datavorous/entrance-exam-dataset huggingface dataset:
In optical communication systems operating at 1200 nm, only 2% of the source frequency is available for TV transmission having a bandwidth of 5 MHz. The number of TV channels that can be transmitted is 10n. The value of n is

Options:

The correct answer is:
6

Solution:
The frequency of optical communication is given by,
v=cλ or v=3×1081200×10-9
v=25×1013 Hz
But only 2% of the source frequency is available for TV transmission.
v′=2.5×1014 ×2%
v′=2.5×1014 ×2100
v′=5×1012 Hz
∴ Number of channels=v′bandwidth
No of channels=5×10125×106=106=1 million

Topic: Physics, Communication System, JEE Main
In order that the function $f(x)=(x+1)^{1 / x}$ is continuous at $x=0, f(0)$ must be defined as

Options:
A $f(0)=0$
B $f(0)=e$
C $f(0)=1 / e$
D $f(0)=1$

The correct answer is:
$f(0)=0$

Solution:
Here $\lim _{x \rightarrow 0^+} f(x)=k, \lim _{x \rightarrow 0^-} f(x)=-k$ and $f(0)=k$
But $f(x)$ is continuous at $x=0$, therefore $k$ must be zero.

Topic: Mathematics, Continuity and Differentiability, JEE Main
In order to determine the Young's Modulus of a wire of radius 0.2 cm (measured using a scale of least count =0.001 cm) and length 1 m (measured using a scale of least count =1 mm), a weight of mass 1 kg (measured using a scale of least count =1 g ) was hanged to get the elongation of 0.5 cm (measured using a scale of least count 0.001 cm). What will be the fractional error in the value of Young's Modulus determined by this experiment?

Options:
A 0.14%
B 0.9%
C 9%
D 1.4%

The correct answer is:
1.4%

Solution:
Y=StressStrain=FLAl=mgLπR2l

ΔYY=Δmm+ΔLL+2ΔRR+Δll

ΔYY×100=10011000+11000+20.0010.2+0.0010.5

=110+110+1+15=1410=1.4%

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2021 (16 Mar Shift 2)
In order to double the frequency of the fundamental note emitted by a stretched string, the length is reduced to $\frac{3}{4}$ th of the original length and the tension is changed. The factor by which the tension is to be changed is

Options:
A $\frac{3}{8}$
B $\frac{2}{3}$
C $\frac{8}{9}$
D $\frac{9}{4}$

The correct answer is:
$\frac{9}{4}$

Solution:
Frequency of stretched string.
$\mathrm{n}=\frac{1}{2 l} \frac{\sqrt{T}}{m}$
$\begin{aligned} & n \propto \frac{\sqrt{T}}{l} \\ & \frac{n_1}{n_2}=\frac{l_2}{I_1} \sqrt{\frac{T_1}{T_2}} \Rightarrow \frac{n}{2 n}=\frac{\frac{3}{4}}{l} \sqrt{\frac{T_1}{T_2}} \\ & \frac{1}{2}=\frac{3}{4} \sqrt{\frac{T_1}{T_2}} \Rightarrow \sqrt{\frac{T_1}{T_2}}=\frac{2}{3} \\ & \frac{T_1}{T_2}=\frac{4}{9} \Rightarrow T_2=\frac{9}{4} T_1\end{aligned}$
$\therefore$ Tension should be changed by the factor $\frac{9}{4}$.

Topic: Physics, Waves and Sound, JEE Main
In order to increase the volume of a gas by \( 10 \% \), the pressure of the gas should be

Options:
A increased by \( 10 \% \).
B increased by \( 1 \% \).
C decreased by \( 10 \% \).
D decreased by \( 1 \% \).

The correct answer is:
decreased by \( 10 \% \).

Solution:
According to Boyle's law,
p∝1V
Hence, in order to increase the volume of a gas by 10%, the pressure of the gas should be decreased by 10%.

Topic: Chemistry, States of Matter, JEE Main
In order to just complete the vertical circular motion, the ratio of kinetic energy of a particle at highest point to that at lowest point is

Options:
A 5
B 2
C 0.5
D 0.2

The correct answer is:
0.2

Solution:
K.E =12 mv2
At lowest point in vertical circular motion Vl=5rg and at highest point Vh=rg
â´ K.EhK.El=15=0.2

Topic: Physics, Motion In Two Dimensions, JEE Main
In order to oxidize a mixture of one mole of each of FeC2O4,Fe2(C2O4)3,FeSO4 and Fe2(SO4)3 in acidic medium, the number of moles of KMnO4 is:

Options:
A 1
B 2
C 3
D 1.5

The correct answer is:
2

Solution:
Eq.KMnO4⇒Eq.FeSO4+Eq.Fe2SO43+EqFeC2O4+Eq.Fe2C2O43
KMnO4⇒MnO4-
5e-+8H++MnO4-⇒Mn+2+4H2O V.F=5
So Eq.KMnO4⇒Eq.FeSO4+Eq.FeC2O4+Eq.Fe2C2O43
n×5=1×1+1×0+1×3+1×6
n×5=1+3+6
n=105=2 mole

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2019 (08 Apr Shift 1)
In order to prepare a buffer solution of pH5.74, sodium acetate is added to acetic acid. If the concentration of acetic acid in the buffer is 1.0 M, the concentration of sodium acetate in the buffer is ______M. (Round off to the Nearest Integer). Given: pKaacetic acid=4.74

Options:

The correct answer is:
10

Solution:
pH=pKa+log[CB][WA]
5.74=4.74+log[CB]1
⇒[CB]=10M

CB =Conjugate base.

WA =Weak acid

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2021 (18 Mar Shift 1)
In order to refine blister copper it is melted in a furnace and is stirred with green logs of wood. The purpose is

Options:
A To expel the dissolved gases in blister copper
B To bring the impurities to surface and oxidize them
C To increase the carbon content of copper
D To reduce the metallic oxide impurities with hydrocarbon gases liberated from the wood.

The correct answer is:
To reduce the metallic oxide impurities with hydrocarbon gases liberated from the wood.

Solution:
The blister copper is melted in reverberatory furnace and stirred with poles of green wood. Any oxide of copper formed is reduced to metallic copper by the reducing hydrocarbons of the wood. The process is called fire refining or poling.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
In organic reactions, sodium in liquid ammonia is used as

Options:
A reducing agent
B hydrating agent
C oxidising agent
D precipitating agent

The correct answer is:
reducing agent

Solution:
reducing agent

Topic: Chemistry, Redox Reactions, JEE Main
In orthorhombic unit cell the value of a, b and c are respectively 4.2 Ã,â8.6 Ã and 8.3 Ã . Given the molecular mass of the solute is 155 g mol-1 and density is 3.3 g/cc , the number of formula units per unit cell is (Report your answer by rounding up to nearest integer)

Options:
A 2
B 3
C 4
D 6

The correct answer is:
4

Solution:
Z= VÃ N A ÃÏ M
V = Volume of unit cell = aÃbÃc
=4.2 Ã 8.6 Ã 8.3 Ã 10-24Ã 6.023 Ã 1023 Ã 3.3155=3.84 â 4

Topic: Chemistry, Solid State, JEE Main
In P4O6 and P4O10 , the number of oxygen atoms bonded to each phosphorus atoms are respectively‐

Options:
A 3 and 3
B 4 and 4
C 3 and 4
D 4 and 3

The correct answer is:
3 and 4

Solution:
In P4O6 each phosphorus is linked to three oxygen atoms; whereas in P4O10 each phosphorus atom is linked to four oxygen atoms.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
In ÎPQR, the equation of the internal angle bisector of angle Q is y=x and the equation of side PR is 3x-y=2. If coordinates of P are 3,2 and 2PQ=RQ, then the coordinates of Q are

Options:
A 3,3
B 7,7
C -2,-2
D 5,5

The correct answer is:
7,7

Solution:
Point of intersection of y=x & 3x-y=2 is S1,1
Now, PS:SR=PR:RQ=1:2
Hence, using section formula R=-3,-1
Let, reflection of P about the line y=x is P'
âP'=2,3 which lies on the side QR
â equation of QR is 4x-5y+7=0
Point of intersection of QR & y=x is
7,7=Q

Topic: Mathematics, Straight Lines, JEE Main
In Periplaneta americana, the blood vascular system is of ____A____type. Blood vessels are____B____. The visceral organs are located in ____C____. ____C____is filled with a fluid which is composed of __D____ and____E___.

Options:
A

A	B	C	D	E

Closed	Absent	Enterocoel	Plasma	Haemocytes

B

A	B	C	D	E
Open	Poorly developed	Haemocoel	RBC	Haemolymph

C

A	B	C	D	E
Open	Absent	Haemocoel	Haemocytes	Lymph

D

A	B	C	D	E

Open	Poorly developed	Haemocoel	Plasma	Haemocytes

The correct answer is:

A	B	C	D	E

Open	Poorly developed	Haemocoel	Plasma	Haemocytes

Solution:
The circulatory system of cockroach: It is of open type. Blood vessels are poorly developed and open into haemocoel or open spaces. Body fluid present in haemocoel is also known as haemolymph. Blood is colourless as there is no respiratory pigment. Blood doesn't perform the function of transporting gases. The circulatory system consists of heart, aorta, haemocoel and haemolymph. Haemolymph is composed of colourless plasma and haemocytes. Cockroach possesses a 13- chambered tubular, muscular, elongated and neurogenic heart. Blood circulation in cockroach is maintained by wing-shaped alary muscles (involuntary). Each chamber of the heart receives blood from the sinus through 1 pair of slit-like openings called ostia.
The flow of blood can be represented as:
Blood in heart → Pumped into the sinuses → Re-enter the heart by ostia → Again pumped into the sinuses.

Topic: Biology, Structural Organisation in Animals, NEET
In phosphorous estimation, $0.31 \mathrm{~g}$ of an organic compound gave $0.444 \mathrm{~g}$ of magnesium pyrophosphate (molar mass $=222 \mathrm{~g} \mathrm{~mol}^{-1}$). The percentage of phosphorous in the compound is

Options:
A $40$
B $30$
C $60$
D $20$

The correct answer is:
$40$

Solution:
No solution. Refer to answer key.

Topic: Chemistry, General Organic Chemistry, JEE Main
In photoelectric effect, if the energy required to overcome the attractive forces on the electron, (work functions) of Li, Na and \( \mathrm{Rb} \) are \( 2.41 \mathrm{eV}, 2.30 \mathrm{eV} \) and \( 2.09 \mathrm{eV} \) respectively, the work function of ' \( \mathrm{K} \) ' could approximately be in eV

Options:
A \( 2.52 \)
B \( 2.20 \)
C \( 2.35 \)
D \( 2.01 \)

The correct answer is:
\( 2.20 \)

Solution:
As the size of atom (element) increases, energy required to overcome the attractive force on outermost electron decreases.
Li, Na, K and Rb are the members of same group and have the following order of size
Li<Na<K<Rb
Thus, the energy in case of K is intermediate of Rb and Na, i.e.,
2.30<K>2.09
or the energy for K is 2.20 eV.

Topic: Physics, Dual Nature of Matter, JEE Main
In photoelectric effect if the intensity of light is doubled, then maximum kinetic energy of photoelectrons will become

Options:
A double
B half
C four times
D no change

The correct answer is:
no change

Solution:
$K_{\max }$ of photoelectrons doesnot depend upon intensity of incident light.

Topic: Physics, Dual Nature of Matter, JEE Main
In photoelectric effect, on increasing the wavelength of incident light,

Options:
A the number of photoelectrons increases.
B the maximum kinetic energy of photoelectrons decreases.
C the number of photoelectrons decreases.
D the maximum kinetic energy of photoelectrons increases.

The correct answer is:
the maximum kinetic energy of photoelectrons decreases.

Solution:
The energy of the incident photon E=hcλwill decrease if we increase the wavelength λ of the incident light.

Here h is Planck constant and c is the speed of light in vacuum.

As a result, the energy released by the photoelectron reduces, but the number of emitted photoelectrons does not change.

According to Einstein's photoelectric equation, K.Emax=hcλ-ϕ0, here ϕ0 is the work function.

As seen from the above equation, kinetic energy and wavelength are inversely proportional to each other, Thus, with an increase in wavelength of the incident light, the kinetic energy of photoelectrons decreases.

Topic: Physics, Dual Nature of Matter, JEE Main
In photoelectric experiment energy of $2.48 \mathrm{eV}$ irradiates a photo sensitive material. The stopping potential was measured to be $0.5 \mathrm{~V}$. Work function of the photo sensitive material is :

Options:
A $1.68 \mathrm{eV}$
B $2.48 \mathrm{eV}$
C $1.98 \mathrm{eV}$
D $0.5 \mathrm{eV}$

The correct answer is:
$1.98 \mathrm{eV}$

Solution:
$\begin{aligned} & \mathrm{eV} \mathrm{V}_{\mathrm{s}}=\mathrm{h} \nu-\phi \\ & 0.5 \mathrm{~V}=2.48-\phi \\ & \text { work function }(\phi)=2.48 \mathrm{~V}-0.5 \mathrm{~V}=1.98 \mathrm{~V}\end{aligned}$

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2024 (06 Apr Shift 1)
In photosynthesis, the light-independent reactions take place at

Options:
A grana of chloroplast
B intergranal lamella of chloroplast
C stroma of chloroplast
D in the lumen of thylakoid

The correct answer is:
stroma of chloroplast

Solution:
The process of photosynthesis consists of two steps, light dependent step is light reaction which occurs in grana and the light independent step which is dark reaction and occurs in stroma of chloroplast. In Dark reaction fixation of carbondioxide and glucose formation mainly occurs by Calvin cycle.

Topic: Biology, Photosynthesis in Higher Plants, NEET
In photosynthesis, the light-independent reactions take place in

Options:
A Photosystem I
B Photosystem II
C Stroma
D Thylakoid lumen

The correct answer is:
Stroma

Solution:
The process of photosynthesis consists of two steps, light-dependent step is a light reaction that occurs in grana and the light-independent step which is a dark reaction and occurs in the stroma of the chloroplast. In Dark reaction fixation of carbondioxide and glucose formation mainly occurs by the Calvin cycle.

Topic: Biology, Photosynthesis in Higher Plants, NEET
In plant breeding programmes, the entire collection (of plants/seeds) having all the diverse alleles for all genes in a given crop is called

Options:
A selection of superior recombinants
B cross-hybridisation among the selected parents
C evaluation and selection of parents
D germplasm collection

The correct answer is:
germplasm collection

Solution:
It is called germplasm collection or the collection of variability. Selection of superior recombinants means selection of the best plant from the whole lot by visual examination and collecting their seeds for growing in field. Cross hybridisation is the method of combining. The characters of different plants together. The selection and evolution are the main steps of hybridisation.

Topic: Biology, Strategies for Enhancement in Food Production, NEET, NEET 2013 (All India)
In plants, mosaic formation and yellowing of plant parts are the symptoms of _________.

Options:
A bacterial diseases
B mycoplasmal diseases
C viral diseases
D fungal diseases

The correct answer is:
viral diseases

Solution:
A Virus is a submicroscopic infectious agent and they require a living host to complete its cycle so they are obligate parasites also. There are many symptoms that start appearing on the virus-infected plants such as some light or dark spots appear on the leaves in mosaic forms and leaves start to roll and curl. Due to chlorophyll degeneration, the plant parts start becoming yellow and veins also become translucent. The virus also affects the height of the plant so they remain dwarf and necrosis occurs (cells die) so all these symptoms appear when plants are infected with virus-like tobacco mosaic virus, potato mosaic virus, etc. They mostly affect plants of the Solanaceae family.

Topic: Biology, Strategies for Enhancement in Food Production, NEET
In Poiseuilli's method of determination of coefficient of viscosity, the physical quantity that requires greater accuracy in measurement is

Options:
A Pressure difference
B Volume of the liquid collected
C Length of the capillary tube
D Inner radius of the capillary tube

The correct answer is:
Inner radius of the capillary tube

Solution:
As Poiseuilli's equation contains $\mathrm{r}^4$ in it, radius of tube should be known more accurately.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
In polymer Buna-S:'S' stands for:

Options:
A Strength
B Sulphur
C Sulphonation
D Styrene

The correct answer is:
Styrene

Solution:
Buna--S, 'S' stand for styrene.

Buna – S is a more commonly used name for a polymer named styrene butadiene rubber. The structure of Buna – S is given by:

Buna – S is formed using the monomers butadiene and styrene.

Thus in Buna – S, the Bu stands for Butadiene, since butadiene is one of the constituent monomers of the given polymer. Also, Na stands for sodium and S represents the monomer Styrene used for the synthesis of the given polymer.

Topic: Chemistry, Polymers, JEE Main, JEE Main 2021 (24 Feb Shift 2)
In potassium ferrocyanide, there are ______ pairs of electrons in the t2g set of orbitals.

Options:

The correct answer is:
3

Solution:
Potassium ferrocyanide, K4FeCN6, has a central iron ion that is coordinated with six cyanide ligands in an octahedral geometry. The oxidation state of the iron ion in potassium ferrocyanide is +2, hence has d2sp3 hybridised state. On applying C.F.T. it shows that all the 6 electrons of d subshell are present in the form of three pairs in t2g orbitals. Hence answer is 3.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2023 (10 Apr Shift 1)
In presence of $\mathrm{AlCl}_3$, benzene and $n$-propyl bromide react in Friedal-Craft's reaction to form

Options:
A n-propyl benzene
B 1, 2-dinormal propyl benzene
C 1, 4-dinormal propyl benzene
D Isopropyl benzene

The correct answer is:
Isopropyl benzene

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
In presence of concentrated alkali OH-, trimethyl acetaldehyde undergoes the

Options:
A Aldol condensation
B Witting reaction
C Cannizzaro reaction
D Perkin reaction

The correct answer is:
Cannizzaro reaction

Solution:
Aldehydes lacking an α -hydrogen undergo self oxidation and reduction (disproportionation reaction), also known as Cannizzaro reaction.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
In presence of cyanide, azide and carbon monoxide, the rate of respiration

Options:
A Decreases
B Increases
C Remains the same
D None of the above

The correct answer is:
Decreases

Solution:
Cyanides, azides and carbon monoxide work as enzyme inhibitors, thus the rate of respiration decreases.

Topic: Biology, Respiration in Plants, NEET
In presence of light & heat toluene chlorinated & react with aqueous $\mathrm{NaOH}$ to give

Options:
A o-Cresol
B p-Cresol
C Mixture of o- Cresol & p-Cresol
D Benzoic acid

The correct answer is:
Benzoic acid

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
In presence of $\mathrm{NaOH}$, phenol react with $\mathrm{CHCl}_3$ to form o-hydroxy benzaldehyde. This reaction is called

Options:
A Riemer-Tiemann's reaction
B Sandmeyer's reaction
C Hoffmann's degradation reaction
D Gattermann's aldehyde synthesis

The correct answer is:
Riemer-Tiemann's reaction

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
In producing chlorine through electrolysis 100 watt power at $125 \mathrm{~V}$ is being consumed. How much chlorine per minute is liberated ? E. C. E. of chlorine is $0.367 \times 10^{-6} \mathrm{~kg} /$ coulomb:

Options:
A $13.6 \mathrm{mg}$
B $17.6 \mathrm{mg}$
C $21.3 \mathrm{mg}$
D $24.3 \mathrm{mg}$

The correct answer is:
$17.6 \mathrm{mg}$

Solution:
Mass of chlorine liberated $=2 \mathrm{It}$
$$
\begin{aligned}
& =0.367 \times 10^{-6} \times \frac{100}{125} \times 60 \\
& =0.0176 \times 10^{-3} \mathrm{~kg} \\
& =17.6 \mathrm{mg}
\end{aligned}
$$

Topic: Physics, Current Electricity, JEE Main
In pteridophytes, spores germinate to produce

Options:
A prothallus
B sporangium
C sporophyte
D zygote

The correct answer is:
prothallus

Solution:
In Pteridophyta, the main plant body is the diploid sporophyte. Spore is the first haploid structure produced after meiosis. It germinates to form a small green independent photosynthetic non-vascular plant body which is called prothallus.

Topic: Biology, Plant Kingdom, NEET
In qualitative analysis, the metals of group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains $\mathrm{Ag}^{+}$and $\mathrm{Pb}^{2+}$ at a concentration of $0.10 \mathrm{M}$. Aqueous $\mathrm{HCl}$ is added to this solution until the $\mathrm{Cl}^{-}$concentration is $0.10 \mathrm{M}$. What will be the concentration of $\mathrm{Ag}^{+}$and $\mathrm{Pb}^{2+}$ be at equilibrium? ( $K_{\mathrm{sp}}$ for $\mathrm{AgCl}=1.8 \times 10^{-10}, K_{\mathrm{sp}}$ for $\mathrm{PbCl}_2=1.7 \times 10^{-5}$ )

Options:
A $\left[\mathrm{Ag}^{+}\right]=1.8 \times 10^{-7} \mathrm{M} ;\left[\mathrm{Pb}^{2+}\right]=1.7 \times 10^{-6} \mathrm{M}$
B $\left[\mathrm{Ag}^{+}\right]=1.8 \times 10^{-11} \mathrm{M} ;\left[\mathrm{Pb}^{2+}\right]=8.5 \times 10^{-5} \mathrm{M}$
C $\left[\mathrm{Ag}^{+}\right]=1.8 \times 10^{-9} \mathrm{M} ;\left[\mathrm{Pb}^{2+}\right]=1.7 \times 10^{-3} \mathrm{M}$
D $\left[\mathrm{Ag}^{+}\right]=1.8 \times 10^{-11} \mathrm{M} ;\left[\mathrm{Pb}^{2+}\right]=8.5 \times 10^{-4} \mathrm{M}$

The correct answer is:
$\left[\mathrm{Ag}^{+}\right]=1.8 \times 10^{-9} \mathrm{M} ;\left[\mathrm{Pb}^{2+}\right]=1.7 \times 10^{-3} \mathrm{M}$

Solution:
$K_{\mathrm{sp}}$ for $\mathrm{AgCl}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]$
$$
\begin{aligned}
\therefore \quad\left[\mathrm{Ag}^{+}\right] & =\frac{1.8 \times 10^{-10}}{10^{-1}} \\
& =1.8 \times 10^{-9} \mathrm{M} . \\
K_{\mathrm{sp}} \text { for } \mathrm{PbCl}_2 & =\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{Cl}^{-}\right]^2 \\
\therefore \quad\left[\mathrm{Pb}^{2+}\right] & =\frac{1.7 \times 10^{-5}}{10^{-1} \times 10^{-1}} \\
& =1.7 \times 10^{-3} \mathrm{M}
\end{aligned}
$$

Topic: Chemistry, Ionic Equilibrium, NEET, NEET 2011 (Mains)
In qualitative analysis, the metals of group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains $\mathrm{Ag}^{+}$and $\mathrm{Pb}^{2+}$ at a concentration of $0.10 \mathrm{M}$. Aqueous $\mathrm{HCl}$ is added to this solution until the $\mathrm{Cl}^{-}$concentration is $0.10 \mathrm{M}$. What will be the concentration of $\mathrm{Ag}^{+}$and $\mathrm{Pb}^{2+}$ be at equilibrium? ( $K_{\mathrm{sp}}$ for $\mathrm{AgCl}=1.8 \times 10^{-10}, K_{\mathrm{sp}}$ for $\mathrm{PbCl}_2=1.7 \times 10^{-5}$ )

Options:
A $\left[\mathrm{Ag}^{+}\right]=1.8 \times 10^{-7} \mathrm{M} ;\left[\mathrm{Pb}^{2+}\right]=1.7 \times 10^{-6} \mathrm{M}$
B $\left[\mathrm{Ag}^{+}\right]=1.8 \times 10^{-11} \mathrm{M} ;\left[\mathrm{Pb}^{2+}\right]=8.5 \times 10^{-5} \mathrm{M}$
C $\left[\mathrm{Ag}^{+}\right]=1.8 \times 10^{-9} \mathrm{M} ;\left[\mathrm{Pb}^{2+}\right]=1.7 \times 10^{-3} \mathrm{M}$
D $\left[\mathrm{Ag}^{+}\right]=1.8 \times 10^{-11} \mathrm{M} ;\left[\mathrm{Pb}^{2+}\right]=8.5 \times 10^{-4} \mathrm{M}$

The correct answer is:
$\left[\mathrm{Ag}^{+}\right]=1.8 \times 10^{-9} \mathrm{M} ;\left[\mathrm{Pb}^{2+}\right]=1.7 \times 10^{-3} \mathrm{M}$

Solution:
$K_{\mathrm{sp}}$ for $\mathrm{AgCl}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]$
$$
\begin{aligned}
\therefore \quad\left[\mathrm{Ag}^{+}\right] & =\frac{1.8 \times 10^{-10}}{10^{-1}} \\
& =1.8 \times 10^{-9} \mathrm{M} . \\
K_{\mathrm{sp}} \text { for } \mathrm{PbCl}_2 & =\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{Cl}^{-}\right]^2 \\
\therefore \quad\left[\mathrm{Pb}^{2+}\right] & =\frac{1.7 \times 10^{-5}}{10^{-1} \times 10^{-1}} \\
& =1.7 \times 10^{-3} \mathrm{M}
\end{aligned}
$$

Topic: Chemistry, Ionic Equilibrium, JEE Main
In qualitative analysis, the metals of group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains Ag+ and Pb2+ at a concentration of 0.10 M. Aqueous HCl is added to this solution until the Cl- concentration is 0.10 M. What will the concentration of Ag+ and Pb2+ be at equilibrium?

( Ksp for AgCl =1.8×10-10,Ksp for PbCl2=1.7×10-5 )

Options:
A [Ag+]=1.8×10-7M,[Pb2+]=1.7×10-6M
B [Ag+]=1.8×10-11M,[Pb2+]=8.5×10-5M
C [Ag+]=1.8×10-9M,[Pb2+]=1.7×10-3M
D [Ag+]=1.8×10-11M,[Pb2+]=1.7×10-4M

The correct answer is:
[Ag+]=1.8×10-9M,[Pb2+]=1.7×10-3M

Solution:
Ksp[AgCl]=[Ag+][Cl-]
[Ag+]=1.8Ã10-1010-1=1.8Ã10-9M
K sp PbCl 2 = Pb 2+ Cl − 2
[Pb2+]=1.7Ã10-510-1Ã10-1=1.7Ã10-3M

Topic: Chemistry, Ionic Equilibrium, JEE Main
In qualitative test for identification of presence of phosphorous, the compound is heated with an oxidising agent. Which is further treated with nitric acid and ammonium molybdate respectively. The yellow coloured precipitate obtained is :

Options:
A \(\mathrm{Na}_3 \mathrm{PO}_4 \cdot 12 \mathrm{MoO}_3\)
B \(\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4 \cdot 12 \mathrm{MoO}_3\)
C \(\mathrm{MoPO}_4 .21 \mathrm{NH}_4 \mathrm{NO}_3\)
D \(\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4 \cdot 12\left(\mathrm{NH}_4\right)_2 \mathrm{MoO}_4\)

The correct answer is:
\(\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4 \cdot 12 \mathrm{MoO}_3\)

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (08 Apr Shift 2)