datavorous/entrance-exam-dataset huggingface dataset:
Paragraph:
Wave property of electrons implies that they will show diffraction effects. Davisson and Germer demonstrated this by diffracting electrons from crystals. The law governing the diffraction from a crystal is obtained by requiring that electron waves reflected from the planes of atoms in a crystal interfere constructively (see in figure).
Question:
Electrons accelerated by potential $V$ are diffracted from a crystal. If $d=1 Å$ and $i=30^{\circ}, V$ should be about $\left(\mathrm{h}=6.6 \times 10^{-34} \mathrm{Js}, \mathrm{m}_{\mathrm{e}}=9.1 \times 10^{-31} \mathrm{~kg}, \mathrm{e}=1.6 \times 10^{-19} \mathrm{C}\right)$

Options:
A
$2000 \mathrm{~V}$

B
$50 \mathrm{~V}$

C
$500 \mathrm{~V}$

D
$1000 \mathrm{~V}$

The correct answer is:
$50 \mathrm{~V}$

Solution:
$$
\begin{aligned}
& 2 d \cos i=n \lambda \\
& 2 d \cos i=\frac{h}{\sqrt{2 m e V}} \\
& v=50 \text { volt }
\end{aligned}
$$

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2008
Paragraph:
Wave property of electrons implies that they will show diffraction effects. Davisson and Germer demonstrated this by diffracting electrons from crystals. The law governing the diffraction from a crystal is obtained by requiring that electron waves reflected from the planes of atoms in a crystal interfere constructively (see in figure).
Question:
If a strong diffraction peak is observed when electrons are incident at an angle 'i' from the normal to the crystal planes with distance ' $d$ ' between them (see figure), de Broglie wavelength $\lambda_{\mathrm{dB}}$ of electrons can be calculated by the relationship ( $n$ is an integer)

Options:
A
$d \sin i=n \lambda_{d B}$

B
$2 d \cos i=n \lambda_{d B}$

C
$2 d \sin i=n \lambda_{d B}$

D
$d \cos i=n \lambda_{d B}$

The correct answer is:
$d \cos i=n \lambda_{d B}$

Solution:
$2 d \cos i=n \lambda_{d B}$

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2008
Paragraph:
When a metal rod $M$ is dipped into an aqueous colourless concentrated solution of compound $\mathrm{N}$, the solution turns light blue. Addition of aqueous $\mathrm{NaCl}$ to the blue solution gives a white precipitate $\mathrm{O}$. Addition of aqueous $\mathrm{NH}_3$ dissolves $\mathrm{O}$ and gives an intense blue solution.Question:
The metal $\operatorname{rod} M$ is

Options:
A
$\mathrm{Fe}$

B
$\mathrm{Cu}$

C
$\mathrm{Ni}$

D
$\mathrm{Co}$

The correct answer is:
$\mathrm{Cu}$

Solution:
Appearance of blue colour on addition of the metal rod $M$ to the solution of $N$ is an indication that metal may be $\mathrm{Cu}$. Later formation of white precipitate on addition of $\mathrm{NaCl}$ and dissolution of this precipitate in aqueous ammonia confirm that while preipitate is of $\mathrm{AgCl}$. This implies that the solution $N$ is of $\mathrm{AgNO}_3$. This confirm that the metal $M$ is only $\mathrm{Cu}$.

Topic: Chemistry, d and f Block Elements, JEE Main
Paramagnetic substances:
A. align themselves along the directions of external magnetic field.
B. attract strongly towards external magnetic field.
C. has susceptibility little more than zero.
D. move from a region of strong magnetic field to weak magnetic field.
Choose the most appropriate answer from the options given below:

Options:
A A, B, C Only
B $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}$
C A, C Only
D B, D Only

The correct answer is:
A, C Only

Solution:
A, C only

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2024 (08 Apr Shift 1)
Parathyroid hormone secretion increases in response to

Options:
A increased secretion of insulin
B a decrease in blood calcium levels
C increased secretion of parathyroid-releasing hormone from the hypothalamus.
D increased production of parathyroid-stimulating hormone from the anterior pituitary

The correct answer is:
a decrease in blood calcium levels

Solution:
Thyroid is the largest endocrine in our body. It is butterfly shaped and situated in the neck region. It consists of two lobes on either side of trachea that are connected by isthmus. Each lobe consists of many follicles that secrete thyroid hormones that control metabolism of the body. In between follicles are parafollicular cells that secrete calcitonin. On the posterior side of thyroid, two pairs of parathyroid glands are seen and these secrete PTH (Parathormone). Calcitonin and Parathormone maintain calcium metabolism. Calcitonin causes blood calcium level to decrease. This is done by deposition of calcium in bones, increasing absorption of calcium by kidneys etc. Parathormone is antagonist to calcitonin. Its main function is to increase calcium in plasma and thus decrease in plasma levels triggers its release.

Topic: Biology, Chemical Coordination and Integration, NEET
Parenchyma cells are concerned with

Options:
A Secretion
B Support
C Carry oxygen
D Storage of surplus food

The correct answer is:
Storage of surplus food

Solution:
Parenchyma is the most simple and unspecialized tissue which is concerned mainly with the vegetative activities of the plant. The main functions of parenchyma cells are:

(i) Storage of food materials, e.g., carrot, beetroot, etc.

(ii) Chlorenchyma helps in photosynthesis. Aerenchyma helps in the floating of the aquatic plants (hydrophytes) and also helps in gaseous exchange during respiration and photosynthesis, e.g., Hydrilla.

(iii) In the turgid state, they give rigidity to the plant organs.

(iv) In an emergency, they behave like meristematic cells and help in the healing of the various plant injuries.

(v) Sometimes they store secretory substances (ergastic substances) such as tannins, resins and gums; they are called idioblasts.

Topic: Biology, Anatomy of Flowering Plants, NEET
Parthenocarpic fruits are

Options:
A dry and indehiscent.
B fleshy and seedless.
C multi seeded.
D provided with haploid embryo.

The correct answer is:
fleshy and seedless.

Solution:
Fruits are mostly formed after fertilization, there are some species in which fruits develop without fertilization such fruits do not bear any seed like seedless grapes, banana, seedless watermelon, Seedless capsicum, seedless pears and some citrus varieties. These plants mainly propagate asexually. Seedless fruits have an increased proportion of edible part than in normal fruits

Topic: Biology, Morphology of Flowering Plants, NEET
Partial pressures of $A, B, C$ and $D$ on the basis of gaseous system $A+2 B \rightleftharpoons C+3 D$ are $\mathbf{A}=0.20 ; \mathbf{B}=0.10 ; \mathbf{C}$ $=0.30$ and $D=0.50 \mathrm{~atm}$. The numerical value of equilibrium constant is

Options:
A 11.25
B 18.75
C 5
D 3.75

The correct answer is:
18.75

Solution:
$A+2 B \rightleftharpoons C+3 D$
$K=\frac{[p C][p D]^3}{[p A][p B]^2}=\frac{0.30 \times 0.50 \times 0.50 \times 0.50}{0.20 \times 0.10 \times 0.10}=18.75$

Topic: Chemistry, Chemical Equilibrium, JEE Main
Particle $A$ moves along $X$ -axis with a uniform velocity of magnitude $10 \mathrm{m} / \mathrm{s}$. Particle $B$ moves with uniform velocity $20 \mathrm{m} / \mathrm{s}$ along a direction making an angle of $60^{\circ}$ with the positive direction of $X$ -axis as shown in the figure. The relative velocity of $B$ with respect to that of $A$ is

Options:
A $10 \mathrm{m} / \mathrm{s}$ along $X$ -axis
B $10 \sqrt{3} \mathrm{m} / \mathrm{s}$ along $Y$ -axis (perpendicular to $X$ -axis)
C $10 \sqrt{5} \mathrm{m} / \mathrm{s}$ along the bisection of the velocities of A and $B$
D $30 \mathrm{m} / \mathrm{s}$ along negative $X$ -axis

The correct answer is:
$10 \sqrt{3} \mathrm{m} / \mathrm{s}$ along $Y$ -axis (perpendicular to $X$ -axis)

Solution:
$$
\begin{aligned}
&\text { The component of velocity of } B \text { along } x \text { -direction }\\
&\begin{aligned}
\mathbf{v}_{B x} &=20 \cos 60^{\circ}=20 \times \frac{1}{2}=10 \mathrm{m} / \mathrm{s} \\
\mathbf{v}_{A} &=10 \hat{\mathbf{i}} \\
\mathbf{v}_{B} &=10 \hat{\mathbf{i}}+10 \sqrt{3} \hat{\mathbf{j}} \\
\mathbf{v}_{B A} &=\mathbf{v}_{B}-\mathbf{v}_{A}=10 \hat{\mathbf{i}}+10 \sqrt{3} \hat{\mathbf{j}}-10 \hat{\mathbf{i}}=10 \sqrt{3} \hat{\mathbf{j}}
\end{aligned}
\end{aligned}
$$

Topic: Physics, Motion In Two Dimensions, JEE Main
Particle A of mass m1 moving with velocity 3i^+j^ms−1 collides with another particle B of mass m2 which is at rest initially. Let v→1 and v→2 be the velocities of particles A and B after collision respectively. If m1=2m2 and after collision v→1−i^+3j^ms−1, the angle between v→1 and v→2 is :

Options:
A 15°
B 60°
C −45°
D 105°

The correct answer is:
105°

Solution:
m1u→1+m2u→2=m1v→1+m2v→2

2m2(3i^+j^)+m2×0=2m2(i^+3j^)+m2×v→2

23i^+2j^=2i^+23j^+v→2

v→2=(3-1)i^-(3-1)j^

v→1=i^+3j^

Angle between v→1 and v→2 is 105°

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2020 (06 Sep Shift 2)
$\beta$ - particle is emitted in radioactivity by

Options:
A
conversion of proton to neutron

B
from outermost orbit

C
conversion of neutron to proton

D
$\beta$-particle is not emitted

The correct answer is:
conversion of neutron to proton

Solution:
${ }_0 n^1 \rightarrow_{+1} p^1+_{-1} e^0$

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2002
particle is moving in an elliptical orbit as shown in figure. If $\vec{p}, \vec{L}$ and $\vec{r}$ denote the linear momentum, angular momentum and position vector of the particle (from focus $\mathrm{O}$ ) respectively at a point $\mathrm{A}$, then the direction of $\vec{\alpha}=\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{L}}$ is along

Options:
A + ve $x$ axis
B - ve $x$ axis
C + ve y axis
D - ve y axis

The correct answer is:
+ ve $x$ axis

Solution:
$\vec{\alpha}=\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{L}}$
$\vec{L}$ is along Z-axis

Topic: Physics, Rotational Motion, JEE Main
Particles of masses $2 M, m$ and $M$ are respectively at points $A, B$ and $C$ with $A B=\frac{1}{2}(B C) \cdot m$ is much-much smaller than $M$ and at time $t=0$, they are all at rest as given in figure. At subsequent times before any collision takes place.

Options:
A
$m$ will remain at rest

B
$m$ will move towards $M$

C
$m$ will move towards $2 M$

D
$m$ will have oscillatory motion

The correct answer is:
$m$ will move towards $2 M$

Solution:
Force of Gravitation, $F_g=\frac{G M m}{r^2}$
Let $A B=r$
So, force on $B$ due to $A$
$$
\begin{aligned}
&=F_{B A}=\frac{G(2 M m)}{(A B)^2} \text { towards } B A . \\
&=\frac{G(2 M m)}{r^2}=2 F_g
\end{aligned}
$$
and force on $B$ due to $C$
$$
\begin{aligned}
&=F_{B C}=\frac{G M m}{(B C)^2} \text { towards } B C \\
&\text { As, }(B C)=2 A B \\
&\Rightarrow F_{B C}=\frac{G M m}{(2 A B)^2}=\frac{G M m}{4(A B)^2}=\frac{G M m}{4 r^2}=\frac{F_g}{4}
\end{aligned}
$$
As $F_{B A}>F_{B C}$,
hence, $m$ will move towards $(B A)$ i.e., $(2 M)$.

Topic: Physics, Gravitation
Particles of masses m, 2m, 3m,â¦.,nm are placed on the same line at distance L, 2L, 3L,..nL from O. The distance of centre of mass from O is

Options:
A 2n+1 L4
B L2n+1
C nn2+1 L2
D 2n+1L3

The correct answer is:
2n+1L3

Solution:
L
âµ xCM=mL+2mÃ2L+â¦+nmÃnLm+2m+â¦+nm
=L12+22+â¦+n21+2+â¦+n
=Lân2ân=Lnn+1 2n+16nn+12
=2n+1L3

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Particles of masses m, 2m, 3m,â¦., nm are placed on the same line at distance L, 2L, 3L,..nL from O. The distance of the centre of mass from O is

Options:
A 2n+1 L4
B L2n+1
C nn2+1 L2
D 2n+1L3

The correct answer is:
2n+1L3

Solution:
L
âµ xCM=mL+2mÃ2L+â¦+nmÃnLm+2m+â¦+nm
=L12+22+â¦+n21+2+â¦+n
=Lân2ân=Lnn+1 2n+16nn+12
=2n+1L3

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Particles of masses $m, 2 m, 3 m, \ldots n m$ grams are placed on the same line at distance $l, 2 l, 3 l \ldots n l$ $\mathrm{cm}$ from a fixed point. The distance of centre of mass of the particles from the fixed point in centimetre is

Options:
A $\frac{(2 n+1) l}{3}$
B $\frac{l}{n+1}$
C $\frac{n\left(n^2+1\right) l}{2}$
D $\frac{2 l}{n\left(n^2+1\right)}$

The correct answer is:
$\frac{(2 n+1) l}{3}$

Solution:
Distance of centre of mass
$$
\begin{aligned}
x_{c m} & =\frac{m_1 x_1+m_2 x_2+\ldots m_n x_n}{m_1+m_2+\ldots m_n} \\
& =\frac{m l+2 m 2 l+\ldots n m \times n l}{m+2 m+3 m+\ldots n m} \\
& =\frac{m l\left(1+4+9+\ldots n^2\right)}{m(1+2+3+\ldots n)} \\
1+4+9 & +\ldots n^2=\Sigma n^2 \\
& =\frac{n(n+1)(2 n+1)}{6} \\
1+2+3+ & \ldots n=\Sigma n=\frac{n(n+1)}{2} \\
\therefore \quad x_{c m} & =\frac{\ln (n+1)(2 n+1)}{6}=\frac{l(2 n+1)}{3}
\end{aligned}
$$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Particulate matters in air can be removed most efficiently by the use of

Options:
A catalytic converters in exhaust of automobiles.
B scrubbers.
C electrostatic precipitators.
D Euro-III emission norms.

The correct answer is:
electrostatic precipitators.

Solution:
The most widely used of which is the electrostatic precipitator, which can remove over 99 percent particulate matter present in the exhaust from a thermal power plant. It has electrode wires that are maintained at several thousand volts, which produce a corona that releases electrons. These electrons attach to dust particles giving them a net negative charge. The collecting plates are grounded and attract the charged dust particles. The velocity of air between the plates must be low enough to allow the dust to fall.

Topic: Biology, Environmental Issues, NEET
Passing H2S in to a mixture of Mn2+,Ni2+,Cu2+ and Hg2+ ions in an acidified aqueous solution precipitates

Options:
A MnS and CuS
B CuS and HgS
C NiS ad HgS
D MnS and NiS

The correct answer is:
CuS and HgS

Solution:
H2S in presence of aqueous solution precipitates sulphide of Cu and Hg (IInd group) apart from Pb+2,Bi+3,Cd+2,As+3,Sb+3 and Sn+2, Mn2+,Ni2+ are of IVth group so cannot be precipitated by H2S .

Topic: Chemistry, Practical Chemistry, JEE Main
Pb and Sn are extracted from their chief ore respectively by

Options:
A Carbon reduction and self reduction
B Self reduction and carbon reduction
C Electrolysis and self reduction.
D Self reduction and electrolysis.

The correct answer is:
Self reduction and carbon reduction

Solution:
PbO and PbSO 4 get reduced by PbS itself which is already present in mixture so because the reduction took place by mixture itself, hence is known as self reduction.
2PbO+PbS â Î 3Pb+SO 2 â
PbS O 4 â+âPbS â Î â2Pbâ+2S O 2 â
SnO 2 + C→Sn + CO 2

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
$\mathrm{PbCl}_{2}$ is insoluble in cold water. Addition of $\mathrm{HCl}$ increases its solubility due to

Options:
A formation of soluble complex anions like $\left[\mathrm{PbCl}_{3}\right]^{-}$
B oxidation of Pb(II) to PB(iV)
C formation of $\left[\mathrm{Pb}(\mathrm{H}_{2}\mathrm{O})_{6}\right]^{2}$
D formation of polymeric lead complexes

The correct answer is:
formation of soluble complex anions like $\left[\mathrm{PbCl}_{3}\right]^{-}$

Solution:
$\mathrm{PbCl}_{2}$ react with HCl as follows $\mathrm{PbCl}_{2}(s)+\mathrm{Cl}^{-} \stackrel{\text { Cold }}{\longrightarrow}\left[\mathrm{PbCl}_{3}\right]^{-}(a q)$
$\mathrm{PbCl}_{2}(s)+2 \mathrm{Cl}^{-} \frac{\text { Excess }}{\text { of HCl }}\left[\mathrm{PbCl}_{4}\right]^{2-}(a q)$
Thus, addition of excess of $\mathrm{Cl}^{-}$ ions change the $\mathrm{PbCl}_{2}$ as soluble complex of $\left[\mathrm{PbCl}_{4}\right]^{-2}$. Hence, becomes soluble.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
PbO2 is obtained from

Options:
A the reaction of PbO with HCl

B thermal decomposition of PbNO32 at 200°C

C the reaction of Pb3O4 with HNO3

D the reaction of Pb with air at room temperature

The correct answer is:
the reaction of Pb3O4 with HNO3

Solution:
(a) PbO+HCl⟶PbCl2+H2O
(not correct option)
(b) 2 PbNO32⟶2PbO+4NO2+O2
(not correct option)
(c) Pb3O4+4HNO3⟶2 PbNO32
+PbO2+2H2O (correct option)
(d) Pb+ air (contains O2,H2O and CO2 ) room
temperature
Protective layer of varying cornposition, mainly PbCO3 is formed only on the surface. (not correct option)

Topic: Chemistry, p Block Elements (Group 13 & 14), KVPY, KVPY 2019 (SA)
$\mathrm{PCl}_3$ reacts with water to form

Options:
A $\mathrm{PH}_3$
B $\mathrm{H}_3 \mathrm{PO}_3, \mathrm{HCl}$
C $\mathrm{POCl}_3$
D $\mathrm{H}_3 \mathrm{PO}_4$

The correct answer is:
$\mathrm{H}_3 \mathrm{PO}_3, \mathrm{HCl}$

Solution:
$\mathrm{PCl}_3+3 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_3 \mathrm{PO}_3+3 \mathrm{HCl}$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
PCl5 causes cleavage of ether linkage R - O - R ′ forming RCl, R'Cl and POCl3.

C5H12O on reaction with PCl5 forms 2-chloropropane and 1-chloroethane as main compounds.

Thus, C5H12O is named as

Options:
A 1-ethoxypropane
B 2-ethoxypropane
C 1-ethylpropane
D 2-ethylpropane

The correct answer is:
2-ethoxypropane

Solution:
R - O - R ′ → PCl 5 R - Cl + R ′ Cl + POCl 3
O is replaced by two Cl-atoms one each on alkly group.
Thus, C5H12O is

Topic: Chemistry, General Organic Chemistry, NEET
PCl5 dissociates as PCl5g⇌PCl3g+Cl2g 5 moles of PCl5 are placed in a 200 litre vessel which contains 2 moles of N2 and is maintained at 600 K. The equilibrium pressure is 2.46 atm. The equilibrium constant Kp for the dissociation of PCl5 is___×10-3. (nearest integer) (Given: R=0.082 L atm K-1 mol-1; Assume ideal gas behaviour)

Options:

The correct answer is:
1107

Solution:
PCl5 ⇌ PCl3+Cl25mole - - Intial moles5-x x x EquilibriumTotal number of moles=5+x+nN2=7+xAs number of moles of Nitrogen gas added is 2ntotal=PVRT=2.46X 2000.082X600=107+x=10x=3Kp=310x2.46310x2.46210x2.46=1107

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2022 (24 Jun Shift 2)
PCl5g→PCl3g+Cl2g

In the above first order reaction the concentration of PCl5 reduces from initial concentration 50 mol L-1 to 10 mol L-1 in 120 minutes at 300 K. The rate constant for the reaction at 300 K is x×10-2 min-1. The value of x is __________.

[Given log5=0.6989]

Options:

The correct answer is:
1

Solution:
PCl5gI order 300 KPCl3g+Cl2g

t=0, [PCl5]= 50Mt=120 min, [PCl5]= 10M

t=120 min10M

⇒K=2.303tlogA0At

⇒K=2.303120log5010

⇒K=2.303120×0.6989=0.013413 min-1

=1.3413×10-2 min-1

1.34⇒ Nearest integer =1

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2021 (20 Jul Shift 2)
PCl5+H2O→A+HCl

A+H2O→B+HCl

Find number of ionisable hydrogen in B

Options:

The correct answer is:
3

Solution:
PCl5 is a yellowish white powder and in moist air, it hydrolyses into POCl3 and finally gets converted to phosphoric acid.

PCl5+H2O→POCl3+2HCl

POCl3+3H2O→H3PO4+3HCl

Phosphoric acid has 3 ionisable hydrogen atoms.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2022 (25 Jun Shift 2)
PCl5 is well known but NCl5 is not.Because,

Options:
A N - does not have vacant d-orbital
B P - does not have 2d orbitals
C Back bonding in NCl5 is not possible
D N atom is more electronegative so does not forms 5 bonds

The correct answer is:
N - does not have vacant d-orbital

Solution:
Due to the absence of vacant d orbitals nitrogen cannot form penta halides, so NCl5 is not formed but PCl5 is formed as phosphorous has vacant d orbitals

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2022 (24 Jun Shift 2)
PCl5⇌PCl3+Cl2 Kc=1.844
3.0 moles of PCl5 is introduced in a 1L closed reaction vessel at 380 K. The number of moles of PCl5 at equilibrium is —×10-3
(Round off to the Nearest Integer)

Options:

The correct answer is:
1400

Solution:
PCl5( g)⇌PCl3( g)+Cl2( g) Kc=1.844t=0 3 0 0t=teq 3-x x x

⇒PCl3Cl2PCl5=x23-x=1.844
⇒x2+1.844-5.532=0
⇒x=-1.844+(1.844)2+4×5.5322
≡1.60
⇒ Moles of PCl5=3-1.60=1.40

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2021 (27 Jul Shift 1)
$\mathrm{PCl}_5 \rightleftharpoons \mathrm{PCl}_3+\mathrm{PCl}_2$
If the equilibrium constant $\left(K_C\right)$ for the above reaction at $500 \mathrm{~K}$ is 1.79 and the equilibrium concentration of $\mathrm{PCl}_5$ and $\mathrm{PCl}_3$ are $1.41 \mathrm{M}$ and $1.59 \mathrm{M}$, respectively, then the concentration of $\mathrm{Cl}_2$ is approximately.

Options:
A $1.26 \mathrm{M}$
B $3.59 \mathrm{M}$
C $0.59 \mathrm{M}$
D $1.59 \mathrm{M}$

The correct answer is:
$1.59 \mathrm{M}$

Solution:
Equilibrium constant $\left(K_C\right)=\frac{\left[\mathrm{PCI}_3\right]\left[\mathrm{CI}_2\right]}{\left[\mathrm{PCI}_5\right]}$
$\begin{aligned}
1.79 & =\frac{1.59 \times\left[\mathrm{CI}_2\right]}{1.41} \\
{\left[\mathrm{CI}_2\right] } & =\frac{1.79 \times 1.41}{1.59}=1.587 \approx 1.59 \mathrm{~m}
\end{aligned}$

Topic: Chemistry, Chemical Equilibrium, JEE Main
PCl 5 vapour decomposes on heating according to the reaction:
PCl 5 g â PCl 3 g + Cl 2 g
The density of a sample of a partially dissociated PCl 5 at 1.0 atm and 500âK was found 4.8âg/L. Calculate the degree of dissociation and Î G â for the reaction at 500âK.
(Given: R=0.082â LatmK â1 mol â1 , R=8.314â JK â1 mol â1 , lnx=2.303 log 10 x ) If your answer is K, mark 2K as answer after rounding off to nearest integer.

Options:

The correct answer is:
47

Solution:
M avg = Ï RT P = 4.8 Ã 0.0 8 2 Ã 5 0 0 1 = 1 9 6.8
PCl 5 g â PCl 3 g + Cl 2 g 1 - Î± Î± Î± Total moles = 1 + Î±
M avg = M PCl 5 1 + Î± = 1 9 6.8 = 2 0 8.5 1 + Î±
â´ Î± = 0.0 6
K P = Î± 2 1â Î± 2 P= 0.06 2 1â 0.06 2 =3.6Ã 10 â3
Î G â = - RT ln K
= - 8.3 1 4 Ã 5 0 0 Ã 2.303 log 3.6 Ã 1 0 - 3
=23.4âkJ

Topic: Chemistry, Chemical Equilibrium, JEE Main
PCl 5 vapour decomposes on heating according to the reaction:

PCl 5 g â PCl 3 g + Cl 2 g

The density of a sample of a partially dissociated PCl 5 at 1.0 atm and 500âK was found 4.8âg/L. Calculate the degree of dissociation and Î G â for the reaction at 500âK.
(Given: R=0.082â LatmK â1 mol â1 , R=8.314â JK â1 mol â1 , lnx=2.303 log 10 x )

Options:

The correct answer is:
23.40

Solution:
M avg = Ï RT P = 4.8 Ã 0.0 8 2 Ã 5 0 0 1 = 1 9 6.8
PCl 5 g â PCl 3 g + Cl 2 g 1 - Î± Î± Î± Total moles = 1 + Î±
M avg = M PCl 5 1 + Î± = 1 9 6.8 = 2 0 8.5 1 + Î±
â´ Î± = 0.0 6
K P = Î± 2 1â Î± 2 P= 0.06 2 1â 0.06 2 =3.6Ã 10 â3
Î G â = - RT ln K
= - 8.3 1 4 Ã 5 0 0 Ã 2.303 log 3.6 Ã 1 0 - 3
=23.4âkJ

Topic: Chemistry, Chemical Equilibrium, JEE Main
p-cresol reacts with chloroform in alkaline medium to give the compound A which adds hydrogen cyanide to form, the compound B. The latter on acidic hydrolysis gives chiral carboxylic acid. The structure of the carboxylic acid is

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
p-cresol reacts with chloroform in alkaline medium to give the compound A which adds hydrogen cyanide to form, the compound B. The latter on acidic hydrolysis gives chiral carboxylic acid. The structure of the carboxylic acid is

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2005
Pentan-3-one is not obtained from

Options:
A 2, 2-dichloro pentane
B 3 , 3 -dichloro pentane
C pentan-3-ol
D pent-2-yne

The correct answer is:
2, 2-dichloro pentane

Solution:
2, 2-dichloropentane gives pentan-2-one, not pentan-3-one as

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Peptides are

Options:
A Esters
B Salts
C Amides
D Ketones

The correct answer is:
Amides

Solution:
Peptides are amides. A peptide is a short chain of amino acids joined by peptide bonds, which are a type of amide covalent chemical bond.

Topic: Chemistry, Biomolecules, JEE Main
Peptization is a:

Options:
A Process of converting a colloidal solution into precipitate
B Process of converting soluble particles to form colloidal solution
C Process of bringing colloidal molecule into solution
D Process of converting precipitate into colloidal solution

The correct answer is:
Process of converting precipitate into colloidal solution

Solution:
The process of converting a precipitate into colloidal solution by shaking it with dispersion medium in the presence of small amount of electrolyte. The electrolyte used is called peptizing agent.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2019 (12 Apr Shift 1)