datavorous/entrance-exam-dataset huggingface dataset:
On mixing a certain alkane with chlorine and irradiating it with ultravioletlight, it forms only one monochloroalkane. This alkane could be

Options:
A
pentane

B
isopentane

C
neopentane

D
propane

The correct answer is:
neopentane

Solution:
In neopentane all the $\mathrm{H}$ atoms are same $\left(1^0\right)$.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2003
On mixing ethyl acetate with aqueous sodium chloride, the composition of the resultant solution is

Options:
A
$\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5+\mathrm{NaCl}$

B
$\mathrm{CH}_3 \mathrm{Cl}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{COONa}$

C
$\mathrm{CH}_3 \mathrm{COCl}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}+\mathrm{NaOH}$

D
$\mathrm{CH}_3 \mathrm{COONa}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$

The correct answer is:
$\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5+\mathrm{NaCl}$

Solution:
No Solution Available

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main, JEE Main 2004
On mixing, heptane and octane form an ideal solution. At $373 \mathrm{~K}$, the vapour pressures of the two liquid components (heptane and octane) are $105 \mathrm{kPa}$ and $45 \mathrm{kPa}$ respectively. Vapour pressure of the solution obtained by mixing $25.0 \mathrm{~g}$ of heptane and $35 \mathrm{~g}$ of octane will be (molar mass of heptane $=100 \mathrm{~g} \mathrm{~mol}^{-1}$ an dof octane $=114 \mathrm{~g} \mathrm{~mol}^{-1}$ ).

Options:
A
$72.0 \mathrm{kPa}$

B
$36.1 \mathrm{kPa}$

C
$96.2 \mathrm{kPa}$

D
$144.5 \mathrm{kPa}$

The correct answer is:
$72.0 \mathrm{kPa}$

Solution:
Mole fraction of Heptane $=\frac{25 / 100}{\frac{25}{100}+\frac{35}{114}}=\frac{0.25}{0.557}=0.45$
$\mathrm{X}_{\text {Hep tane }}=0.45$.
$\therefore$ Mole fraction of octane $=0.55=\mathrm{X}_{\text {octane }}$
Total pressure $=\sum X_i P_i^0$
$=(105 \times 0.45)+(45 \times 0.55) \mathrm{kP}_{\mathrm{a}}$
$=72.0 \mathrm{KPa}$

Topic: Chemistry, Solutions, JEE Main, JEE Main 2010
On moving a charge of 20 coulombs by $2 \mathrm{~cm}, 2 \mathrm{~J}$ of work is done, then the potential difference between the points is

Options:
A
0.1 V

B
8 V

C
2 V

D
0.5 V

The correct answer is:
0.1 V

Solution:
$$
\mathrm{W}=\mathrm{qV} \Rightarrow \mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}=2 / 20=0.1 \mathrm{~V}
$$
Here $W$ is the work done in moving charge $q$ from point $A$ to $B$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2002
On ozonolysis, an alkene X forms two carbonyl compounds namely butan-2-one and methanal. The total number of methyl groups present in alkene X is

Options:

The correct answer is:
2

Solution:
Alkene X i.e., 2-methylbut-1-ene contains two methyl − CH 3 groups.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
On passing a gas, 'X', through Nessler's reagent, a brown precipitate is obtained. The gas 'X ' is

Options:
A H2S
B CO2
C NH3
D Cl2

The correct answer is:
NH3

Solution:
Nessler's reagent is an aqueous solution of potassium iodide, mercuric chloride, and potassium hydroxide. Its chemical formula is K2HgI4. It is used to detect the presence of ammonia. It turns into a pale brown colour in the presence of ammonia.

Nessler’s Reagent Reaction:

2K2HgI4⏟Nessler's reagent+NH3+3KOH→HgO.HgNH2I⏟Iodide of Millon's basebrown precipitate+7KI+2H2O

Topic: Chemistry, Practical Chemistry, JEE Main, JEE Main 2024 (29 Jan Shift 2)
On passing $C$ ampere of electricity through a electrolyte solution for $t$ second, $m$ gram metal deposits on cathode. The equivalent weight $E$ of the metal is

Options:
A $E=\frac{C \times t}{m \times 96500}$
B $E=\frac{C \times m}{t \times 96500}$
C $E=\frac{96500 \times m}{C \times t}$
D $E=\frac{C \times t \times 96500}{m}$

The correct answer is:
$E=\frac{96500 \times m}{C \times t}$

Solution:
Faraday's constant (F) is a fundamental physical constant relating the amount of electric charge transferred per mole of electrons transferred in an electrochemical reaction. Its value is approximately 96500 coulombs per mole (C/mol).
The equation $E=96500 \times m /(C \times t)$ represents one way to express Faraday's law of electrolysis.
In the equation:
- $E$ is the equivalent weight of the metal ( $\mathrm{g} / \mathrm{eq})$
- $m$ is the mass of the metal deposited on the cathode ( $\mathrm{g}$ )
- $\mathrm{C}$ is the current passed through the electrolyte (A)
- $\mathrm{t}$ is the time the current flows ( $\mathrm{s}$ )
So, the equivalent weight (E) of a metal can be calculated by dividing the product of the Faraday's constant (F) and the mass ( $\mathrm{m}$ ) of the metal deposited on the cathode by the product of the current $(\mathrm{C})$ and the time $(\mathrm{t})$ the current flows.

Topic: Chemistry, Electrochemistry, JEE Main
On Pauling scale which of the following does not have electronegativity $\geq 3.0$

Options:
A Oxygen
B Nitrogen
C Chlorine
D Bromine

The correct answer is:
Bromine

Solution:
According to Pauling scale Br does not have electronegativity of greater than 3.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
On placing a thin film of mica of thickness $12 \times 10^{-5} \mathrm{~cm}$ in the path of one of the interfering waves in Young's double slit experiment using monochromatic light, the fringe pattern shifts through a distance equal to width of a bright fringe. If wavelength used is $\lambda=6 \times 10^{-5} \mathrm{~cm}$, the refractive index of mica is

Options:
A $1.4$
B $1.1$
C $1.3$
D $1.5$

The correct answer is:
$1.5$

Solution:
Path difference due to the mica film: $(\mu-1) \mathrm{t}$
The shift in the pattern be $\Delta$ yand the corresponding path difference can be written as: $\Delta x=\frac{\Delta y D}{d}=(\mu-1) t$
$\therefore \Delta y=\frac{d}{D}(\mu-1) t$
Fringe width is $\beta=\frac{\lambda D}{d}$, so,
$\Delta y=\frac{\beta}{\lambda}(\mu-1) t$
If the shift is equal to the fringe width $\beta$
$\begin{aligned} & \beta=\frac{\beta}{\lambda}(\mu-1) t \\ & \Rightarrow(\mu-1) t=\lambda \\ & \therefore \mu=\frac{\lambda}{t}+1=\left(\frac{6 \times 10^{-5}}{12 \times 10^{-5}}\right)+1=1.5\end{aligned}$

Topic: Physics, Wave Optics, JEE Main
On producing the waves of frequency \( 1000 \mathrm{~Hz} \) in a Kundt's tube, the total distance between \( 6 \) successive nodes is \( 85 \mathrm{~cm} \). Speed of sound in the gas filled in the tube is

Options:
A \( 330 \mathrm{~m} \mathrm{~s}^{-1} \)
B \( 340 \mathrm{~m} \mathrm{~s}^{-1} \)
C \( 350 \mathrm{~m} \mathrm{~s}^{-1} \)
D \( 300 \mathrm{~m} \mathrm{~s}^{-1} \)

The correct answer is:
\( 340 \mathrm{~m} \mathrm{~s}^{-1} \)

Solution:
In Kundt's tube, the distance between two successive nodes is λ2, where λ is the wavelength of the wave in the tube.

Speed of sound in any medium is
v=fλ, where f is the frequency of the wave.

Here, distance between 6 successive nodes is
5λ2=85 cm
λ=34 cm=0.34 m
Now, speed of sound in the tube
v=fλ
=1000×0.34
=340 m s-1

Topic: Physics, Waves and Sound, JEE Main
On $R$, the relation $\rho$ be defined by 'xpy holds if and only if $x-y$ is zero or irrational'. Then,

Options:
A $\rho$ is reflexive and transitive but not symmetric
B $\rho$ is reflexive and symmetric but not transitive
C $\rho$ is symmetric and transitive but not reflexive
D $\rho$ is equivalence relation

The correct answer is:
$\rho$ is reflexive and symmetric but not transitive

Solution:
We have, $x p y \Rightarrow x-y$ is zero or irrational. Now, $x-x=0$
$\Rightarrow(x, x) \in \rho$
$\therefore \rho$ is reflexive.
Again, if $x-y$ is either zero or irrational, then $y-x$ will also be either zero or inational.
$(x, y) \in \rho$
$\Leftrightarrow \quad(y, x) \in \rho$
$\therefore \rho$ is symmetric.
Again, $(2, \sqrt{3}) \in \rho$ and $(\sqrt{3}, 4) \in \rho$
$But (2,4) \notin \rho$
$\therefore p$ is not transitive.

Topic: Mathematics, Sets and Relations, JEE Main
On $R$, the set of real numbers, a relation $\rho$ is defined as $a \rho b$ if and only if $1+a b>0$. Then,

Options:
A $\rho$ is an equivalence relation
B $\rho$ is reflexive and transitive but not symmetric
C $\rho$ is reflexive and symmetric but not transitive
D $\rho$ is only symmetric

The correct answers are:
$\rho$ is reflexive and symmetric but not transitive

Solution:
We know that,
$1+a^{2}>0, a \in R$
$\Rightarrow \quad(a, a) \in \rho$
$\therefore \rho$ is reflexive
Again, $\quad$ let $(a, b) \in \rho$
$\Rightarrow \quad 1+a b>0$
$\Rightarrow \quad 1+b a>0$
$\Rightarrow \quad(b, a) \in \rho$
$\therefore \rho$ is symmetric Now, $\quad(1,-0 \cdot 1) \in \rho$
and $(-0 \cdot 1,-9) \in \rho$
but $\quad(1,-9) \notin \rho$
$\therefore \rho$ is not transitive.

Topic: Mathematics, Sets and Relations, JEE Main
On reaction of Lead Sulphide with dilute nitric acid which of the following is not formed?

Options:
A Nitric oxide
B Nitrous oxide
C Lead nitrate
D Sulphur

The correct answer is:
Nitrous oxide

Solution:
$\mathrm{PbS}+\mathrm{HNO}_3 \rightarrow \mathrm{Pb}\left(\mathrm{NO}_3\right)_2+\mathrm{NO}+\mathrm{S}+\mathrm{H}_2 \mathrm{O}$
Nitrous oxide $\left(\mathrm{N}_2 \mathrm{O}\right)$ is not formed during the reaction.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2024 (09 Apr Shift 1)
On reductive ozonolysis yields

Options:
A 6-oxoheptanal
B 6-oxoheptanoic acid
C 6-hydroxyheptanal
D 3-hydroxypentanal

The correct answer is:
6-oxoheptanal

Solution:
Ozonolysis of alkenes involves the addition of an ozone molecule to an alkene, to form an ozonide. Formed ozonide undergoes cleavage by \(\mathrm{Zn} / \mathrm{H}_2 \mathrm{O}\) to give smaller oxidised product.
Given compound undergo ozonolysis to give 6-oxoheptanal.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
On saturating aqueous solution of $\mathrm{Cu}(\mathrm{II}), \mathrm{Pb}$ (II) and $\mathrm{Zn}$ (II) ions with $\mathrm{H}_2 \mathrm{~S}$ will precipitate

Options:
A Only $\mathrm{CuS}$
B Only $\mathrm{PbS}$
C Both $\mathrm{CuS}$ and $\mathrm{PbS}$
D $\mathrm{CuS}, \mathrm{PbS}$ and $\mathrm{ZnS}$

The correct answer is:
$\mathrm{CuS}, \mathrm{PbS}$ and $\mathrm{ZnS}$

Solution:
$\mathrm{H}_2 \mathrm{~s}$ gas is passed through a saturating aqueous solution containing Cu(I), $\mathrm{Pb}$(II) and $\mathrm{Zn}$(II) ion.
$\begin{aligned} & \mathrm{Pb}^{+2}+\mathrm{H}_2 \mathrm{~S} \longrightarrow \mathrm{PbS} \downarrow \\ & \mathrm{Cu}^{+2}+\mathrm{H}_2 \mathrm{~S} \longrightarrow \mathrm{CuS} \downarrow \\ & \mathrm{Zn}^{+2}+\mathrm{H}_2 \mathrm{~S} \longrightarrow \mathrm{ZnS} \downarrow\end{aligned}$

Topic: Chemistry, Practical Chemistry, JEE Main
On set $A=\{1,2,3\},$ relations $R$ and $S$ are given by $R=\{(1,1),(2,2),(3,3),(1,2),(2,1)\}$
$S=\{(1,1),(2,2),(3,3),(1,3),(3,1)\}$
Then,

Options:
A $R \cup S$ is an equivalence relation
B $R \cup S$ is reflexive and transitive but not symmetric
C $R \cup S$ is reflexive and symmetric but not transitive
D $R \cup S$ is symmetric and transitive but not reflexive

The correct answer is:
$R \cup S$ is reflexive and symmetric but not transitive

Solution:
We have, $R=\{(1,1),(2,2),(3,3),(1,2),(2,1)\}$
$S=\{(1,1),(2,2),(3,3),(1,3),(3,1)\}$
$\therefore R \cup S=\{(1,1),(2,2),(3,3),(1,2),(2,1),(1,3),(3,1)\}$
Since, $(2,1) \in R \cup S,(1,3) \in R \cup S$
but $(2,3) \in R \cup S$
$\therefore \mathrm{R} \cup \mathrm{S}$ is reflexive and symmetric but not transitive.

Topic: Mathematics, Sets and Relations, JEE Main
On shaking $\mathrm{H}_2 \mathrm{O}_2$ with acidified potassium dichromate and ether, ethereal layer becomes

Options:
A Green
B Red
C Blue
D Black

The correct answer is:
Blue

Solution:
This is due to the formation of $\mathrm{CrO}_5$
$\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+\mathrm{H}_2 \mathrm{SO}_4+4 \mathrm{H}_2 \mathrm{O}_2 \rightarrow \mathrm{K}_2 \mathrm{SO}_4+\underset{\text { Blue }}{2 \mathrm{CrO}_5}+5 \mathrm{H}_2 \mathrm{O}$

Topic: Chemistry, Hydrogen, JEE Main
On subjecting 10 ml mixture of N2 and CO to repeated electric spark to form CO2 and NO, 7 ml of O2 was required for combustion. What was the mole percent of CO in the mixture ? (All volumes were measured under identical conditions)

Options:
A 4
B 6
C 40
D 60

The correct answer is:
60

Solution:
N 2 + O 2 â 2 NO CO + 1 2 O 2 â CO 2
Let x ml be the volume of N2 in 10 ml mixture.
Total Volume of O2 required = x + 1 0 - x 2 = 7 (given)
â x = 4 ml
Volume of CO = 1 0 - 4 = 6 ml
Mol % of CO = 6 Ã 1 0 0 1 0 = 6 0

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
On the basis of information given below mark the correct option. Information:
(A) In bromoethane and chloroethane mixture intermolecular interactions of \( \mathrm{A}-\mathrm{A} \) and \( \mathrm{B}-\mathrm{B} \) type are nearly same as
\( \mathrm{A}-\mathrm{B} \) type interactions.
(B) In ethanol and acetone mixture, \( \mathrm{A}-\mathrm{A} \) or \( \mathrm{B}-\mathrm{B} \) type intermolecular interactions are stronger than \( \mathrm{A}-\mathrm{B} \) type
interactions.
(C) In chloroform and acetone mixture, \( \mathrm{A}-\mathrm{A} \) or \( \mathrm{B}-\mathrm{B} \) type intermolecular interactions are weaker than \( \mathrm{A}-\mathrm{B} \) type
interactions.

Options:
A Solution (B) and (C) will follow Raoult's law.
B Solution (A) will follow Raoult's law.
C Solution (B) will show negative deviation from Raoult's law.
D Solution (C) will show positive deviation from Raoult's law.

The correct answer is:
Solution (A) will follow Raoult's law.

Solution:
Each constituent of a binary liquid solution will obey Rault's law if the A-A and B-B forces of attraction are more or less the same as the A-B force of attraction. In such a case, the total vapour pressure of the solution is the sum of the partial vapour pressures of components of the solution.

The bromoethane and chloroethane mixture follows the Raoult's law because the intermolecular interactions of A-A and B-B type are nearly the same as A-B type interactions.

Topic: Chemistry, Solutions, JEE Main
On the basis of kinetic theory of gases, the gas exerts pressure because its molecules:

Options:
A continuously stick to the walls of container.
B suffer change in momentum when impinge on the walls of container.
C continuously lose their energy till it reaches wall.
D are attracted by the walls of container.

The correct answer is:
suffer change in momentum when impinge on the walls of container.

Solution:
From the assumption of KTG, the molecules of gas collide with the walls and suffers momentum change which results in force on the wall and hence pressure.

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2021 (24 Feb Shift 2)
On the basis of kinetic theory of gases, the gas exerts pressure because its molecules:

Options:
A continuously stick to the walls of container.
B suffer change in momentum when impinge on the walls of container.
C continuously lose their energy till it reaches wall.
D are attracted by the walls of container.

The correct answer is:
suffer change in momentum when impinge on the walls of container.

Solution:
From the assumption of KTG, the molecules of gas collide with the walls and suffers momentum change which results in force on the wall and hence pressure.

Topic: Physics, Kinetic Theory of Gases, JEE Main
On the basis of Lanthanoid contraction, explain the following Stability of the complexes of lanthanoids.

Options:
A Stability of complexes increases.
B Stability of complexes decreases.
C Stability of complexes remains constant.
D No specific trend can be observed.

The correct answer is:
Stability of complexes increases.

Solution:
Correct Option is : (A)
Stability of complexes increases.

Topic: Chemistry, d and f Block Elements, NEET
On the basis of Lanthanoid contraction, explain the following Trends in acidic character of lanthanoid oxides.

Options:
A Acidic character increases from La to Lu.
B Acidic character decreases from La to Lu.
C Acidic character remains constant.
D No specific trend can be observed.

The correct answer is:
Acidic character increases from La to Lu.

Solution:
Correct Option is : (A)
Acidic character increases from La to Lu.

Topic: Chemistry, d and f Block Elements, NEET
On the basis of Lanthanoid contraction, explain the following Trends in the stability of oxo salts of lanthanoids from La to Lu.

Options:
A Stability of oxo salts increases.
B Stability of oxo salts decreases.
C Stability of oxo salts remains constant.
D No specific trend can be observed.

The correct answer is:
Stability of oxo salts decreases.

Solution:
Correct Option is : (B)
Stability of oxo salts decreases.

Topic: Chemistry, d and f Block Elements, NEET
On the basis of the following $\mathrm{E}^{\circ}$ values, the strongest oxidizing agent is
$$
\begin{array}{ll}
{\left[\mathrm{Fe}(\mathrm{CN})_\theta\right]^{4-} \rightarrow\left[\mathrm{Fe}(\mathrm{CN})_\theta\right]^{3-}+\mathrm{e}^{-1} ;} & \mathrm{E}^0=-0.35 \mathrm{~V} \\
\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+}+\mathrm{e}^{-1} ; & \mathrm{E}^0=-0.77 \mathrm{~V}
\end{array}
$$

Options:
A $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$
B $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$
C $\mathrm{Fe}^{2+}$
D $\mathrm{Fe}^{3+}$

The correct answer is:
$\mathrm{Fe}^{3+}$

Solution:
For the strongest oxidizing agent, the oxidizing potential should be least. Here, the oxidizing potential of $\mathrm{Fe}^{+2}$ is less than that of $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$. Therefore, $\mathrm{Fe}^{+2}$ is stronger oxidizing agent than $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$. Also, the stronger oxidizing agent should easily reduce itself. Here, $\mathrm{Fe}^{+3}$ is easily reduced than $\mathrm{Fe}^{+2}$. Therefore, among all the four, $\mathrm{Fe}^{+3}$ is the stronger oxidizing agent.

Topic: Chemistry, Electrochemistry, NEET, NEET 2008 (Mains)
On the basis of the following thermochemical data: $\left(\Delta \mathrm{fG}^{\circ} \mathrm{H}_{\text {(aq) }}^{+}=0\right)$
$$
\begin{aligned}
& \mathrm{H}_2 \mathrm{O}(\ell) \rightarrow \mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) ; \Delta \mathrm{H}=57.32 \mathrm{~kJ} \\
& \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(\ell) ; \Delta \mathrm{H}=-286.20 \mathrm{~kJ}
\end{aligned}
$$
The value of enthalpy of formation of $\mathrm{OH}^{-}$ion at $25^{\circ} \mathrm{C}$ is:

Options:
A
$-22.88 \mathrm{~kJ}$

B
$-228.88 \mathrm{~kJ}$

C
$+228.88 \mathrm{~kJ}$

D
$-343.52 \mathrm{~kJ}$

The correct answer is:
$-228.88 \mathrm{~kJ}$

Solution:
By adding the two given equations, we have
$$
\mathrm{H}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{OH}_{(\mathrm{aq})}^{-} ; \Delta \mathrm{H}=-228.88 \mathrm{Kj}
$$
Here $\Delta \mathrm{H}_{\mathrm{f}}^{\circ}$ of $\mathrm{H}_{(\mathrm{aq})}^{+}=0$
$\therefore \Delta \mathrm{H}_{\mathrm{f}}^{\circ}$ of $\mathrm{OH}^{-}=-228.88 \mathrm{~kJ}$

Topic: Chemistry, Thermodynamics (C), JEE Main
On the basis of the following thermochemical data: $\left(\Delta \mathrm{fG}^{\circ} \mathrm{H}_{\text {(aq) }}^{+}=0\right)$
$$
\begin{aligned}
& \mathrm{H}_2 \mathrm{O}(\ell) \rightarrow \mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) ; \Delta \mathrm{H}=57.32 \mathrm{~kJ} \\
& \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(\ell) ; \Delta \mathrm{H}=-286.20 \mathrm{~kJ}
\end{aligned}
$$
The value of enthalpy of formation of $\mathrm{OH}^{-}$ion at $25^{\circ} \mathrm{C}$ is:

Options:
A
$-22.88 \mathrm{~kJ}$

B
$-228.88 \mathrm{~kJ}$

C
$+228.88 \mathrm{~kJ}$

D
$-343.52 \mathrm{~kJ}$

The correct answer is:
$-228.88 \mathrm{~kJ}$

Solution:
By adding the two given equations, we have
$$
\mathrm{H}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{OH}_{(\mathrm{aq})}^{-} ; \Delta \mathrm{H}=-228.88 \mathrm{Kj}
$$
Here $\Delta \mathrm{H}_{\mathrm{f}}^{\circ}$ of $\mathrm{H}_{(\mathrm{aq})}^{+}=0$
$\therefore \Delta \mathrm{H}_{\mathrm{f}}^{\circ}$ of $\mathrm{OH}^{-}=-228.88 \mathrm{~kJ}$

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2009
On the basis of the information available from the reaction $\frac{4}{3} A I+O_2 \rightarrow \frac{2}{3} A_2 O_3, \Delta G=-827 \mathrm{kJmol}^{-1}$ of $O_2$ the minimum emf required to carry out an electrolysis of $\mathrm{Al}_2 \mathrm{O}_3$ is $\left(F=96500 \mathrm{C} \mathrm{mol}^{-1}\right)$

Options:
A $8.56 \mathrm{~V}$
B $2.14 \mathrm{~V}$
C $4.28 \mathrm{~V}$
D $6.42 \mathrm{~V}$

The correct answer is:
$2.14 \mathrm{~V}$

Solution:
$V=827 \times 10^{3 /} 4 \times 96500=2.14 \mathrm{~V}$

Topic: Chemistry, Electrochemistry, JEE Main
On the ellipse x28+y24=1, let P be a point in the second quadrant such that the tangent at P to the ellipse is perpendicular to the line x+2y=0. Let S and S' be the foci of the ellipse and e be its eccentricity. If A is the area of the triangle SPS', then the value of 5-e2·A is

Options:
A 12
B 6
C 14
D 24

The correct answer is:
6

Solution:
We have an ellipse x28+y24=1, we have a general point P(8cosθ,2sinθ)

Equation of tangent at point p

x8cosθ+y2sinθ-1=0

Slope =-12cotθ

Given that, tangent at P is perpendicular to the line x+2y=0.

So, product of their slopes are perpendicular ⇒-12cotθ×-12=-1

⇒-12cotθ=2

⇒cotθ=-22

⇒cosθ=-223,sinθ=13

So, point P-83,23

A=12×2ae×23

Where, e=a2-b2a2=8-48=12

∴A=12×2×22×12×23=43So, 5-e2A=5-12×43=6

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2021 (26 Aug Shift 1)
On the interval $[0,1]$ the function $x^{25}(1-x)^{75}$ takes its maximum value at the point

Options:
A 0
B $1 / 4$
C $1 / 2$
D $1 / 3$

The correct answer is:
$1 / 4$

Solution:
Given, $f(x)=x^{25}(1-x)^{75}$
$$
\begin{aligned}
\Rightarrow f^{\prime}(x) &=25 x^{24}(1-x)^{75}-75 x^{25}(1-x)^{74} \\
&=25 x^{24}(1-x)^{74}(1-4 x) \\
\therefore \quad f^{\prime}(x) &=0 \\
\Rightarrow \quad x &=0,1,1 / 4
\end{aligned}
$$
If $x < 1 / 4$, then
$$
f^{\prime}(x)=25 x^{24}(1-x)^{74}(1-4 x)>0 .
$$
and if $x>1 / 4$, then
$$
f^{\prime}(x)=25 x^{24}(1-x)^{74}(1-4 x) < 0 .
$$
Thus, $f^{\prime}(x)$ changes its sign from positive to negative as $x$ passes through $1 / 4$ from left to right. Hence, $f(x)$ attains its maximum at $x=1 / 4$.

Topic: Mathematics, Application of Derivatives, JEE Main
On the interval $(1,3)$, the function $f(x)=3 x+\frac{2}{x}$ is

Options:
A Strictly decreasing
B Strictly increasing
C Decreasing in $(2,3)$ only
D Neither increasing nor decreasing

The correct answer is:
Strictly increasing

Solution:
$f(x)=3 x+\frac{2}{x} \Rightarrow f^{\prime}(x)=3-\frac{2}{x^2}$
Clearly $f^{\prime}(x)\gt0$ on the interval $(1,3)$
$\therefore f(x)$ is strictly increasing.

Topic: Mathematics, Application of Derivatives, JEE Main
On the interval \( [0,1] \) the function \( x^{25}(1-x)^{75} \) takes its maximum value at the point

Options:
A \( 0 \)
B \( 1 / 4 \)
C \( 1 / 2 \)
D \( 1 / 3 \)

The correct answer is:
\( 1 / 4 \)

Solution:
Let f (x) = x25 (1 - x)75, x Ïµ [0,1]

∴ f ' (x) = 25 x24 (1 - x)75 - 75x25 (1- x)74

= 25 x24 (1 - x)74 [(1 - x) - 3x]

= 25 x24 (1 - x)74 (1 - 4x)

For maximum value of f (x), put f'(x) = 0

∴25x24 (1 - x)74 (1 - 4x) = 0

∴ x=0,1,14

Also, at x = 0 , y = 0
at x = 1, y = 0
and x = 1/4, y > 0
f (x) attains maximum at x = 1 / 4.

Topic: Mathematics, Application of Derivatives, JEE Main
On the set $R$ of real numbers, the relation $\rho$ is defined by $x \rho y,(x, y) \in R$

Options:
A If $|x-y| < 2$, then $\rho$ is reflexive but neither symmetric nor transitive.
B II $x-y < 2$, then $\rho$ is reflexeve and symmetric bu not transitive.
C $If(x \mid \geq y,$ then $\rho$ is reflexive and transitive but not symmetric.
D If $x>|y|$, then $p$ is transitive but neither reflexve nor symmetric.

The correct answer is:
If $x>|y|$, then $p$ is transitive but neither reflexve nor symmetric.

Solution:
On the set $R$ of real numbers
For reflexive,
$x \rho x \Rightarrow(x, x) \in R$
$\Rightarrow x > |x|$ which is not true. $\Rightarrow \rho$ is not reflexive.
For symmetric,
$(x, y) \in R \Rightarrow x > |y|$
and $(y, x) \in R \Rightarrow y > |x|$
So, $x > |y| \neq y > |x|$ $\Rightarrow \rho$ is not symmetric.
For transitive, $(x, y) \in R \Rightarrow x > |y|(y, z) \in R \Rightarrow y > |z|$
$\Rightarrow x > |z| \Rightarrow(x, z) \in R$
$\Rightarrow \rho$ is transitive.

Topic: Mathematics, Sets and Relations, JEE Main