Grass-G/jee-main-questions huggingface dataset:
Subject: Chemistry —  — 

Question: The half life periods of a reaction at initial concentrations
[IMAGE] and [IMAGE] are 200 sec and 40
sec respectively. The order of reaction is



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

Question: The most reactive chlorine towards $H_{2}O$ is


Answer: 3

Solution:
[IMAGE] Reaction Intermediate -- Carbocation:
III will give most stable carbocation Þ stabilised by +m effect of
[IMAGE] I will give stable carbocation 3º carbocation so order of reaction = III
> I > II.
Subject: Chemistry — Inorganic Chemistry — P-block Halogen
Difficulty: Moderate

Question: XeF4 and XeF6 are expected to be

1) Oxidizing
2) Reducing
3) Unreactive
4) Strongly basic

Correct option: 1

Solution:
XeF4 & XeF6 both are strong oxidising agents. These can oxidise
H2 to H+, Cl- to Cl2, I- to I2 and Ce(III) to Ce(IV).
XeF4 rFkk XeF6 nksuksa izcy vkWDlhdkjd
Cl^- dks Cl2 esa, I^- dks I2 esa rFkk Ce(III) dks Ce(IV) esa
vkWDlhd`r dj ldrs gSA
XeF4 + 4I^- [IMAGE] 2I2 + 4F^- + Xe.
Subject: Chemistry — Inorganic Chemistry — S-block
Difficulty: Easy

Question: The commercial name for calcium oxide is

1) Milk of lime
2) Slaked lime
3) Limestone
4) Quick lime

Correct option: 4

Solution:
Calcium oxide has the chemical formula $\begin{matrix}
E \propto a^{2} \\
a \propto E
\end{matrix}$ and is commercially clled quick lime.
Subject: Chemistry — Atomic Structure — Quantum theory of light & Photoelectric Effect
Difficulty: Moderate

Question: A bulb of 40 W is producing a light of wavelength 620 nm with
80 of efficiency then the number of photons emitted by the bulb in 20
seconds are (1eV = 1.6 × 10-19 J, hc = 12400 eV

1) 2 × 1018
2) 1018
3) 1021
4) 2 × 1021

Correct option: 4

Solution:
$Power = \frac{nhC}{\lambda \times t} \Rightarrow 40 \times \frac{80}{100} = \frac{n \times 6.62 \times 10^{- 34} \times 3 \times 10^{8}}{620 \times 10^{- 9} \times 20} \Rightarrow n = 2 \times 10^{21}$
Subject: Chemistry — Organic Chemistry — Organic Chemistry
Difficulty: Moderate

Question: [IMAGE] has how many isomeric forms that contain
a benzene ring


Answer: 5

Solution:
[IMAGE] has 5 isomers
Subject: Chemistry —  — 

Question: In adjoining diagram depicts the temperature behavior of the rate
constant, k for two reactions A and B. Which of the following is correct
at low temperature?

1) k of reaction [IMAGE] of reaction B and
[IMAGE] of reaction [IMAGE] of reaction B
2) k of reaction [IMAGE] of reaction B and
[IMAGE] of reaction [IMAGE] of reaction
B
3) k of reaction [IMAGE] of reaction B and
[IMAGE] of reaction [IMAGE] of reaction B
4) k of reaction [IMAGE] of reaction B and
[IMAGE] of reaction [IMAGE] of reaction
B.


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

Question: Molecular shapes of $SF_{4},CF_{4}$ and $XeF_{4}$ are

1) The same with 2,0 and 1 lone pairs of electrons respectively
2) The same, with 1,1 and 1 lone pairs of electrons respectively
3) Different, with 0,1 and 2 lone pairs of electrons respectively
4) Different, with 1,0 and 2 lone pairs of electrons respectively

Correct option: 4

Solution:
SF_4 shape is see saw, CF_4 shape is tetrahedral, XeF_4
shape is square planar.
Subject: Chemistry — Carbonyl Compound (Aldehyde and Ketone) — Carboxylic acid (Preparation Methods)
Difficulty: Tough

Question: The acid D obtained through the following sequence of reactions
is

1) Succinic acid
2) Malonic acid
3) Maleic acid
4) Oxalic acid

Correct option: 1

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

Question: In the depression of freezing point experiment, it is found that:
I. The vapour pressure of the solution is less than that of pure
solvent.
II. The vapour pressure of the solution is more than that of pure
solvent.
III. Only solute molecules solidify at the freezing point.
IV. Only solvent molecules solidify at the freezing point.
III. dsoy v.kq gh fgekad ij Bksl esa ifjofrZr gksrs gSA
IV. dsoy v.kq gh fgekad ij Bksl

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

Correct option: 3

Solution:
The freezing point of a solvent depresses as a non-volatile
solute is added to a solvent. According to Raoult's law, when a
non-volatile solute is added to a solvent the vapour pressure of the
solvent decreases. At freezing point it will be only the solvent
molecules which will solidify.
?kV A vuqlkj tc vok" tks rks ok"i nkc ?kV aA fgekad ij dsoy
foyk;d
Subject: Chemistry — Organic Chemistry — Organic Chemistry
Difficulty: Easy

Question: Which of the following will show aromatic behaviour?


Correct option: 2

Solution:
[IMAGE] (1) Molecule is planar.
(2) [IMAGE] Electrons are present.
Subject: Chemistry — Physical Chemistry — Chemical Equilibrium
Difficulty: Easy

Question: For the following three reactions a, b and c, equilibrium
constants are given:
fuEu rhu vfHkfØ;kvksa a, b rFkk c ds fy, mu fLFkjkad fn, x, gSa:
(a) CO(g) + H2O(g)[IMAGE] CO2(g) + H2(g); K1
(b) CH4(g) + H2O(g)[IMAGE] CO(g) + 3H2(g); K2
(c) CH4(g) + 2H2O(g)[IMAGE] CO2(g) + 4H2(g); K3
Which of the following relations is correct ?

1) K2 K3 = K1
2) K3 = K1K2
3) K3 K23 = K12
4) $K\sqrt{K_{2}} = K_{3}$

Correct option: 2

Solution:
c = a + b
Subject: Chemistry — Physical Chemistry — Ionic Equilibrium
Difficulty: Moderate

Question: Determine solubility of PbI2 in a 0.1 M solution of Pb(NO3)2.is
( ___ × 10-4). (Ksp of PbI2 = 7.1 × 10-9)
Pb(NO3)2 ds 0.1 M PbI2 ( ___ × 10-4).(PbI2


Answer: 1.4

Solution:
x + 0.1 2x
Ksp = (x + 0.1)(2x)2 = 8 × 10-9
$\Rightarrow$ 4x2 = 8 × 10-8
$\Rightarrow$ x = 1.4 × 10-4 M
Subject: Chemistry — Physical Chemistry — Electrochemistry
Difficulty: Moderate

Question: Molar conductivity of 0.04 solution at 298 k is 200 Scm2mole-1.
A conductivity cell which is filled with MgCl2 have area of
cross-section of electrode 4cm2 & distance between electrode is 8 cm. If
potential difference between electrode is 10V then find current flow in
miliampere.


Answer: 40

Solution:
$\lambda_{m} = \frac{K \times 1000}{M}$
$200 = \frac{K \times 1000}{0.04}$
$K = \frac{200 \times 0.04}{1000} = 8 \times 10^{- 3}S{cm}^{- 1}$
$K = \left( \frac{\mathcal{l}}{9} \right) \times \frac{1}{R}$
$8 \times 10^{- 3} = \left\lbrack \frac{8}{4} \right\rbrack \times \frac{1}{R}$
$R = \frac{2}{8 \times 10^{- 3}} = \frac{1}{4} \times 10^{3}\Omega$
Again $\Rightarrow V = IR$
So
$I = \frac{V}{R} = \frac{10}{\left( \frac{1}{4} \times 10^{3} \right)} = 4 \times 10^{- 2}A$
Subject: Chemistry — Organic Chemistry — Hydrocarbon
Difficulty: Tough

Question: The major product in the following reaction is


Correct option: 3

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

Question: The conversion of PbO_2 to Pb(NO_3)_2 is

1) Oxidation
2) Reduction
3) Neither oxidation nor reduction
4) Both oxidation and reduction

Correct option: 2

Solution:
${{Pb}^{+ 4}O}_{2} \rightarrow {Pb}^{+ 2}\left( NO_{3} \right)_{2}$. in
this reaction reduction occurs.
Subject: Chemistry — Organic Chemistry — Hydrocarbon
Difficulty: Moderate

Question: A compound decolourises KMnO4 but does not give ammonical AgNO3
ppt test,d;kSfxd KMnO4 dks fo?kfVr djrk

1) CH3COCH3
2) C2H6
3) C2H4
4) C2H2

Correct option: 3

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

Question: Which of the following does not represent the correct order of
the property indicated against it?

1) F > Cl > Br > I (Electro negativity)
2) Li > B > Be > C (Second ionisation enthalpy)
3) O < S > Se > Te (Electron affinity)
4) Ti < Zr[IMAGE] Hf (Atomic size)

Correct option: 2

Solution:
Li > B > C > Be (Second ionisation enthalpy)
Subject: Chemistry — Physical Chemistry — Some Basic Concepts of Chemistry
Difficulty: Easy

Question: Aspirin has the formula C_9H_8O_4. How many atoms of oxygen
are there in a tablet weighing 360 mg ?

1) 1.204 × 10^23
2) 1.08 × 10^22
3) 1.204 × 10^24
4) 4.81 × 10^21

Correct option: 4

Solution:
Number of oxygen atoms$= \frac{360 \times 10^{- 3}}{180}$× 6.02
× 10^23 × 4 = 4.81 × 10^21
Subject: Chemistry — Physical Chemistry — 
Difficulty: Moderate

Question: Water is added to the solution M such that the mole fraction of
water in the solution becomes 0.9. The boiling point of this solution is


Answer: 376.2

Solution:
$\Delta T_{b} = K_{b} \cdot m$
$= 0.52 \times \frac{0.1}{0.9 \times 18} \times 1000 = \frac{520}{9 \times 18} = 3.20$
$T_{b} = 373 + 3.20 = 376.2K$
Subject: Chemistry — Organic Chemistry — Hydrocarbons
Difficulty: Moderate

Question: Bromination of cyclohexene under conditions given below yields


Correct option: 2

Solution:
The given reaction follows the mechanism as shown below.
[IMAGE] This reaction undergoes free radical substitution.
In case of bromination of cyclohexene in the presence of light, the
product formed is shown below.
Subject: Chemistry — Physical Chemistry — Ionic Equilibrium
Difficulty: Moderate

Question: In a solution containing 0.01 M HCl and 0.1 M H_2CO_3, ratio
of produced from strong acid & weak acid respectively is x: 1.
Report your answer as x/500. Given: of H_2CO_3 are 4 × 10^-7 & 4 ×
10^-11 respectively.


Answer: 5

Solution:
$4 \times 10^{- 7} = \frac{0.01 \times \left\lbrack {HCO}_{3}^{-} \right\rbrack}{0.1}$
$\Rightarrow$ [H+]~from weak acid~ = [HCO_3^-] = 4 × 10^-6 M
$\therefore$ Ratio =
$\frac{0.01}{4 \times 10^{- 6}} = \frac{2500}{1} = x$
So, answer $= \frac{x}{500} = x$
Subject: Chemistry — Organic Chemistry — Steric Inhibition of Resonance (SIR Effect)
Difficulty: Easy

Question: Select the correct statement about this compound.

1) All three C-N bond length are same.
2) C1-N and C3-N bonds length are same but shorter than C5-N bond
length.
3) C1-N and C5-N bonds length are same but longer than C3-N bond
length.
4) C1-N and C3-N bonds length are different but both are longer than
C5-N bond length.

Correct option: 3

Solution:
Due to SIR effect.
Subject: Chemistry — Inorganic Chemistry — p block
Difficulty: Moderate

Question: Atoms in a P_4 molecule of white phosphorus are arranged
regularly in space in which of the following way

1) At the corners of tetrahedron
2) At the corners of a cube
3) At the corners of a four membered ring
4) At the centre and corners of an equilateral triangle

Correct option: 1

Solution:
Subject: Chemistry — Equivalent Concept — Volume strength of H_2O_2, Hardness of water
Difficulty: Moderate

Question: What will the concentration of in a sample of 1 liter
hard water if after treatment with washing soda 10 g insoluble CaCO3 is
precipitated.

1) 0.2 M
2) 0.1 M
3) 0.3 M
4) 0.4 M

Correct option: 2

Solution:
${nCaCl}_{2} = {nCaCO}_{3}$
 = 0.1mole$$
Subject: Chemistry — Mole concept — Stoichiometry, Equation based calculations
Difficulty: Moderate

Question: When 100g of ethylene polymerises entirely to polyethene, the
weight of polyethene formed as per the equation n(C2H4) → (-CH2-CH2-)n
is

2) 100g
4) 100ng

Correct option: 2

Solution:
By applying POAC for C atoms
moles of ethylene × 2 = mole of polythene × n × 2
$\frac{100g}{28} \times 2 = \frac{wt.\ of\ polethene}{28 \times n} \times n \times 2$
wt. of polyethene = 100 g
Subject: Chemistry — Physical Chemistry — Mole Concept
Difficulty: Tough

Question: In a gaseous mixture, if an alkane (C_xH_2x + 2) and an alkene
(C_yH_2y) are taken in 2: 1 mole ratio, the average molecular weight
of mixture is observed to be 20. If the same alkane and alkene are taken
in 1: 2 mole ratio, the average molecular weight of mixture is observed
to be 24. Then, the value of are respectively:
 (C_xH_2x + 2) (C_yH_2y)
 2: 1, 20 
 1: 2 -, 
 24

1) 2, 1 (2) 1, 2
3) 2, 3
4) 3, 2

Answer: 2

Solution:
[IMAGE] 28x + 14y = 56...... (1)
[IMAGE] 14x + 28y = 70....... (2)
Subject: Chemistry — Physical Chemistry — Some Basic Concepts of Chemistry
Difficulty: Easy

Question: x + y [IMAGE] x_2 y_3
2 moles of X and 4 moles of Y are mixed in a container. Difference of
final moles of Y and X_2Y_3 is



Solution:
2X + 3Y [IMAGE] X_2 Y_2
2 4 0
LR ER
2 - 2 4 - 3 1 mole
Subject: Chemistry — Physical Chemistry — Atomic Structure
Difficulty: Easy

Question: are INTEGER ANSWER TYPE Questions.
(The answer of each of the questions is 2 digits
21 Energy of an emitted photon with wavelength 620 nm in electron volt
is:, 620 nm


Answer: 02

Solution:
Subject: Chemistry — Organic Chemistry — ALCOHOL
Difficulty: Easy

Question: The increasing order of acidity among phenol, p-methylphenol,
m-nitrophenol and p-nitrophenol is. Chose the correct integer as your
option.

1) phenol<p-methylphenol<m-nitrophenol <p-nitrophenol
2) p-nitrophenol<p-methylphenol<m-nitrophenol <phenol
3) phenol <p-methylphenol<m-nitrophenol <p-nitrophenol
4) p-methylphenol< phenol<m-nitrophenol <p-nitrophenol

Correct option: 4

Solution:
Subject: Chemistry — Real Gas — Vander waal equation and virial equation of state
Difficulty: Moderate

Question: At 273 K temp. and 9 atm pressure, the compressibility for a gas
is 0.9. The volume of 1 mili-mole of gas at this temperature and
pressure is

1) 2.24 litre
2) 0.020 mL
3) 2.24 mL
4) 22.4 mL

Correct option: 3

Solution:
$Z = \frac{PV}{nRT};V = \frac{0.9 \times 0.0821 \times 273}{9} = 2.24litre/mol$
Volume of 1 mili-mole of gas = 2.24 mL
Subject: Chemistry — Physical Chemistry — 
Difficulty: Moderate

Question: The rate of chemisorption

1) decreases with increase of pressure
2) increases with increase of pressure
3) is independent of pressure
4) is independent of temperature

Correct option: 2

Solution:
On increasing pressure more molecule will into contact with the
surface of solid adsorbent.
Subject: Chemistry — Periodic table — Atomic and Ionic radius
Difficulty: Moderate

Question: Match the correct atomic radius with the element:
S.No. Element Code Atomic radius (pm)
(i) Be (p) 74
(ii) C (q) 88
(iii) O (r) 111
(iv) B (s) 77
(v) N (t) 66

1) (i) - r, (ii) - q, (iii) - t, (iv) - s, (v) - p
2) (i) - t, (ii) - s, (iii) - r, (iv) - p, (v) - q
3) (i) - r, (ii) - s, (iii) - t, (iv) - q, (v) - p
4) (i) - t, (ii) - p, (iii) - r, (iv) - s, (v) - q

Correct option: 3

Solution:
On moving left to right in a period, atomic radii decreases due
to increase in Z_eff and addition of electrons to the same outermost
shell.
Subject: Chemistry — S-block — Miscellaneous (Silicones, Silicates, Zeolites & Alums)
Difficulty: Moderate

Question: Bleaching powder turns Red litmus to blue and finally white, it
is due to:
(A) OH-
(B) HCl
(C) OCl-
(D) Cl


Correct option: 2

Solution:
(i) CaOCl2 + H2O → Ca(OH)2 + Cl2
(ii) Cl2 + H2O → HOCl+ HCl
HOCl → HCl + 
Cl2 + H2O → 2HCl +
Subject: Chemistry — Physical Chemistry — Ionic Equilibrium
Difficulty: Moderate

Question: The solubility of Ag2CO3 in water ­­is 1.26 × 10-4 mole/litre.
What is its solubility in 0.02 M Na2CO3 solution ? Assume no hydrolysis
of CO32- ion. (Take = 1.26)
ty esa Ag2CO3 1.26 x 10-4 eksy@yhVj
esa bl ekusa fd CO32- vk;u


Answer: 10

Solution:
Na2CO3 [IMAGE] 2Na+ + CO32-
2 × 0.01 0.01
Ag2CO3 [IMAGE] 2 Ag+ + CO32-
2 x (x + 0.01) $\approx$ 0.01
Ksp = 4x2 (0.01)
Ksp = 4s3 = 4 × 10-12 = 4x2 (0.01) $\Rightarrow$ x = 10-5
Subject: Chemistry — Physical Chemistry — Some Basic Concepts of Chemistry
Difficulty: Tough

Question: 81 gm mixture of MgCO_3(s) and NH_2COONH_4(s) is heated to
constant mass.
MgCO_3(s) [IMAGE] MgO(s) + CO_2(g)
NH_2COONH_4(s) [IMAGE] 2NH_3(g) + CO_2(g)
If vapour density of gaseous mixture evolved was found to
be$\frac{61}{4}$then find mole of MgCO_3 in theoriginal sample.

1) 50
2) 60
3) 75
4) None of these

Correct option: 1

Solution:
Let no. of moles of MgCO_3 = a
let no. of moles of NH_2COONH_4 = b
MgCO_3(s) [IMAGE] MgO(s) + CO_2(g) a - -
- - a
NH_2COONH_4(s) [IMAGE] 2NH_3(g) + CO_2(g)
b - -
- 2b b
a × 84 + b × 75 = 81.......(1)
$\frac{(a + b) + 4 + (2b)17}{a + b + 2b} = \frac{61}{4} \times 2$........(2)
$a\ = b = \frac{1}{2}$ MgCO_3 = 50
Subject: Chemistry — Inorganic Chemistry — Environmental Chemistry
Difficulty: Easy

Question: ____ is a secondary pollutant.

1) CO
2) CO_2
3) PAN
4) NO_2

Correct option: 3

Solution:
Explanation: Secondary pollutants are not emitted directly from
any source; they form by reacting with molecules present in the
atmosphere. Some secondary pollutants are nitric acid, PAN etc.
Subject: Chemistry — Periodic table — Atomic and Ionic radius
Difficulty: Easy

Question: Choose the correct order of atomic radii of Fluorine and Neon
(in pm) out of the options given below

1) 72, 160
2) 160, 160
3) 72, 72
4) 160, 72

Correct option: 1

Solution:
Vanderwaal's radius is considered in case of neon, which
normally has a large value. Also, in neon, the valence shell is
completely filled with electrons. As a result, there are interelectronic
repulsions and thus electron cloud expands. So, atomic radii of neon is
greater than that of fluorine.
Subject: Chemistry — Organic Chemistry — Carbonyl compound_Final
Difficulty: Moderate

Question: (X) is the product of cross aldol condensation between
benzaldehyde (C6H5CHO) and acetone. What is its structure ?

2) C6H5-CH=C-(CH3)2
3) C6H5-CO-CH2-C=(CH3)2
4) None of these

Correct option: 1

Solution:
Subject: Chemistry — Physical Chemistry — Thermodynamics
Difficulty: Easy

Question: [IMAGE] Standard entropy change for the above reaction is

1) $60.3{JK}^{- 1}{mol}^{- 1}$
2) $80.7\ JK^{- 1}{mol}^{- 1}$
3) $- 70\ JK^{- 1}mol^{- 1}$
4) $\ + 10.7\ JK^{- 1}{mol}^{- 1}$

Correct option: 2

Solution:
$80.7{JK}^{- 1}{mol}^{- 1}$
Subject: Chemistry — Hydrocarbon — Hydrocarbon
Difficulty: Moderate

Question: Which gas is obtained when iodoform is heated with Ag powder ?
tc s Ag pw.kZ ds lkFk xeZ rks
izkIr gksrh

1) $CH_{4}$
2) $C_{2}H_{4}$
3) $C_{2}H_{6}$
4) $C_{2}H_{2}$

Correct option: 4

Solution:
Subject: Chemistry — Physical Chemistry — Gaseous state
Difficulty: Easy

Question: The compressibility factor for a gas is 0.9 at 273K and 9 atm.
Find the volume of $10^{–3}$ mol of gas at this T and P
$R = \left( \frac{0.0821atmL}{molK} \right)$

1) 2.24 L
2) 22.4 L
3) 2.24 m
4) 22.4 m

Correct option: 3

Solution:
$Z = \frac{PV}{nRT}$
Or $0.9 = \frac{9 \times V}{10^{- 3} \times 0.0821 \times 273}$
Or $V = 2.24 \times 10^{- 3}L\text{ or }2.24mL$
Subject: Chemistry — Organic Chemistry — Organic Chemistry
Difficulty: Easy

Question: Which statement is true for cyclohexane

1) It has two possible isomers
2) It has three conformations
3) Boat conformation is most stable
4) Chair and boat conformations differ in energy by

Correct option: 4

Solution:
Chair and boat conformations differ in energy by
Subject: Chemistry — Physical Chemistry — Chemical Equilibrium
Difficulty: Moderate

Question: For the reaction$H_{2}(g) + I_{2}(g) \rightleftharpoons 2HI(g)$
K_c = 66.9 at 350ºC and Kc = 50.0 at 448ºC. The reaction has

1) $\Delta$H = + ve
2) $\Delta$H = - ve
3) $\Delta$H = zero
4) $\Delta$H sign can not be determined

Correct option: 2

Solution:
$H_{2}(g) + I_{2}(g) \rightleftharpoons 2HI(g)$
$log\frac{K_{2}}{K_{1}} = \frac{\Delta H}{2.303R}\left\lbrack \frac{1}{T_{1}} - \frac{1}{T_{2}} \right\rbrack$
$log\frac{50}{66.9} = \frac{\Delta H}{2.303R}\left\lbrack \frac{1}{623} - \frac{1}{721} \right\rbrack$
After calculation negative value of $\Delta$H is obtained.
Subject: Chemistry — Inorganic Chemistry — P-Block Element
Difficulty: Tough

Question: On heating KClO_3, we get

1) Cl_2O
2) ClO_2
3) ClO_3
4) Cl_2 O_7

Correct option: 2

Solution:
3KClO3 + 3H_2SO4 [IMAGE] 3KHSO_4 +
HClO_4 + 2ClO_2 + H_2O.
Subject: Chemistry — Physical Chemistry — Thermodynamics
Difficulty: Easy

Question: The work done in ergs for the reversible expansion of one mole
of an ideal gas from a volume of 10 litres to 20 litres at 25ºC is

1) $2.303 \times 298 \times 0.082log2$
2) $298 \times 10^{7} \times 8.31 \times 2.303log2$
3) $2.303 \times 298 \times 0.082log0.5$
4) $8.31 \times 10^{7} \times 298 - 2.303\log\log\ 0.5\ $

Correct option: 2

Solution:
$W = 2.303\frac{V_{2}}{V_{1}}\ $
$= 2.303 \times 1 \times 8.314 \times 10^{7} \times 298\log\log\ \frac{20}{10}\ $
$= 298 \times 10^{7} \times 8.314 \times 2.303log2$.
Subject: Chemistry — Physical Chemistry — Thermodynamics
Difficulty: Tough

Question: The enthalpy of neutralization of a weak monoprotic acid (HA) in
1M solution with a strong base is -55.95 kJ/mol. If acid HA requires
1.4 complete ionization and enthalpy of
neutralization of the strong monobasic acid with a strong monoacidic
base is -57.3 kJ/mol. What is the ionization of the weak acid in
molar solution ?


Answer: 3.57

Solution:
$HA - H^{+} + A^{-};\Delta_{r}H = 1.4kJ/mol$
$\Delta H_{neutalization\ } = \Delta H_{ionization\ } + \Delta_{r}H\ of\ \left( H^{+} + {OH}^{-} \rightarrow H_{2}O \right)$
$- 55.95 = \Delta H_{ionization\ } - 57.3$
$\Delta H_{ionization\ }\ for\ 1MHA = 1.35kJ/mol$ heat utilized by 1 M acid for ionization
$= \frac{1.35}{1.4} \times 100 = 96.43\%$
So, acid is$\ 100 - 96.43 = 3.57\%$ ionized.
Subject: Chemistry — Physical Chemistry — Surface Chemistry
Difficulty: Moderate

Question: Softening of hard water is done using sodium aluminium silicate
(zeolite). This causes

1) adsorption of Ca2+ and Mg2+ ions of hard water replacing Na+ ions.
2) adsorption of Ca2+ and Mg2+ ions of hard water replacing Al3+ ions
3) both (A) and (B)
4) none of these

Correct option: 1

Solution:
Softening of hard water: Ion exchange resins used for
softening of hard water is based upon selective and competitive
adsorption of ions on resins.
${Na}_{2}Z + {Ca}^{+ 2} \longrightarrow CaZ + 2{Na}^{+}$
The organic polymers containing groups like -COOH, -SO3H and -NH2
etc. possess the property of selective adsorption of ions from
solution. These are quite useful in the softening of water.
Subject: Chemistry — Chemical Kinetics — Experimental Method to calculate order and rate law of
Difficulty: Moderate

Question: At 373 K, a gaseous reaction A→ 2B + C is found to be of first
order. Starting with pure A, the total pressure at the end of 10 min was
176 mm of Hg and after a long time when A was completely dissociated, it
was 270 mm of Hg. The pressure of A at the end of 10 minutes was

1) 94 mm of Hg
2) 47 mm of Hg
3) 43 mm of Hg
4) 90 mm of Hg


Solution:
[IMAGE] Pressure of A after 10 minute = 90 - 43 = 47
Subject: Chemistry — Physical Chemistry — 
Difficulty: Moderate

Question: The elevation in boiling point of a solution of 13.44 g of CuCl2
in 1kg of water using the following information, will be (Molecular
weight of CuCl2 = 134.4 and Kb = 0.52 K molal-1):
1kg ty esa 13.44 g CuCl2 DoFkukad fuEu lwpuk djrs gq,] gksxk dk v.kqHkkj (CuCl2 = 134.4 o Kb = 0.52 K
molal-1)


Answer: 0.16

Solution:
$\Delta T_{0} = iK_{0}m$ i (vant Haff factor) of ¼okUVgkWQ
xq.kkad½ CuCI2 = 3
$\Delta T_{b} = 3 \times 0.52 \times \frac{13.44}{134.4 \times 1} = 0.156 = 0.16$
Subject: Chemistry — Physical Chemistry — Periodic Table
Difficulty: Easy

Question: Which of the following statement is correct

1) Due to lanthanide contraction size of 3d series elements
[IMAGE] 4d series element.
(2) Due to lanthanide contraction size of 4d series elements
[IMAGE]  5d series element.
3) Due to lanthanide contraction size of 3d series elements < 5d
series element.
4) Due to lanthanide contraction size of 5d series element > 4d
series elements.

Correct option: 2

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

Question: When 20 g of naphthoic acid (C11H8O2) is dissolved in 50 g of
benzene (Kf = 1.72 K kg mol-1), a freezing point depression of 2 K is
observed. The van't Hoff factor (i) is:
tc 20 g us¶FkkWbd vEy (C11H8O2) dks 50 g csathu eas ?kksyk ( (Kf
= 1.72 K kg mol-1) rc fgekad esa voueu 2 K A okUVgkWQ
xq.kkad (i) gksxk


Answer: 0.5

Solution:
$\Delta T_{f} = i \times K_{f} \times \ molality\ $(eksyyrk)
$\Rightarrow 2 = i \times 1.72 \times \frac{20}{172} \times \frac{1000}{50}$
$\Rightarrow 2 = 4i$
$\Rightarrow i = 1/2 = 0.5$
Subject: Chemistry — Physical Chemistry — Atomic Structure
Difficulty: Easy

Question: Calculate the ratio of the energy of a photon of wavelength 3000 
to that of a photon of wavelength respectively.
3000 - s - s


Answer: 2

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

Question: In P_4 O_10 molecule
(A) There are 4 P-P bond
(B) There are 8 P-O bond
(C) The bond angle is 180°
(D) The phosphorus atom is sp^3 hybridised
P_4O_10

1) 4 P-P
2) 8 P-O
3) 180° (4) sp^3

Answer: 4

Solution:
P_4O_10 molecule has no P-P bonds, 16 P-O bonds POP angle
is about 109°.28 & P atom is sp^3
P_4O_10 P-P 16P-O, POP 
109°.28 P sp^3