datavorous/entrance-exam-dataset huggingface dataset:
Equimolar solutions of the following compounds are prepared separately in water. Which will have the lowest $\mathrm{pH}$ value ?

Options:
A
$\mathrm{BeCl}_2$

B
$\mathrm{SrCl}_2$

C
$\mathrm{CaCl}_2$

D
$\mathrm{MgCl}_2$

The correct answer is:
$\mathrm{BeCl}_2$

Solution:
Metal halide on hydrolysis with water form corresponding hydroxides.
The basic strength of hydroxide increases as we move down in a group. This is because of the increase in size which results in decrease of ionization energy which weakens the strength of $\mathrm{M}-\mathrm{O}$ bonds in MOH and thus increases the basic strength.
$$
\begin{aligned}
& \mathrm{Be}(\mathrm{OH})_2 \quad \mathrm{Mg}(\mathrm{OH})_2 \\
& \text { Amphoteric Weak base } \\
& \mathrm{Ca}(\mathrm{OH})_2 \mathrm{Sr}(\mathrm{OH})_2 \mathrm{Ba}(\mathrm{OH})_2 \\
&
\end{aligned}
$$
Hence, $\mathrm{Be}(\mathrm{OH})_2$ will have lowest $\mathrm{pH}$.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2013 (23 Apr Online)
Equimolar solutions of the following were prepared in water separately. Which one of the solutions will record the highest $\mathrm{pH}$ ?

Options:
A $\mathrm{CaCl}_2$
B $\mathrm{SrCl}_2$
C $\mathrm{BaCl}_2$
D $\mathrm{MgCl}_2$

The correct answer is:
$\mathrm{BaCl}_2$

Solution:
No Solution Available

Topic: Chemistry, s Block Elements, NEET, NEET 2008 (Mains)
Equivalent amounts of H2 and I2 are heated in a closed vessel till equilibrium is obtained. If 80% of the hydrogen can be converted to HI, the K c at this temperature is

Options:
A 64
B 16
C 0.25
D 4

The correct answer is:
64

Solution:
Let here be 1M each of H2 and I2
H2 + I2 â 2HI
1 1 0
(1-0.8) (1-0.8) 2Ã0.8
=0.2 =0.2 =1.6
K c  =  HI 2 H 2 I 2
= 1.6×1.6 0.2×0.2
K c  = 64

Topic: Chemistry, Chemical Equilibrium, JEE Main
Equivalent amounts of $\mathrm{H}_{2}$ and $\mathrm{I}_{2}$ are heated in a closed vessel till equilibrium is obtained. If $80 \%$ of the hydrogen can be converted to $\mathrm{HI}$, the $\mathrm{Kc}$ at this temperature is

Options:
A 64
B 16
C $0.25$
D 4

The correct answer is:
64

Solution:
$\mathrm{K}_{\mathrm{c}}=\frac{[\mathrm{HI}]^{2}}{\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right]}=\frac{(2 \mathrm{x})^{2}}{(1-\mathrm{x})^{2}}$
Given $\mathrm{x}=80 \%=0.80$
$\therefore \mathrm{K}_{\mathrm{c}}=\frac{(2 \times 0.80)^{2}}{(1-0.80)^{2}}=64$

Topic: Chemistry, Chemical Equilibrium, JEE Main
Equivalent capacitance between $A$ and $B$ in the figure is

Options:
A $20 \mu \mathrm{F}$
B $8 \mu \mathrm{F}$
C $12 \mu \mathrm{F}$
D $16 \mu \mathrm{F}$

The correct answer is:
$8 \mu \mathrm{F}$

Solution:
$\frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}=\frac{1}{4}+\frac{1}{4}=\frac{2}{4}$
$C=2 \mu F$
$2 \mu \mathrm{F}$ will be parallel to $4 \mu \mathrm{F}$. so the resultant capacitance

$\begin{aligned} \frac{1}{C^{\prime \prime}} &=\frac{1}{4 \mu F}+\frac{1}{4 \mu F}=\frac{2}{4} \mu F \\ C^{-} &=2 \mu F \\ C_{A Q} &=2 \mu F+6 \mu F \\ &=8 \mu F \end{aligned}$

Topic: Physics, Capacitance, JEE Main
Equivalent resistance between the adjacent corners of a regular n-sided polygon of uniform wire of resistance R would be :

Options:
A n-1Rn2
B n-1R2n-1
C n2Rn-1
D n-1Rn

The correct answer is:
n-1Rn2

Solution:
Given the total resistance of the wire is R. If the polygon has n numbers of side, then the resistance r, as shown in the following figure, of each side can be written as

r= Rn.................(1)

Among all the resistances, n-1 resistances are connected in series combination, whereas, their equivalent resistance is connected in parallel with the nth resistance. The equivalent circuit diagram for the above set of resistances can be depicted as follows:

From the above equivalent circuit diagram, the net equivalent resistance Req for the entire combination can be calculated as follows-

1Req= 1n-1r+1r= 1n-1Rn+1Rn= nn-1R+nR= n2n-1RReq= n-1Rn2

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (01 Feb Shift 2)
Equivalent resistance of the following network is ________ Ω.

Options:

The correct answer is:
1

Solution:
In the above diagram, 6Ω is short circuit. Therefore, we can remove it.

In the left part, 2 Ω and 2 Ω are in parallel combination, which will give us 1 Ω which again in turn is in series combination with the 2 Ω(adjacent to 3 Ω resistor), therefore net contribution of this part will be 1 Ω+2 Ω=3 Ω.

Now, all three 3 Ω are in parallel across point A & B, therefore Req=3×13=1 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2024 (31 Jan Shift 1)
Equivalent statement of the statement “If 9>10 then 32=5” will be

Options:
A If 32=5 then 9>10
B 9<10 and 32≠5
C 9<10 or 32=5
D None of these

The correct answer is:
None of these

Solution:
We know that, pâqâ¡~pâ¨q
â´ equivalent statement will be 9â¯10 or 32=5
or 9â¤10 or 32=5

Topic: Mathematics, Mathematical Reasoning, JEE Main
Equivalent weight of KMnO4 when it is convect into MnSO4 is Where M = molar mass of KMnO4.

Options:
A M/5
B M/6
C M/3
D M/2

The correct answer is:
M/5

Solution:
MnO4-+5e-→Mn2+
So equivalent weight of KMnO4
=Molar mass of KMnO4MNo. of mol of electrons gained by 1 mol of KMnO4
=M5 .

Topic: Chemistry, Redox Reactions, JEE Main
Escape velocity from the earth's surface is

Options:
A equal to that from the moon's surface.
B greater than that from the moon's surface
C less than that from the moon's surface.
D equal to that from the sun's surface.

The correct answer is:
greater than that from the moon's surface

Solution:
Escape velocity from the surface of earth is given as vescape=2gR, where g is gravitational acceleration on the surface of earth and R is radius of earth. As we know that, radius of moon, Rmoon=Rearth4 and gravitational acceleration of moon, gmoon=gearth6.

⇒vescapemoon=2×gearth6×Rearth4=vescapeearth10

Escape velocity from surface of earth is more than that from surface of moon.

Topic: Physics, Gravitation, JEE Main
Escape velocity of a body from earth is 11.2 km s-1. If the radius of a planet be one-third the radius of earth and mass be one-sixth that of earth, the escape velocity from the plate is:

Options:
A 11.2 km s-1
B 8.4 km s-1
C 4.2 km s-1
D 7.9 km s-1

The correct answer is:
7.9 km s-1

Solution:
Given: RP=RE3, MP=ME6

Escape velocity is given by, v=2GMR

Therefore,

vE=2GMERE ...(i)

vP=2GMPRP...(ii)

⇒vEvP=2

⇒vP=vE2=11.22=7.9 km s-1

Topic: Physics, Gravitation, JEE Main, JEE Main 2024 (30 Jan Shift 2)
Escape velocity of a rocket is $11.2 \mathrm{~km} / \mathrm{sec}$. It is released at an angle of $45^{\circ}$. Its escape velocity is

Options:
A $11.2 \mathrm{~m} / \mathrm{sec}$
B $11.2 \sqrt{2} \mathrm{~km} / \mathrm{sec}$
C $11.2 \mathrm{~km} / \mathrm{sec}$
D $22.3 \mathrm{~km} / \mathrm{sec}$.

The correct answer is:
$11.2 \mathrm{~km} / \mathrm{sec}$

Solution:
The escape velocity is independent of the angle of projection. So, the escape velocity of the rocket remains same.

Topic: Physics, Gravitation, AIIMS, AIIMS 2001
Ester and acetamide are distinguished by

Options:
A Hydrolysis with strong acids or alkali
B Derivatives of fatty acids
C Both (a) and (b)
D None of these

The correct answer is:
Both (a) and (b)

Solution:
Ester and acetamide are distinguished by hydrolysis with strong acids or alkali and derivatives of fatty acids.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Ester containing α -hydrogens undergo self condensation in presence of a strong base such as sodium ethoxide to form β -ketoesters. This reaction is called

Options:
A Aldol condensation
B Claisen condensation
C Diekmann condensation
D Crossed-Claisen condensation

The correct answer is:
Claisen condensation

Solution:
Ester containing Î± -hydrogens undergo self condensation in presence of a strong base such as sodium ethoxide to form Î² -ketoesters. This reaction is called Claisen condensation.
CH3-C||O-OC2H5+H-CÎ±H2COOC2H5âC2H5O-Na+CH3-C||O-CÎ±H2COOC2H5A Î²-keto ester+C2H5OH
Crossed-Claisen condensation is feasible only when one of the reactants has no Î± -hydrogen and is thus incapable of undergoing self condensation. For example

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Ethane nitrile \( \stackrel{\mathrm{Na} / \mathrm{EtOH}}{\longrightarrow} \mathrm{A} \stackrel{\mathrm{HNO}_{3}}{\longrightarrow} \mathrm{B} \stackrel{\mathrm{PCC}}{\longrightarrow} \mathrm{C} \)
The product \( \mathrm{C} \) in the above series is

Options:
A \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHO} \).
B \( \mathrm{CH}_{3} \mathrm{COCH}_{3} \).
C \( \mathrm{CH}_{3} \mathrm{CONH}_{2} \).
D \( \mathrm{CH}_{3} \mathrm{CHO} \).

The correct answer is:
\( \mathrm{CH}_{3} \mathrm{CHO} \).

Solution:
In the presence of Na/C2H5OH, the reduction of cyanide takes place. It is an electron transfer reduction called Mendius reaction. Ethane nitrile gives ethanamine A in this reaction.

The primary amine A undergoes diazotisation with nitrous acid and forms alcohol B.

Alcohol B on oxidation with a mild oxidising agent pyridinium chlorochromate (PCC) gives acetaldehyde C.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Ethanol is converted into ethoxy ethane by:

Options:
A heating excess of ethanol with conc. \( \mathrm{H}_{2} \mathrm{SO}_{4} \) at \( 140^{\circ} \mathrm{C} \)
B heating ethanol with excess of conc. \( \mathrm{H}_{2} \mathrm{SO}_{4} \) at \( 443 \mathrm{~K} \)
C treating with conc. \( \mathrm{H}_{2} \mathrm{SO}_{4} \) at room temperature
D treating with conc. \( \mathrm{H}_{2} \mathrm{SO}_{4} \) at \( 273 \mathrm{~K} \)

The correct answer is:
heating excess of ethanol with conc. \( \mathrm{H}_{2} \mathrm{SO}_{4} \) at \( 140^{\circ} \mathrm{C} \)

Solution:
Ethanol is converted to ethoxy ethane by heating with conc. H2SO4 at 140o413K, i.e.
Step 1:
Step 2: Another CH3CH2OH attacks on CH3CH2⊕ ion as follows:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Ethanol is prepared industrially by

Options:
A Hydration of ethylene
B Fermentation of sugars
C Both the above
D None of these

The correct answer is:
Both the above

Solution:
Hydration of alkenes

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Ethanol, when reacted with $\mathrm{PCl}_5$ gave $\mathrm{A}, \mathrm{POCl}_3$ and $\mathrm{HCl}$. $A$ reacts with silver nitrite to form $B$ (major product) and $\mathrm{AgCl} . A$ and $B$ are respectively

Options:
A $\mathrm{C}_2 \mathrm{H}_5 \mathrm{Cl}$ and $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OC}_2 \mathrm{H}_5$
B $\mathrm{C}_2 \mathrm{H}_6$ and $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OC}_2 \mathrm{H}_5$
C $\mathrm{C}_2 \mathrm{H}_5 \mathrm{Cl}$ and $\mathrm{C}_2 \mathrm{H}_5 \mathrm{NO}_2$
D $\mathrm{C}_2 \mathrm{H}_6$ and $\mathrm{C}_2 \mathrm{H}_5 \mathrm{NO}_2$

The correct answer is:
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{Cl}$ and $\mathrm{C}_2 \mathrm{H}_5 \mathrm{NO}_2$

Solution:
According to the conditions of question

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Ethelene can be separated from acetylene by passing the mixture through :

Options:
A fuming $\mathrm{H}_2 \mathrm{SO}_4$
B pyrogallol
C ammoniacal $\mathrm{Cu}_2 \mathrm{Cl}_2$
D Charcoal powder

The correct answer is:
ammoniacal $\mathrm{Cu}_2 \mathrm{Cl}_2$

Solution:
Hints : $\mathrm{H}-\mathrm{C} \equiv \mathrm{C}-\mathrm{H}+\mathrm{Cu}_2 \mathrm{Cl}_2 \rightarrow \mathrm{Cu}^{+} \mathrm{C}^{-} \equiv \mathrm{C}^{-} \mathrm{Cu}^{+} \downarrow$
Red ppt.
$$
\mathrm{H}_2 \mathrm{C}=\mathrm{CH}_2+\mathrm{Cu}_2 \mathrm{Cl}_2 \rightarrow \mathrm{No} . \mathrm{ppt}
$$

Topic: Chemistry, Hydrocarbons, JEE Main
Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as \(\mathrm{CH}_3 \mathrm{COOH}(\ell)+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(\ell) \rightleftharpoons \mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5(\ell)+\mathrm{H}_2 \mathrm{O}(\ell)\)
(i) Write the concentration ratio (reaction quotient), \(Q_e\), for this reaction (Note: water is not in excess and is not a solvent in this reaction)
(ii) At \(293 \mathrm{~K}\), if one starts with \(1.00 \mathrm{~mol}\) of acetic acid and \(0.18 \mathrm{~mol}\) of ethanol, there is \(0.171 \mathrm{~mol}\) of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.
(iii) Starting with \(0.5 \mathrm{~mol}\) of ethanol and \(1.0 \mathrm{~mol}\) of acetic acid and maintaining it at \(293 \mathrm{~K}, 0.214 \mathrm{~mol}\) of ethyl acetate is found after sometime. Has equilibrium been reached ?

Options:

None

Solution:
(i) \(\mathrm{Q}_{\mathrm{c}}=\frac{\left[\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5\right]\left[\mathrm{H}_2 \mathrm{O}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]\left[\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right]}\)
Let \(\mathrm{V}\) be the volume of reaction vessel.
Reaction quotient \(\left(\mathrm{Q}_{\mathrm{c}}\right)=\frac{(0.214 / \mathrm{V})(0.214 / \mathrm{V})}{(0.786 / \mathrm{V})(0.286 / \mathrm{V})}=0.203\)
As \(\mathrm{Q}_{\mathrm{c}} \neq \mathrm{K}_{\mathrm{c}}\), equilibrium has not been attained.

Topic: Chemistry, Equilibrium
Ethyl acetoacetate shows, which type of isomerism

Options:
A Chain
B Optical
C Metamerism
D Tautomerism

The correct answer is:
Tautomerism

Solution:
$\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5$ shows tautomeric isomerism.

Topic: Chemistry, General Organic Chemistry, JEE Main
Ethyl amine on acetylation gives which of the following?

Options:
A $\mathrm{N}$-ethyl acetamide
B Acetamide
C Methyl acetamide
D N,N-dimethyl acetamide

The correct answer is:
$\mathrm{N}$-ethyl acetamide

Solution:
Explanation: Ethyl amine on acetylation gives $\mathrm{N}$-ethyl acetamide, as shown in given reaction.
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{NH}_2+\mathrm{CH}_3 \mathrm{COCl} \rightarrow \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{NHCOCH}_3+\mathrm{HCl}$

Topic: Chemistry, Amines, JEE Main
Ethyl amine on acetylation gives

Options:
A $\mathrm{N}$-ethyl acetamide
B Acetamide
C Methyl acetamide
D None

The correct answer is:
$\mathrm{N}$-ethyl acetamide

Solution:
\(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{NH}_2+\mathrm{CH}_3 \mathrm{COCl} \rightarrow \underset{\text{N Ethylethanamide}}{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{NHCOCH}_3+\mathrm{HCl}}\)

Topic: Chemistry, Amines, JEE Main
Ethyl chloride on heating with AgCN forms a compound X. The functional isomer of X is -

Options:
A C2H5CN
B C2H5NH2
C C2H5NC
D None of these

The correct answer is:
C2H5CN

Solution:
C 2 H 5 Cl + AgCN → C 2 H 5 NC + AgCl X
Thus, X is ethyl isocyanide. As the question is asking about the functional isomer of X, the functional isomer of C2H5NC will be C2H5CN, i.e. ethyl cyanide.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Ethyl chloride on heating with silver cyanide forms a compound $X$. The functional isomer of $X$ is

Options:
A $\mathrm{C}_2 \mathrm{H}_5 \mathrm{NC}$
B $\mathrm{C}_2 \mathrm{H}_5 \mathrm{CN}$
C $\mathrm{H}_3 \mathrm{C}-\mathrm{NH}-\mathrm{CH}_3$
D $\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2$

The correct answer is:
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{CN}$

Solution:
The functional isomer of ethyl isocyanide is ethyl cyanide $\mathrm{C}_2 \mathrm{H}_5-\mathrm{C} \equiv \mathrm{N}$

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Ethyl isocyanide on hydrolysis in acidic medium generates

Options:
A
propanoic acid and ammonium salt

B
ethanoic acid and ammonium salt

C
methylamine salt and ethanoic acid

D
ethylamine salt and methanoic acid

The correct answer is:
ethylamine salt and methanoic acid

Solution:
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{~N} \equiv \mathrm{C}+\mathrm{H}_2 \mathrm{O} \stackrel{\mathrm{H}^{+}}{\longrightarrow} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{NH}_2+\mathrm{HCOOH}$. Therefore it gives only one mono chloroalkane.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main, JEE Main 2003
Ethyl isocyanide on hydrolysis in acidic medium generates

Options:
A
propanoic acid and ammonium salt

B
ethanoic acid and ammonium salt

C
methylamine salt and ethanoic acid

D
ethylamine salt and methanoic acid

The correct answer is:
ethylamine salt and methanoic acid

Solution:
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{~N} \equiv \mathrm{C}+\mathrm{H}_2 \mathrm{O} \stackrel{\mathrm{H}^{+}}{\longrightarrow} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{NH}_2+\mathrm{HCOOH}$. Therefore it gives only one mono chloroalkane.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Ethyl isocyanide on hydrolysis in acidic medium generates

Options:
A Propanoic acid and ammonium salt
B Ethanoic acid and ammonium salt
C Methylamine salt and ethanoic acid
D Ethylamine and methanoic acid

The correct answer is:
Ethylamine and methanoic acid

Solution:
Ethyl isocyanide on hydrolysis form primary amines.
CH 3 CH 2 NC + H 2 O → H + CH 3 CH 2 NH 2 + HCOOH

Topic: Chemistry, Amines, JEE Main
Ethyl methyl ketone is obtained by heating calcium salts of

Options:
A formic acid $+$ propionic acid
B acetic acid + propionic acid
C acetic acid only
D acetic acid + methanoic acid

The correct answer is:
acetic acid + propionic acid

Solution:
Ethyl methyl ketone is an unsymmetric ketone, so two different acids (in the form of calcium salt) are used. The structure of ethyl methyl ketone is
$$
\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COCH}_{3}
$$
The structure shows the presence of 4 carbon atoms, thus the calcium salts must be of acetic acid and propionic acid to form $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COCH}_{3}$.
$$
\begin{array}{r}
\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COO}\right)_{2} \mathrm{Ca}+\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2} \mathrm{Ca} \stackrel{\text { Distill }}{\longrightarrow} \\
\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COCH}_{3}+\mathrm{CaCO}_{3}
\end{array}
$$

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Ethylamine C2H5NH2 can be obtained from N-ethylphthalimide on treatment with:

Options:
A CaH2
B NH2NH2
C H2O
D NaBH4

The correct answer is:
NH2NH2

Solution:

Topic: Chemistry, Amines, JEE Main, JEE Main 2019 (10 Apr Shift 1)
Ethylene dibromide on heating with metallic sodium in ether yields

Options:
A ethene
B ethyene
C 2-butene
D 1-butene

The correct answer is:
2-butene

Solution:
→ Δ Ether CH 3 −CH = CH 2-Butene − CH 3  + 4NaBr

Topic: Chemistry, Hydrocarbons, JEE Main
Ethylene difluoride on hydrolysis gives

Options:
A Glycol
B Fluoroethanol
C Difluoroethanol
D Freon

The correct answer is:
Glycol

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Ethylene glycol gives oxalic acid on oxidation with

Options:
A acidified $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$
B acidified $\mathrm{KMnO}_{4}$
C alkaline $\mathrm{KMnO}_{4}$
D periodic acid

The correct answer is:
alkaline $\mathrm{KMnO}_{4}$

Solution:
Alkaline Potassium permanganate is a strong oxidising agent. It oxidises ethylene glycol to oxalic acid.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Ethylene glycol is used as an antifreeze in a cold climate. Mass of ethylene glycol which should be added to $4 \mathrm{~kg}$ of water to prevent it from freezing at $-6^{\circ} \mathrm{C}$ will be :
$\left[\mathrm{K}_{\mathrm{t}}\right.$ for water $=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$, and molar mass of ethylene glycol $=62 \mathrm{~g} \mathrm{~mol}^{-1}$ )

Options:
A
$204.30 \mathrm{~g}$

B
$400.00 \mathrm{~g}$

C
$304.60 \mathrm{~g}$

D
$804.32 \mathrm{~g}$

The correct answer is:
$804.32 \mathrm{~g}$

Solution:
$$
\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \times \mathrm{m}=\mathrm{K}_{\mathrm{f}} \times \frac{\mathrm{w}_2 \times 1000}{\mathrm{w}_1 \times \mathrm{m}_2}
$$
$w_1 \& w_2=$ wt of solvent \& solute respecting $m_2=m w$ of solute
$$
\Delta \mathrm{T}_{\mathrm{f}}=0^{\circ}-\left(-6^{\circ}\right)=6=1.86 \times \frac{\mathrm{w}_2 \times 1000}{4000 \times 62}
$$
Therefore $w_2=800 \mathrm{~g}$

Topic: Chemistry, Solutions, JEE Main, JEE Main 2011
Ethylene is prepared by the dehydration of

Options:
A Ethyl alcohol
B Methyl alcohol
C Acetic acid
D Oxalic acid

The correct answer is:
Ethyl alcohol

Solution:
$\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{OH} \xrightarrow[\text { Dehydration }]{\text { Conc. } \mathrm{H}_2 \mathrm{SO}_4} \mathrm{CH}_2=\mathrm{CH}_2+\mathrm{H}_2 \mathrm{O}$

Topic: Chemistry, Hydrocarbons, JEE Main
Ethylidene dichloride is obtained by the reaction of excess of $\mathrm{HCl}$ with

Options:
A ethylene
B acetylene
C propene
D methane

The correct answer is:
acetylene

Solution:
[Alkenes due to presence of only $1^{\circ}$ of $\begin{array}{lll}\text { unsaturation, } & \text { form } & \text { only } & \text { monochloro }\end{array}$ derivative.]

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Euchlorine is

Options:
A obtained by heating perchlorate with conc. HCl
B a chloride of europium
C a mixture of Cl2 and Cl2O7
D a micture of Cl2 and ClO2

The correct answer is:
a micture of Cl2 and ClO2

Solution:
When KClO3 is heated with conc. HCl, a mixture of Cl2 and ClO2 is formed. The mixture is known as euchlorine.
2KClO3+4HCl→2KCl+2ClO2+Cl2+2H2O

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Euchromatin is:

a. Loosely packed b. Stains light c. Transcriptionally active d. Early replicating

Choose the option with the correct choice of characters.

Options:
A a and c only
B a, b, c, and d
C b, c, and d only
D a, b, and c only

The correct answer is:
a, b, c, and d

Solution:
Chromatin is composed of DNA and histone proteins. Chromatin is of two types i.e.euchromatin and heterochromatin. In euchromatin, the chromatin fibers are loosely packed and they undergo the process of condensation and decondensation in the cell cycle. They often remain active in transcription(RNA synthesize through DNA). Due to a less compact structure, they show light staining. They try to replicate in the early S phase. So all the above options are correct.

Topic: Biology, Molecular Basis of Inheritance, NEET
Eutrophication is often seen in

Options:
A deserts.
B fresh water lakes.
C ocean.
D mountains.

The correct answer is:
fresh water lakes.

Solution:
Eutrophication is often seen in fresh water lakes. The main causes of eutrophication in these water systems are sediments and thermal stratification, with agricultural run-off from fertilizers and manure, containing nitrates and phosphates and discharge of partially treated or untreated sewages or phosphate containing detergents.

Topic: Biology, Environmental Issues, NEET
Eutrophication of water bodies leading to killing of fishes is mainly due to non-availability of

Options:
A light.
B essential minerals.
C oxygen.
D food.

The correct answer is:
oxygen.

Solution:
During eutrophication of water bodies, the BOD level increases due to rapid growth of microbes. Eutrophication is the natural ageing of a lake by nutrient enrichment of its water. Depending on the climate, the size of the lake and other factors, the natural ageing of a lake may span thousands of years but it is accelerated due to human activties.

The rise in BOD levels due to increased organic matter decreases the oxygen levels beyond a critical concentration where the survival of aqutic species like species becomes difficult.

Topic: Biology, Environmental Issues, NEET, NEET 2015 (Phase 2)
Eutrophication of water bodies leading to killing of fishes is mainly due to non-availability of

Options:
A light.
B essential minerals.
C oxygen.
D food.

The correct answer is:
oxygen.

Solution:
During eutrophication of water bodies, BOD level increases due to rapid growth of microbes.

Topic: Biology, Environmental Issues, NEET
Evaluate \( \cot \left(\sum_{n=1}^{19} \cot ^{-1}\left(1+\sum_{p=1}^{n} 2 p\right)\right) \)

Options:
A \( \frac{21}{19} \)
B \( \frac{19}{21} \)
C \( \frac{23}{22} \)
D \( \frac{22}{23} \)

The correct answer is:
\( \frac{21}{19} \)

Solution:
cot⁡∑n=119cot-1⁡1+∑p=1n2p
=cot∑n=119tan-111+nn+1
=cot∑n=119tan-1n+1-n1+nn+1
=cot⁡∑n=119(tan-1n+1-tan-1n)
=cot⁡tan-120-tan-11
=1tan(tan-120-tan-11)
=120-11+(20).1=2119

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main