propylformate
I II III IV

1) I > II > III > IV
2) III > IV > II > I
3) IV > III > II > I
4) I > II > IV > III

Correct option: 3

Solution:
Boiling point µ molecular weight.
Subject: Chemistry — Organic Chemistry — Organic compounds containing Nitrogen
Difficulty: Easy

Question: Aniline on treatment with excess of bromine water gives

1) Aniline bromide
2) 0 - bromoaniline
3) p - bromoaniline
4) 2, 4, 6 - tribromoaniline

Correct option: 4

Solution:
Subject: Chemistry — Inorganic Chemistry — Periodic table
Difficulty: Moderate

Question: $A^{-}(g) \rightarrow A^{2 +}(g)$ $\Delta H = 1100\ KJ/mol$
$A(g) \rightarrow A^{2 +}(g)$ $\Delta H = 1200\ KJ/mol$
Electron gain enthalphy of A is $P \times 10^{2}KJ/mol$. What is the
value of P ?

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.

Answer: 1

Solution:
$A(g) + e^{-} \rightarrow A^{-}(g)$
$\Delta H = + E \cdot G \cdot E$....(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)
$- E \cdot G \cdot E + 1200 = 1100$
$- E \cdot G \cdot E = - 100\ KJ/mol$
$P = 1$.
Subject: Chemistry — Organic Chemistry — IUPAC Nomenclature
Difficulty: Moderate

Question: The IUPAC name of neopentane is

1) 2,2-dimethylpropane
2) 2-methylpropane
3) 2,2 -dimethylbutane
4) 2 -methylbutane

Correct option: 1

Solution:
Subject: Chemistry — Isomerism — Specific rotation, optical purity, enantiomeric excess and
Difficulty: Tough

Question: 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).

1) 80, +2.7 unit
2) 20, -27 unit
3) 20, +2.7 unit
4) 80, -27 unit

Correct option: 3

Solution:
Subject: Chemistry — Organic Chemistry — Preparation of hydrocarbon (Handout)
Difficulty: Moderate

Question: Acetylene may be prepared by electrolysis of

1) potassium oxalate
2) potassium acetate
3) potassium maleate
4) potassium succinate

Correct option: 3

Solution:
[IMAGE] + 2CO2 + 2NaOH + 2H2O
Subject: Chemistry — Inorganic Chemistry — D & F -Block Elements
Difficulty: Moderate

Question: The equivalent weight of [IMAGE] in acidic medium


Answer: 49

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

Question: 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?


Answer: 4

Solution:
Only four structural isomers are possible for diphenyl methane.
Subject: Chemistry — Inorganic Chemistry — Environmental Chemistry
Difficulty: Moderate

Question: Besides carbondioxide and nitrous oxide, the other greenhouse
gas is

1) methane
2) carbon monoxide
3) both a and b
4) none of these

Correct option: 3

Solution:
Explanation: Greenhouse gases such as carbondioxide, nitrogen
oxide, methane, carbon monoxide, water vapours and CFCs
(chlorofluorocarbons) causes greenhouse effect on plants.
Subject: Chemistry — Physical Chemistry — Chemical Kinetics
Difficulty: Moderate

Question: 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 time t
and at the end of the reaction respectively, then there is 50 inversion
when

1) $r_{0} = 2r_{t} - r_{\infty}$
2) $r_{0} = r_{t} - r_{\infty}$
3) $r_{0} = r_{t} - 2r_{\infty}$
4) $r_{0} = r_{t} + r_{\infty}$

Correct option: 1

Solution:
Given,
$\left( r_{\infty} - r_{0} \right) = a,\left( r_{\infty} - r_{t} \right) = (a - x)$
At 50 Inversion
$\frac{a}{2} = (a - x)$
$\frac{\left( r_{\infty} - r_{0} \right)}{2} = \left( r_{\infty} - r_{t} \right)$
$\left( r_{\infty} - r_{0} \right) = 2r_{\infty} - 2r_{t}$
$r_{0} = 2r_{t} - r_{\infty}$
Subject: Chemistry —  — 

Question: What is the number of Chiral Carbon in the compound C.



Solution:
[IMAGE] Compound (C) has one chiral C-atom.
Subject: Chemistry — Physical Chemistry — Redox reaction
Difficulty: Tough

Question: $MnO_{4}^{2 -}$(1 mole) in neutral aqueous medium
isdisproportionateto X mole of $MnO_{4}^{-}$ and Y mole of $MnO_{2}$.
3X+6Y =


Answer: 4

Solution:
$MnO_{4}^{2 -}$in neutral aqueous medium isdisproportionate to
$\frac{2}{3}$ mole of $MnO_{4}^{-}$ and $\frac{1}{3}$ mole of $MnO_{2}$.
Subject: Chemistry — Physical Chemistry — Chemical and Ionic Equilibrium
Difficulty: Moderate

Question: 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

1) 2.496
2) 1.87
3) 2
4) 4

Correct option: 1

Solution:
$\frac{22}{100}$ × 3.2 = 0.704
[IMAGE] at equil. moles of HI = 3.2 - 0.704 = 2.496
Subject: Chemistry — Organic Chemistry — Biomolecules
Difficulty: Easy

Question: Sucrose is

1) A reducing sugar
2) Not a reducing sugar
3) Partial reducing sugar
4) Mixed sugar

Correct option: 2

Solution:
Sucrose is not a reducing sugar.
Subject: Chemistry — Organic Chemistry — Hydrocarbon
Difficulty: Moderate

Question: The major product obtained in the reaction


Correct option: 3

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

Question: 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


Correct option: 2

Solution:
(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)
Subject: Chemistry — Organic Chemistry — Carbonyl compound_Final
Difficulty: Tough

Question: , Product (X) in this reaction is


Correct option: 1

Solution:
Subject: Chemistry — Organic Chemistry — Aldehydes, Ketones & Carboxylic Acids
Difficulty: Moderate

Question: An alkene of molecular formula [IMAGE] on
ozonolysis gives 2, 2 dimethyl propanal & 2 butanon, then the alkene is

1) 2,2,4 -trimethyl -3 -hexene
2) 2,2,6 -trimethyl- 3 -hexene
3) 2,3,4 -trimethyl- 2 -hexene
4) 2,2,4 -trimethyl- 2 -hexene

Correct option: 1

Solution:
[IMAGE] On the basis of product formation, it would be alkene.
Subject: Chemistry — Organic Chemistry — Hydrocarbon
Difficulty: Moderate

Question: The Corey-House alkane synthesis is carried out by treating an
alkyl halide with

1) Lithium metal
2) Copper metal
3) Lithium metal followed by reaction with cuprous iodide and then
treating the product with an alkyl halide
4) Cuprous iodide followed by reaction with alkyl halide
dksjh gkmLk,Ydsu la"ys'k.k ds lkFk,d,Ydkby gsykbM

Correct option: 3

Solution:
Subject: Chemistry —  — 

Question: In which reaction correct product is given ?



Solution:
[IMAGE] This is Wolf Kischner Reduction
Subject: Chemistry — Physical Chemistry — Ionic Equilibrium
Difficulty: Easy

Question: 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

1) H^+ ions released by HOCN
2) Sum of H^+ ions released by all monoprotic acids
3) Sum of H^+ ions released only the first dissociation of all the
acids.
4) Overall H^+ ion concentration in the solution.

Correct option: 4

Solution:
For HOCN^-, K_a =
$\frac{\left\lbrack H^{+} \right\rbrack\left\lbrack {OCN}^{-} \right\rbrack}{\lbrack HOCN}$
 = total H^+ concentration of solution.
Subject: Chemistry — Atomic Structure — Quantum mechanical model of atom, Shrodinger wave equation
Difficulty: Moderate

Question: 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

1) a & c
2) b & c
3) Only b
4) a, b

Correct option: 4

Solution:
(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.
Subject: Chemistry — Physical Chemistry — Electrochemistry
Difficulty: Moderate

Question: Given standard electrode potentials
The standard electrode
potential$E^{0}$for$Fe^{+ + +} + e^{-} \rightarrow Fe^{+ +}$ is


Answer: 0.77

Solution:
$\Delta G^{\circ} = - nE^{0}F$
$Fe^{2 +} + 2e^{-} \rightarrow Fe$....(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)
${Fe}^{3 +} + e^{-} \rightarrow {Fe}^{2 +}$
$\Delta G^{\circ} = 0.108F - 0.880F = - 0.772F$
$E^{0}$for the reaction$= - \frac{\Delta G^{\circ}}{nF}$
$= - \frac{( - 0.772F)}{1 \times F} = + 0.772V$
Subject: Chemistry — Mole concept — Stoichiometry, Equation based calculations (Elementary level
Difficulty: Moderate

Question: When 100g of ethylene polymerises entirely to polyethene, the
weight of polyethene formed as per the equation [IMAGE] is

2) 100g
4) 100ng

Correct option: 2

Solution:
By applying POAC for C atoms
moles of ethylene × 2 = mole of polythene × n × 2
[IMAGE] wt. of polyethene = 100 g
Subject: Chemistry — Inorganic Chemistry — p block
Difficulty: Moderate

Question: Lead pipes are corroded quickly by

1) Dil. H_2SO_4
2) Conc. H_2SO_4
3) Acetic acid
4) Water

Correct option: 3

Solution:
Organic acids dissolve lead in presence of oxygen
Subject: Chemistry — Organic Chemistry — Alcohols, Phenols and Ethers
Difficulty: Moderate

Question: Phenolphthalein is obtained by heating phthalic anhydride with
conc. H_2 SO_4 and

1) Benzyl alcohol
2) Benzene
3) Phenol
4) Benzoic acid

Correct option: 3

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

Question: 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 ty dk
?kuRo Øe'k 1.86 K kg mole-1 o 0.997 g cm-3 gaS


Answer: 0.228

Solution:
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 $= \frac{0.05 \times 1000}{498.5} = 0.1$
Determination of van`t Hoff factor, i
${CH}_{3}COOH \rightarrow {CH}_{3}{COO}^{-} + H^{+}$
No. of moles at start 1 0 0
No. of moles at equb. 1-0.23 0.23 0.23
$\Delta T_{r} = (1 - 0.23 + 0.23 + 0.23) \times 1.86 \times 0.1 = 0.228K$
Hindi ty dk Hkkj$= 500 \times 0.997 = 498.5g$,lhfVd vEy ks 
$= \frac{\ Wt.\ of\ {CH}_{3}COOH\ (ingm)\ }{\ Mol.wt.of\ {CH}_{3}COOH} = \frac{3 \times 10^{- 3} \times 10^{3}}{60} = 0.05$
498.5 g ty esa CH3COOH ds 0.05 eksy gksxasA
1000 g ty $= \frac{0.05 \times 1000}{498.5} = 0.1$
okWUVgkWQ xq.kkad fy, eku.k
${CH}_{3}COOH \rightarrow {CH}_{3}{COO}^{-} + H^{+}$
çkjEHk ksa 1 0 0
 ks 1-0.23 0.23 0.23
$\Delta T_{r} = (1 - 0.23 + 0.23 + 0.23) \times 1.86 \times 0.1 = 0.228K$
Subject: Chemistry — Organic Chemistry — Hydrocarbons
Difficulty: Moderate

Question: In which reactions major product is propane ?
(a) Propene

2) b and d
3) a and d
4) b and c

Correct option: 4

Solution:
(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
Subject: Chemistry — Physical Chemistry — Atomic Structure
Difficulty: Moderate

Question: 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


Answer: 2

Solution:
$\lambda = \frac{h}{mv}$
$mv = \frac{h}{\lambda}$
$m^{2}v^{2} = \frac{h^{2}}{m\lambda^{2}}$
$\frac{1}{2}{mv}^{2} = \frac{h^{2}}{2m\lambda^{2}}$
$KE = \frac{h^{2}}{2m\lambda^{2}} = \frac{\left( 6.6 \times 10^{- 34} \right)^{2}}{2 \times 10^{- 4} \times \left( 3.31 \times 10^{- 29} \right)^{2}} = 2 \times 10^{- 6}$
Subject: Chemistry — Physical Chemistry — Ionic Equilibrium
Difficulty: Moderate

Question: 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


Answer: 2

Solution:
Ksp = 4 × 10-12 = = =
xx. (10-4)1+x
$\therefore$ x = 2
Subject: Chemistry — Physical Chemistry — Chemical and Ionic Equilibrium
Difficulty: Easy

Question: 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

1) 11.25
2) 18.75
3) 5
4) 3.75

Correct option: 2

Solution:
A + 2 B = C + 3D
Subject: Chemistry — Inorganic Chemistry — Hydrogen And Its Compound
Difficulty: Easy

Question: Hydrogen molecule differs from chlorine molecule in the
following respect

1) Hydrogen molecule is non-polar but chlorine molecule is polar
2) Hydrogen molecule is polar while chlorine molecule is non-polar
3) Hydrogen molecule can form intermolecular hydrogen bonds but
chlorine
molecule does not
4) Hydrogen molecule cannot participate in coordination bond
formation but chlorine molecule can

Correct option: 4

Solution:
Chlorine has lone pair which it can donate to form co-ordinate bond
while hydrogen cannot.
Subject: Chemistry — Organic Chemistry — Organic compounds containing Nitrogen
Difficulty: Moderate

Question: Number of isomeric primary amines obtained from C_4H_11N are


Answer: 4

Solution:
Four 1º amines are possible
Subject: Chemistry — Physical Chemistry — Thermodynamics
Difficulty: Moderate

Question: At 27º C, one mole of an ideal gas is compressed isothermally
and reversibly from a pressure of 2 atm to 10 atm. The values of E
and q are (R = 2).

1) 0, - 965.84 cal
2) - 965.84 cal,+ 965.84 cal
3) + 865.58 cal - 865.58 cal
4) - 865.58 cal,-865.58 cal

Correct option: 1

Solution:
$W = 2.303\frac{P_{2}}{P_{1}}\ $
$= 2.303 \times 1 \times 2 \times 300\log\log\ \frac{10}{2}\ = 965.84$
At constant temperature, $E = 0$.
$E = q + w;q = - w = - 965.84\ $Cal
Subject: Chemistry — Organic Chemistry — Aldehyde Ketone & Carboxylic acid
Difficulty: Tough

Question: The correct statement about the synthesis of erythritol
(C(CH_2OH)_4) used in the preparation of PETN is

1) The synthesis requires four aldol condensations between methanol
and ethanol.
2) The synthesis requires two aldol condensations and two Cannizzaro
reactions.
3) The synthesis requires three aldol condensations and one
Cannizzaro reaction.
4) Alpha hydrogens of ethanol and methanol are involved in this
reaction.

Correct option: 3

Solution:
The synthesis of erythritol 1.67 ×10^5 J involves first three
aldol reactions between formaldehyde and acetaldehyde and then one
Cannizzaro reaction with formaldehyde. It can be expressed as,
Subject: Chemistry — Physical Chemistry — 
Difficulty: Moderate

Question: The Henry's law constant for the solubility of N2 gas in water
at 298 K is 1.0 × 105 atm. The mole fraction of N2 in air is 0.8. The
number of moles of N2 from air dissolved in 10 moles of water of 298 K
and 5 atm pressure is:
N2 xSl dh ty esa ? fy, gsujh fu;e fLFkjkad (Henry's law
constant) dk eku 298 K ij 1.0 × 105 atm gSA N2 izHkkt
 mifLFkr N2

1) 4 × 10-4
2) 4.0 × 10-5
3) 5.0 × 10-4
4) 4.0 × 10-6

Correct option: 1

Solution:
$P_{N_{2}} = K_{H} \times X_{N_{2}}$
eksy
In 10 mole solubility is eksy
Subject: Chemistry — Physical Chemistry — Redox Reaction
Difficulty: Moderate

Question: Which is the best description of the behavior of bromine in the
reaction given below

1) Oxidized only
2) Reduced only
3) Proton acceptor only
4) Both oxidized and reduced

Correct option: 4

Solution:
[IMAGE] In the above reaction the oxidation number of [IMAGE] increases from zero [IMAGE] to +1 [IMAGE] and decrease from zero [IMAGE] to -1 (in
[IMAGE] ). Thus [IMAGE] is oxidized as well
as reduced & hence it is a redox reaction.
Subject: Chemistry — Physical Chemistry — 
Difficulty: Easy

Question: If is the number of atoms in the unit cell that represents
the closest packing sequence ---A B C A B C---,the number
of tetrahedral voids in the unit cell is equal to

1) Z
2) _2Z

Correct option: 2

Solution:
Number of tetrahedral voids in the unit cell = 2 × number of
atoms = 2 Z
Subject: Chemistry — Organic Chemistry — Chemistry in everyday life
Difficulty: Tough

Question: Which of the following can possibly be used as analgesic without
causing addiction and any modification

1) Morphine
2) N -acetyl paraamino phenol
3) Diazepam
4) Tetra hydro-cat-enol

Correct option: 3

Solution:
"Diazepam" can be used as analgesic without addition and
modification.
Subject: Chemistry — Physical Chemistry — 
Difficulty: Moderate

Question: Structure similar to zinc blende is found in

1) AgCl
2) NaCl
3) CuCl
4) TlCl

Correct option: 3

Solution:
Zinc blende (ZnS) has fcc structure and is an ionic crystal
having 4: 4 co-ordination number.
Subject: Chemistry — Physical Chemistry — Mole Concept
Difficulty: Moderate

Question: 1 g sample of alkaline earth metal react completely with 4.08 g
H2SO_4 and yield an ionic product MSO_4. Then find out the atomic mass
of Alkaline earth metal (M)?,d 4.08 xzke ds lkFk iw.kZr 1 4.08 H2SO4 
 MSO4 M 
 ?


Answer: 24

Solution:
M + H2SO4 [IMAGE] MSO4 + H2
[IMAGE] This is atomic weight of M (M ) =
Subject: Chemistry — P-block VA, VIA, VIIA elements — Hydracids
Difficulty: Easy

Question: Which can do glass etching?

1) HIO4
2) HF
3) HNO3
4) SiF4

Correct option: 2

Solution:
SiO2 + 6HF → H2SiF6 + 2H2O
Subject: Chemistry — Inorganic Chemistry — D & F -Block Elements
Difficulty: Moderate

Question: The correct order of magnetic moments (spin only values in B.M.)
among is


Correct option: 3

Solution:
[IMAGE] Number of unpaired electron =
[IMAGE] Number of unpaired electron =
[IMAGE] Number of unpaired electron =
Magnetic moment [IMAGE] Where, n = number of unpaired electron i.e., greater the number of
unpaired electron greater will be the paramagnetic character.
Subject: Chemistry — Inorganic Chemistry — P-block nitrogen
Difficulty: Moderate

Question: The number of P--O--P bonds in cyclic trimetaphosphoric acid
is:
pØh; VªkbZesVkQkWLQksfjd vEy


Answer: 3

Solution:
[IMAGE] According to the structure of cyclic metaphosphoric acid, (HPO3)3,
there are three P - O - P bonds.
pØh; esVkQWkLQksfjd vEy, (HPO_3)_3 dh lajpuk ij;gk¡ rhu P
- O - P A
Subject: Chemistry — Organic Chemistry — Organic compounds containing Nitrogen
Difficulty: Easy

Question: Aniline on treatment with conc. HNO_3 + conc. H_2SO_4mixture
yields

1) o - and p - nitro anilines
2) m - nitro anilines
3) A black tarry matter
4) No reaction

Correct option: 3

Solution:
[IMAGE] Nitration of aniline without protecting the amino group is not possible
because HNO_3
is strong oxidizing agent which oxidizes aniline.