datavorous/entrance-exam-dataset huggingface dataset:
The slope of the tangent to the curve y=x3+x+54 which also passes through the origin is

Options:

The correct answer is:
28

Solution:
Let point Px1,y1 be the point on this curve such that the tangent of P passes through the origin.

Equation of tangent at P is y-y1=3x12+1x-x1
It passes through 0,0 so y1=3x12+1x1
⇒x13+x1+54=3x13+x1
⇒x13=27⇒x1=3
So, dydxx1,y1=3x12+1=39+1=28

Topic: Mathematics, Application of Derivatives, JEE Main
The slope of the tangent to the curve y=∫xx2cos-1t2dt at x=12 is equal to

Options:
A cos-114-π3
B cos-114+π3
C 2cos-114-π3
D 2cos-114+π3

The correct answer is:
2cos-114-π3

Solution:
dydx=cos-1x4.2x-cos-1x2.1
dydxx=12=cos-114.2-cos-112
=2cos-114-π3

Topic: Mathematics, Application of Derivatives, JEE Main
The slopes of isothermal and adiabatic curves are related as

Options:
A Isothermal curve slope = adiabatic curve slope
B Isothermal curve slope $=\gamma \times$ adiabatic curve slope
C Adiabatic curve slope $=\gamma \times$ isothermal curve slope
D Adiabatic curve slope $=\frac{1}{2} \times$ isothermal curve slope

The correct answer is:
Adiabatic curve slope $=\gamma \times$ isothermal curve slope

Solution:
For Isothermal process $P V=$ constant
$\Rightarrow\left(\frac{d P}{d V}\right)=\frac{-P}{V}=$ Slope of Isothermal curve
For adiabatic $\mathrm{PV}^\gamma=$ constant
$\Rightarrow \frac{d P}{d V}=\frac{-\gamma P}{V}=$ Slop of adiabatic curve slope
Clearly, $\left(\frac{d P}{d V}\right)_{\text {adiabatic }}=\gamma\left(\frac{d P}{d V}\right)_{\text {Isothermal }}$

Topic: Physics, Thermodynamics, JEE Main
The slopes of the tangents to the curve y=x+1x-3 at the points where it cuts the x-axis, are m1 and m2, then the value of m1+m2 is equal to

Options:
A 8
B -2
C 2
D 0

The correct answer is:
0

Solution:
The given curve is y=x2-2x-3, which cuts the x-axis at -1,0 and 3,0
Now, dydx=2x-2
Thus, m1=2-1-2=-4 & m2=23-2=4
∴m1+m2=0

Topic: Mathematics, Application of Derivatives, JEE Main
The slopes of two lines represented by \( x^{2}\left(\tan ^{2} \theta+\cos ^{2} \theta\right)-2 x y \tan \theta+y^{2} \sin ^{2} \theta=0 \) are \( m_{1} \) and \( m_{2} \), then \( \left(m_{1}-m_{2}\right) \) is equal to

Options:

The correct answer is:
2

Solution:
We have

x2tan2θ+cos2θ-2xytanθ+y2sin2θ=0

Then,

m1+m2=2tanθsin2θ

m1m2=tan2θ+cos2θsin2θ

⇒m1m2=sec2θ+cot2θ

Then,

m1-m22=4sin2θcos2θ-4sec2θ-4cot2θ

⇒m1-m22=4sec2θ1sin2θ-1-4cot2θ

⇒m1-m22=4sec2θcos2θsin2θ-4cot2θ

⇒m1-m22=4sin2θ-4cos2θsin2θ

⇒m1-m22=41-cos2θsin2θ

⇒m1-m22=4

⇒m1-m2=2

Topic: Mathematics, Straight Lines, JEE Main
The slow rate of decomposition of fallen logs in nature is due to their

Options:
A Low cellulose content
B Low moisture content
C Poor nitrogen content
D Anaerobic environment around them

The correct answer is:
Low moisture content

Solution:
Low moisture content will lead to decrease in number of decomposers, because decomposers require an optimum moisture for their proper growth and functioning.

Topic: Biology, Ecosystem, NEET, NEET 2008 (Mains)
The smallest and the largest values of
$\tan ^{-1}\left(\frac{1-x}{1+x}\right), 0 \leq x \le1$ are

Options:
A $0, \pi$
B $0, \frac{\pi}{4}$
C $-\frac{\pi}{4}, \frac{\pi}{4}$
D $\frac{\pi}{4}, \frac{\pi}{2}$

The correct answer is:
$0, \frac{\pi}{4}$

Solution:
We have,
$\tan ^{-1}\left(\frac{1-x}{1+x}\right)=\tan ^{-1} 1-\tan ^{-1} x=\frac{\pi}{4}-\tan ^{-1} x$
Since
$0 \leq x \leq 1 \Rightarrow 0 \leq \tan ^{-1} x \leq \frac{\pi}{4}$
$\begin{aligned}
& \Rightarrow 0 \geq-\tan ^{-1} x \geq \frac{-\pi}{4} \Rightarrow \frac{\pi}{4} \geq \frac{\pi}{4}-\tan ^{-1} x \geq 0 \\
& \Rightarrow \frac{\pi}{4} \geq \tan ^{-1}\left(\frac{1-x}{1+x}\right) \geq 0
\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The smallest angle of the triangle whose sides are $6+\sqrt{12}, \sqrt{48}, \sqrt{24}$ is

Options:
A $\frac{\pi}{4}$
B $\frac{\pi}{6}$
C $\frac{\pi}{3}$
D $\frac{\pi}{2}$

The correct answer is:
$\frac{\pi}{6}$

Solution:
Let $a=6+\sqrt{12}, b=\sqrt{48}, c=\sqrt{24}$
Here $\mathrm{c}$ is the smallest side. $\angle \mathrm{C}$ is the smallest angle of the triangle.
Now $\cos C=\frac{a^2+b^2-c^2}{2 a b}$
$=\frac{(48+24 \sqrt{3})+48-24}{4(3+\sqrt{3}) \cdot 4 \sqrt{3}}=\frac{\sqrt{3}}{2}$
so, $\quad \angle C=\pi / 6$

Topic: Mathematics, Properties of Triangles, JEE Main
The smallest bond angle is found in

Options:
A $I F_7$
B $\mathrm{CH}_4$
C $\mathrm{BeF}_2$
D $B F_3$

The correct answer is:
$I F_7$

Solution:
It is shows $s p^3 d^3$-hybridization. Hence the bond angle is about $72^{\circ}$.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The smallest natural number n , such that the coefficient of x in the expansion of x2+1x3n is C23 n , is

Options:
A 58
B 38
C 35
D 23

The correct answer is:
38

Solution:
In the expansion of x2+1x3n the general term is Tr+1=Cr nx2n-r1x3r

=Cr nx2n-2r-3r=Cr nx2n-5r

For coefficient of x, 2n-5r=1

⇒r=2n-15

So, we have the coefficient as C2n-15 n

Using, the given value and Crn=Cn-rn

⇒C2n-15 n=C23=Cn-23 n n

If 2n-15=23⇒n=58 and if 2n-15=n-23⇒n=38

Thus, the minimum value of 'n' is 38.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2019 (10 Apr Shift 2)
The smallest natural value of a for which the function fx=2x+1-a2-x+2a+1ln⁡2x-6 is increasing ∀x∈R, is

Options:

The correct answer is:
1

Solution:
f ′ x = 2 x+1 +a 2 −x + 2a+1 ln2
=ln⁡22-x2x+a2x+1+1

Since fx is an increasing function ∀x∈R
⇒f'x≥0⇒a>0
∴ The smallest natural value of a=1

Topic: Mathematics, Application of Derivatives, JEE Main
The smallest positive integer \( n \) for which \( \left(\frac{1+i}{1-i}\right)^{n}=1 \), is

Options:

The correct answer is:
4.00

Solution:
Given, 1+i1-in=1

⇒1+i1-i×1+i1+in=1

⇒1+i212-i2n=1

⇒12+i2+2i1-i2n=1

⇒2i2n=1

⇒ in=1
Hence, n is an integer multiple of 4.
So, the smallest positive integer value of n is 4.

Topic: Mathematics, Complex Number, JEE Main
The smallest positive integral value of a, such that the function fx=x4-4ax2+10 has more than two local extrema, is

Options:
A 1
B 2
C 4
D 16

The correct answer is:
1

Solution:
f â² x =4 x 3 â8ax
=4xx2-2a
=4xx-2ax+2a
Hence, fx has 3 local extrema if 2a is real i.e., 2a>0âa>0
â´ Value of minimum a=1

Topic: Mathematics, Application of Derivatives, JEE Main
The smallest possible natural number n, for which the equation x2-nx+2014=0 has integral roots, is

Options:

The correct answer is:
91

Solution:
In x2-nx+2014=0
Î±+Î²=n and Î±Î²=2014
Now,
Î±Î²=2014=1Ã2014 or -1Ã-2014
=2Ã1007 or -2Ã-1007
=19Ã106 or -19Ã-106
=38Ã53 or -38Ã-53
For least possible value, n=Î±+Î²
=38+53
=91

Topic: Mathematics, Quadratic Equation, JEE Main
The smallest value of $5 \cos \theta+12$ is

Options:
A 5
B 12
C 7
D 17

The correct answer is:
7

Solution:
$$
\begin{aligned}
& \text { Hints : } 5 \cos \theta+12,-1 \leq \cos \theta \leq 1 \\
& \Rightarrow-5 \leq 5 \cos \theta \leq 5 \\
& \therefore 5 \cos \theta+12 \geq-5+12 \Rightarrow 5 \cos \theta+12 \geq 7
\end{aligned}
$$

Topic: Mathematics, Functions, JEE Main
The smog is essentially caused by the presence of

Options:
A
$\mathrm{O}_2$ and $\mathrm{O}_3$

B
$\mathrm{O}_3$ and $\mathrm{N}_2$

C
Oxides of sulphur and nitrogen

D
$\mathrm{O}_2$ and $\mathrm{N}_2$

The correct answer is:
Oxides of sulphur and nitrogen

Solution:
Insulin

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2004
The sodium salt of an organic acid X produces effervescence with concentrated H2SO4. X reacts with the acidified aqueous CaCl2 solution to give a white precipitate which decolourises acidic solution of KMnO4. X is

Options:
A HCOONa
B CH3COONa
C Na2C2O4
D C6H5COONa

The correct answer is:
Na2C2O4

Solution:
Na2C2O4+H2SO4→Na2SO4+CO+CO2+H2O

Compound X is Sodium oxalate Na2C2O4.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2017 (02 Apr)
The soft-iron is a suitable material for making an electromagnet. This is because soft-iron has

Options:
A low coercivity and high retentivity.
B low coercivity and low permeability.
C high permeability and low retentivity.
D high permeability and high retentivity.

The correct answer is:
high permeability and low retentivity.

Solution:
Electromagnets are made of soft iron because of its high permeability and low retentivity. High permeability means that the core provides complete linkage of magnetic flux of the primary coil to the secondary coil. The determination of the magnetism left in the material after the removal of the magnetizing field is known as retentivity. The required condition for the operation of electromagnets is the fast reversal of polarity, so high retentivity is undesirable. So, soft iron core has high permeability and low retentivity.

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2022 (24 Jun Shift 2)
The sol given below with negatively charged colloidal particles is:

Options:
A AgNO3 added to KI solution
B KI added to AgNO3 solution
C Al2O3·xH2O in water
D FeCl3 added to hot water

The correct answer is:
AgNO3 added to KI solution

Solution:
1FeCl3 added to hot water,

(2) When AgNO3 is added to KI,

(3) When Kl is added to AgNO3,

4Al2O3·2H2O in water

AlOH3/Al3+ positively charged

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2021 (26 Aug Shift 2)
The soldiers of Napolean army while at Alps during freezing winter suffered a serious problem as regards to the tin buttons of their uniforms. White metallic tin buttons got converted to grey powder. This transformation is related to

Options:
A
an interaction with nitrogen of the air at very low temperatures

B
an interaction with water vapour contained in the humid air

C
a change in the partial pressure of oxygen in the air

D
a change in the crystalline structure of tin

The correct answer is:
a change in the crystalline structure of tin

Solution:
No Solution Available

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main, JEE Main 2004
The solid cylinder and a hollow cylinder, both of the same mass and same external diameter are released from the same height at the same time on a inclined plane. Both roll down without slipping. Which one will reach the bottom first?

Options:
A Both together only when angle of inclination of plane is $45^{\circ}$
B Both together
C Hollow cylinder
D Solid cylinder

The correct answer is:
Both together

Solution:
$\mathrm{t}=\frac{1}{\sin \theta} \sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}\left(1+\frac{\mathrm{k}^2}{\mathrm{R}^2}\right)}$
If a solid and hollow body of same shape are allowed to roll down on inclined plane then as $\left(\frac{\mathrm{k}^2}{\mathrm{R}^2}\right)_{\mathrm{S}} < \left(\frac{\mathrm{k}^2}{\mathrm{R}^2}\right)_{\mathrm{H}}$, solid body will reach the bottom first with greater velocity.

Topic: Physics, Rotational Motion, NEET, NEET 2010 (Mains)
The solid cylinder of length 80 cm and mass M has a radius of 20 cm. Calculate the density of the material used if the moment of inertia of the cylinder about an axis CD parallel to AB as shown in figure is 2.7 kg m2.

Options:
A 1.49×102 kg m-3
B 7.5×101 kg m-3
C 14.9 kg m-3
D 7.5×102 kg m-3

The correct answer is:
1.49×102 kg m-3

Solution:
I=Mr22+ML24=ρπr2Lr22+ρπr2LL24

=ρπr2Lr22+L24

⇒ρ=Iπr2Lr22+L24

On putting the values, we get ρ=1.49×102 kg m-3.

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2021 (26 Aug Shift 2)
The solid $\mathrm{NaCl}$ is a bad conductor of electricity since

Options:
A In solid there are no ions
B Solid $\mathrm{NaCl}$ is covalent
C In solid $\mathrm{NaCl}$ there is no velocity of ions
D In solid $\mathrm{NaCl}$ there are no electrons

The correct answer is:
In solid $\mathrm{NaCl}$ there is no velocity of ions

Solution:
Solid $\mathrm{NaCl}$ is a bad conductor of electricity because ions are not free to move.

Topic: Chemistry, Solid State, JEE Main
The solubilities of AgBr in water, in 0.01 M  CaBr 2 , 0.01 M KBr and in 0.05 M  AgNO 3 are S 1 ,S 2 ,S 3 and S 4 respectively. Then the order of solubilities is

Options:
A S 1  >  S 2  >  S 3  >  S 4
B S 1 â>â S 3 â>â S 2 â>â S 4
C S 2  >  S 1  >  S 3  >  S 4
D S 4  >  S 3  >  S 1  >  S 2

The correct answer is:
S 1  >  S 3  >  S 2  >  S 4

Solution:
Solubility of AgBr in 0.01âMâ CaBr 2 S 2 ......(i)
CaBr 2 is a strong electrolyte and dissociates completely to give common ion Br â
CaBr 2 0.01âM â Ca 2+ 0.01âM + 2Br â 0.02âM
AgBr(s)â Ag + (aq)+Br â (aq);ââ K sp
S 2 âââ0 .02+S 2 â S 2 Ã0.02= K sp
~0.02 or S 2 =50â K sp
Similarly, KBr gives common ion Br â
And AgNO 3 gives common ion Ag + and we can calculate
S 3 â=â 100 K sp ......(iii)
And S 4 =20  K sp ......(iv) respectively
From (i), (ii), (iii) and (iv)
S 1 â>â S 3 â>â S 2 â>â S 4
(Note: K sp â>â100â K sp âââµâ K sp order â 10 â12 )

Topic: Chemistry, Chemical Equilibrium, JEE Main
The solubilities of carbonates decrease down the magnesium group due to a decrease in

Options:
A
hydration energies of cations

B
inter-ionic attraction

C
entropy of solution formation

D
lattice energies of solids

The correct answer is:
hydration energies of cations

Solution:
The solubility is governed by $\Delta \mathrm{H}_{\text {solution }}$ i.e. $\Delta \mathrm{H}_{\text {solution }}=\Delta \mathrm{H}_{\text {lattice }}-\Delta \mathrm{H}_{\text {Hydration }}$.
Due to increase in size the magnitude of hydration energy decresaes and hence the solubility.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2003
The solubility (in mol $\mathrm{L}^{-1}$ ) of $\mathrm{AgCl}$ $\left(K_{\mathrm{sp}}=1.0 \times 10^{-10}\right)$ in a $0.1 \mathrm{M} \mathrm{KCl}$ solution will be

Options:
A
$1.0 \times 10^{-9}$

B
$1.0 \times 10^{-10}$

C
$1.0 \times 10^{-5}$

D
$1.0 \times 10^{-11}$

The correct answer is:
$1.0 \times 10^{-9}$

Solution:
Let solubility of $\mathrm{AgCl}=x \mathrm{~mole} / \mathrm{L}$
$$
\begin{aligned}
& \mathrm{AgCl} \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{Cl}^{-} \\
& \text {i.e., } K_{\mathrm{sp}(\mathrm{AgCl})}=x \times x \\
& \mathrm{KCl} \longrightarrow \mathrm{K}^{+}+\mathrm{Cl}^{-} \\
& {\left[\mathrm{Cl}^{-}\right] \text {from } \mathrm{KCl}=0.1 \mathrm{~m}} \\
& \text { Total }\left[\mathrm{Cl}{ }^{-}\right] \text {in solution }=x+0.1 \\
& K_{s p}(\mathrm{AgCl})=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]=x(x+0.1) \\
& 1.0 \times 10^{-10}=x(x+0.1) \\
& 1.0 \times 10^{-10}=x^2+0.1 x \\
& 1.0 \times 10^{-10}=0.1 x \quad\left(\text { as } x^2< < 1\right) \\
& x=1.0 \times 10^{-9} \mathrm{~mol} / \mathrm{L}
\end{aligned}
$$

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2012 (07 May Online)
The solubility in water of a sparingly soluble salt $\mathrm{AB}$ is $1.0 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}$. Its solubility product number will be

Options:
A
$4 \times 10^{-10}$

B
$1 \times 10^{-15}$

C
$1 \times 10^{-10}$

D
$4 \times 10^{-15}$

The correct answer is:
$4 \times 10^{-15}$

Solution:
$\begin{array}{ll}{[\mathrm{A}]=1.0 \times 10^{-5},} & {[\mathrm{~B}]=\left[1.0 \times 10^{-5}\right]} \\ \mathrm{K}_{\mathrm{sp}}=[2 . \mathrm{B}]^2[\mathrm{~A}] \quad=\left[2 \times 10^{-5}\right]^2\left[1.0 \times 10^{-5}\right]=4 \times 10^{-15}\end{array}$

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2003
The Solubility of AB2 is 0.05 g per 100 mL at 25 o  C. Calculate K sp of AB2 at 25 o  C? [Atomic mass of A = 20 amu, atomic mass of B = 40 amu ]

Options:
A 10-3
B 5×10-7
C 10-6
D 5×10-3

The correct answer is:
5×10-7

Solution:
=â0.05âg/100âmL
=â0.5âg/L
= 0.5 100 =5× 10 −3  mol/L
Ksp=4s3=4Ã5Ã10-33=5Ã10-7

Topic: Chemistry, Ionic Equilibrium, JEE Main
The solubility of a gas in water depends on

Options:
A Nature of the gas
B Temperature
C Pressure of the gas
D All of the above

The correct answer is:
All of the above

Solution:
Solubility of a gas depends on nature of gas, temperature and pressure of gas.

Topic: Chemistry, Solutions, JEE Main
The solubility of a sparingly soluble salt AOH2 (mol. wt. 192.3) is 19.23  g/litre at 300 K. The pH of its saturated solution assuming 80% ionisation at 300 K is :

Options:
A 1.0970
B 12.9030
C 13.2041
D 12.0000

The correct answer is:
13.2041

Solution:
S = 19.23 192.3 mol/litre = 0.1 M
A OH 2 â A 2 + + 2 OH - 2 S
â´ OH - = 0.1 Ã 2 Ã 8 0 1 0 0 = 0.16
â´ pOH = 0.7958
â´ pH = 13.2041

Topic: Chemistry, Ionic Equilibrium, JEE Main
The solubility of AgCl will be maximum in which of the following?

Options:
A 0.01 M HCl
B 0.01 M KCl
C Deionised water
D 0.01M AgNO3

The correct answer is:
Deionised water

Solution:
Due to the common ion effect of chloride, the solubility of silver chloride is less in HCl and KCl, compared to solubility in water. In deionized water no common ion effect will take place so maximum solubility

AgCl(s)⇌Ag+(aq)+Cl-(aq)

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2022 (29 Jun Shift 1)
The solubility of AgCN in a buffer solution of pH=3 is x. The value of x is: [Assume : No cyano complex is formed ; KspAgCN=2.2×10-16 and KaHCN=6.2×10-10]

Options:
A 1.9×10-5
B 2.2×10-16
C 0.625×10-6
D 1.6×10-6

The correct answer is:
1.9×10-5

Solution:
Let, the solubility be x

Given that,

KspAgCN=2.2×10-16 and KaHCN=6.2×10-10

AgCN⇌Ag+x+CN-x Ksp=2.2×10-16

H++CN-⇌HCN K=1 Ka=16.2×10-10

Ksp×1 Ka=Ag+CN-×[HCN]H+CN-

2.2×10-16×16.2×10-10=[S][S]10-3

S2=2.26.2×10-9

S2=3.55×10-10

S=3.55×10-10

S=1.88×10-5⇒1.9×10-5

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2021 (25 Feb Shift 1)
The solubility of $\mathrm{Agl}$ in $\mathrm{Nal}$ solution is less than that in pure water because

Options:
A Agl forms complex with $\mathrm{NaI}$
B Of common ion effect
C Solubility product of Aglis less than that of $\mathrm{NaI}$
D The temperature of the solution decreases

The correct answer is:
Of common ion effect

Solution:
Solubility is decreased due to common ion effect.

Topic: Chemistry, Ionic Equilibrium, JEE Main
The solubility of alkali metal salts in water is due to the fact that the cations get hydrated by water molecules.
What is the relative degree of hydration?

Options:
A Csaq+>Rbaq+>Kaq+>Naaq+>Liaq+
B Liaq+>Naaq+>Kaq+>Rbaq+>Csaq+
C Naaq+>Kaq+>Rbaq+>Csaq+>Liaq+
D Csaq+>Naaq+>Liaq+>Kaq+>Rbaq+

The correct answer is:
Liaq+>Naaq+>Kaq+>Rbaq+>Csaq+

Solution:
As hydration energy ∝ charge on cation ∝1size of cation
As on moving down size of ion increases so degree of hydration decreases hence the correct order is
Liaq+>Naaq+>Kaq+>Rbaq+>Csaq+

Topic: Chemistry, s Block Elements, JEE Main
The solubility of $\mathrm{BaSO}_4$ in water is $2.33 \times 10^{-3} \mathrm{gm} /$ litre. $\left.\mathrm{BaSO}_4=233\right)$

Options:
A $1 \times 10^{-5}$
B $1 \times 10^{-10}$
C $1 \times 10^{-15}$
D $1 \times 10^{-20}$

The correct answer is:
$1 \times 10^{-10}$

Solution:
The solubility of $\mathrm{BaSO}_4$ in $g /$ /litre is given $2.33 \times 10^{-3}$
$\because$ in mole/litre. $n=\frac{W}{m . w t}=1 \times 10^{-5}=\frac{2.33 \times 10^{-3}}{233}$
Because $\mathrm{BaSO}_4$ is a compound
$K_{s p}=S^2=\left[1 \times 10^{-5}\right]^2=1 \times 10^{-10}$

Topic: Chemistry, Ionic Equilibrium, JEE Main
The solubility of CaOH2 in water is :
[Given : The solubility product of CaOH2 in water =5.5×10-6]

Options:
A 1.77×10-2
B 1.77×10-6
C 1.11×10-2
D 1.11×10-6

The correct answer is:
1.11×10-2

Solution:
Given that,

The solubility product of CaOH2 in water =5.5×10-6

CaOH2⇌Ca2+aqs+2OH-aq2s

ksp=s2 s2⇒5.5×10-6=4 s3

4s3=5.5×10-6

s=5.541/3×10-2
s=1.11×10-2

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2021 (25 Feb Shift 2)
The solubility of $\mathrm{CaF}_2$ is s moles/litre. Then its solubility product is ________.

Options:
A $s^2$
B $4s^3$
C $3s^2$
D $s^3$

The correct answer is:
$4s^3$

Solution:
$\mathrm{CaF}_2 \rightleftharpoons \underset{(S)}{\mathrm{Ca}^{++}}+\underset{(2 . S)^2}{2 \mathrm{~F}^{-}} ; K_{s p}=4 S^3$

Topic: Chemistry, Ionic Equilibrium, JEE Main
The solubility of CdSO4 in water is 8.0×10-4 mol L-1. Its solubility in 0.01M H2SO4 solution is ___ ×10-6 mol L-1 (Round off to the Nearest integer) (Assume that solubility is much less than 0.01 M )

Options:

The correct answer is:
64

Solution:
In pure water,

Ksp=S2=8×10-42

=64×10-8

In 0.01 M H2SO4

H2SO4(aq)x→2H(aq)+x+SO42-(aq.)(x+0.01)

Ksp=x(x+0.01)

=64×10-8

x+0.01≅0.01 M

So, x(0.01)=64×10-8

x=64×10-6M

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2021 (18 Mar Shift 2)
The solubility of fluorides of alkali metals in water is

Options:
A LiF>NaF>KF>RbF>CsF
B CsF>RbF>KF>NaF>LiF
C LiF>KF>NaF>CsF>RbF
D CsF>KF>NaF>RbF>LiF

The correct answer is:
CsF>RbF>KF>NaF>LiF

Solution:
Lattice energy Î± 1r++r-
With a small anion as F-, the lattice energy decreases more rapidly along a series of salts with increasing cation size and thus solubility increases
LiF<NaF<KF<RbF<CsF

Topic: Chemistry, s Block Elements, JEE Main
The solubility of iodine in water increases in the presence of

Options:
A Alcohol
B Chloroform
C Sodium hydroxide
D Potassium iodide

The correct answer is:
Potassium iodide

Solution:
Iodine has the least affinity for water and is only slightly soluble in it. However , it dissolves in $10 \%$ aq. solution of $\mathrm{KI}$ due to the formation of a complex ion $\Gamma_3$ i.e.
$I_2+K I \rightleftharpoons K I_3$ or $I_2+I^- \rightleftharpoons I_3^-$ (complex ion)

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The solubility of \(\mathrm{BaSO}_4\), in water, is \(2.33 \times 10^{-3}\) gram/litre. Its solubility product will be (molecular weight of \(\mathrm{BaSO}_4=233\))

Options:
A \(1 \times 10^{-15}\)
B \(1 \times 10^{-10}\)
C \(1 \times 10^{-5}\)
D \(1 \times 10^{-20}\)

The correct answer is:
\(1 \times 10^{-10}\)

Solution:
Solubility of \(\mathrm{BaSO}_4\) in \(\mathrm{g} \mathrm{L}^{-1}=2.33 \times 10^{-3}\)
Solubility of \(\mathrm{BaSO}_4\) in \(\mathrm{mol} \mathrm{L}^{-1}\)
\(=\frac{2.33}{233} \times 10^{-3}=10^{-5}\)
\(\begin{array}{lcc}
\text {Now, } & \mathrm{BaSO}_4 \rightleftharpoons & \mathrm{Ba}^{2+}+ & \mathrm{SO}_4^{2-} \\
\text {Solubility } & & s & s
\end{array}\)
\(\begin{aligned}
\therefore \text { Solubility Product } & =\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{SO}_4{ }^{2-}\right]=s^2 \\
& =\left(10^{-5}\right)^2=10^{-10}
\end{aligned}\)

Topic: Chemistry, Ionic Equilibrium, JEE Main
The solubility of \( \mathrm{CdSO}_{4} \) in water is \( 8.0 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \). Its solubility in \( 0.01 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4} \) solution is \( \times 10^{-6} \mathrm{~mol} \mathrm{~L}^{-1} \)
(Round off to the Nearest integer) (Assume that solubility is much less than \( 0.01 \mathrm{M} \) )

Options:

The correct answer is:
64

Solution:
In pure water,

Ksp=S2=8×10-42

=64×10-8

In 0.01 M H2SO4

H2SO4(aq)x→2H(aq)+x+SO42-(aq.)(x+0.01)

Ksp=x(x+0.01)

=64×10-8

x+0.01≅0.01 M

So, x(0.01)=64×10-8

x=64×10-6M

Topic: Chemistry, Ionic Equilibrium, JEE Main
The solubility of metal halides depends on their nature, lattice enthalpy and hydration enthalpy of the individual ions, amongst fluorides of alkali metals, the lowest solubility of LiF in water is due to

Options:
A Ionic nature of lithium fluoride
B High lattice enthalpy
C High hydration enthalpy of lithium ion
D Low ionisation enthalpy of lithium atom

The correct answer is:
High lattice enthalpy

Solution:
The lowest solubility of LiF in water is due to strong packing of Li+ and F- ion, lattice energy of LIF is higher than its hydration energy.

Topic: Chemistry, s Block Elements, JEE Main
The solubility of $\mathrm{Mg}(\mathrm{OH})_2$ is $\mathrm{S}$ moles/litre. The solubility product under the same condition is

Options:
A
$4 \mathrm{~S}^3$

B
$3 S^4$

C
$4 \mathrm{~S}^2$

D
$S^3$

The correct answer is:
$4 \mathrm{~S}^3$

Solution:
No Solution Available

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2002
The solubility of N2 in water at 300âK and 500âtorr partial pressure 0 .01 g L â1 . The solubility (in g L â1 ) at 750âtorr partial pressure is:

Options:
A 0.0075
B 0.005
C 0.02
D 0.015

The correct answer is:
0.015

Solution:
P 2  =  K H X 2
where P 2 → partial presence of gas
X 2 â mole fraction of gas in solution
K H  = Henry's law constant
So, Pâ solubility
p 1 p 2 â=â s 1 s 2 âââ 500 0.01 â=â 750 x
â´âxâ=â0.015âg/L

Topic: Chemistry, Solutions, JEE Main