datavorous/entrance-exam-dataset huggingface dataset:
The internal energy change when a system goes from state $\mathrm{A}$ to $\mathrm{B}$ is $40 \mathrm{~kJ} / \mathrm{mole}$. If the system goes from $\mathrm{A}$ to $\mathrm{B}$ by a reversible path and retums to state $\mathrm{A}$ by an irreversible path what would be the net change in internalenergy?

Options:
A
$>40 \mathrm{~kJ}$

B
$ < 40 \mathrm{~kJ}$
C
Zero

D
$40 \mathrm{~kJ}$

The correct answer is:
Zero

Solution:
For a cyclic process the net change in the internal energy is zero because the change in internal energy does not depend on the path.

Topic: Chemistry, Thermodynamics (C), JEE Main
The internal energy change when a system goes from state $\mathrm{A}$ to $\mathrm{B}$ is $40 \mathrm{~kJ} / \mathrm{mole}$. If the system goes from $\mathrm{A}$ to $\mathrm{B}$ by a reversible path and retums to state $\mathrm{A}$ by an irreversible path what would be the net change in internalenergy?

Options:
A
$>40 \mathrm{~kJ}$

B
$ < 40 \mathrm{~kJ}$
C
Zero

D
$40 \mathrm{~kJ}$

The correct answer is:
Zero

Solution:
For a cyclic process the net change in the internal energy is zero because the change in internal energy does not depend on the path.

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2003
The internal energy of a substance

Options:
A decrease with increase in temperature
B increase with increase in temperature
C remains unaffected with change in temperature
D calculated by $E=m c^2$

The correct answer is:
increase with increase in temperature

Solution:
The internal energy of a substance always increases with increase in temperature.

Topic: Chemistry, Thermodynamics (C), JIPMER, JIPMER 2005
The internal energy of n1 mol of hydrogen of temperature T is equal to the internal energy of n2 mol of helium at temperature 2T. What will be the value of the ratio n1n2?

Options:

The correct answer is:
1.2

Solution:
The internal energy of n moles of an ideal gas at temperature T is given by
U=f⁡2nRT ( f⁡ = degrees of freedom)
U1=U2
f⁡1n1T1=f⁡2n2T2
∴ n1n2=f⁡2T2f⁡1T1=3×25×1=65=1.2
Here, f⁡ 2 = degrees of freedom of He = 3
and, f⁡ 1 = degrees of freedom of H2 = 5

Topic: Physics, Thermodynamics, JEE Main
The internal energy of the air, in a room of volume V, at temperature T and with outside pressure P increasing linearly with time, varies as

Options:
A increases linearly
B increases exponentially
C decreases linearly
D remains constant

The correct answer is:
increases linearly

Solution:
As per first law of Thermodynamics,

dQ=dU+dW

here, dW=0 since V=constant,

∴ dQ=dU=nCvdT⇒dU α dT ...1

Now pV=RT⇒dT=VRdP⇒dT α dP ...2

from 1 & 2,

dU α dP

Topic: Physics, Thermodynamics, TS EAMCET, TS EAMCET 2018 (04 May Shift 1)
The internal energy U, pressure P and volume V of an ideal gas are related as U =3PV+4. The gas is

Options:
A either monoatomic or diatomic
B polyatomic only
C monoatomic only
D diatomic only

The correct answer is:
polyatomic only

Solution:
U=3PV+4

nf2RT=3PV+4

f2PV=3PV+4

f=6+8PV

Since degree of freedom is more than 6 therefore gas is polyatomic

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2021 (26 Feb Shift 2)
The internal resistance of a cell of e.m.f. $12 \mathrm{~V}$ is $5 \times 10^{-2} \Omega$. It is connected across an unknown resistance. Voltage across the cell, when a current of $60 \mathrm{~A}$ is drawn from it, is

Options:
A $15 \mathrm{~V}$
B $12 \mathrm{~V}$
C $9 \mathrm{~V}$
D $6 \mathrm{~V}$

The correct answer is:
$9 \mathrm{~V}$

Solution:
$V=E-$ ir $=12-60 \times 5 \times 10^{-2}=9 \mathrm{~V}$.

Topic: Physics, Current Electricity, JEE Main
The internal resistance of the cell shown in the figure is negligible. On closing the key k, the ammeter reading changes from 0.25 A to 512 A, then what is the value of resistance R1 (in ohm)?

Options:

The correct answer is:
15

Solution:
When k was open
Reading of A =E10 A = 0.25,
When k was closed
Reading of A =E10+R110R1=512
⇒ 12 5 + 0.5R1 =10R1
R1 = 15 Ω

Topic: Physics, Current Electricity, JEE Main
The international space station is maintained in a nearly circular orbit with a mean altitude of $330 \mathrm{~km}$ and a maximum of $410 \mathrm{~km}$. An astronaut is floating in the space station's cabin. The acceleration of astronaut as measured from the earth is.

Options:
A zero
B nearly zero and directed towards the earth
C nearly g and directed along the line of travel of the station.
D nearly $\mathrm{g}$ and directed towards the earth.

The correct answer is:
nearly $\mathrm{g}$ and directed towards the earth.

Solution:
No Solution

Topic: Physics, Gravitation, KVPY, KVPY 2015 (SA)
The internuclear distances in $\mathrm{O}-\mathrm{O}$ bonds for $\mathrm{O}_2^{+}, \mathrm{O}_2, \mathrm{O}_2^{-}$and $\mathrm{O}_2^{2-}$ respectively are :

Options:
A
$1.30 ~Å, 1.49 ~Å, 1.12 ~Å, 1.21 ~Å$

B
$1.49 ~Å, 1.21 ~Å, 1.12 ~Å, 1.30 ~Å$

C
$1.21 ~Å, 1.12 ~Å, 1.49 ~Å, 1.30 ~Å$

D
$1.12 ~Å, 1.21 ~Å, 1.30 ~Å, 1.49 ~Å$

The correct answer is:
$1.12 ~Å, 1.21 ~Å, 1.30 ~Å, 1.49 ~Å$

Solution:
The bond length follows the order
$\mathrm{O}_2^{+} < \mathrm{O}_2 < \mathrm{O}_2^{-} < \mathrm{O}_2^{2-}$
According to this the possible values are
$1.12 ~Å, 1.21 ~Å, 1.30 ~Å, 1.49 ~Å$

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The internuclear distances in $\mathrm{O}-\mathrm{O}$ bonds for $\mathrm{O}_2^{+}, \mathrm{O}_2, \mathrm{O}_2^{-}$and $\mathrm{O}_2^{2-}$ respectively are :

Options:
A
$1.30 ~Å, 1.49 ~Å, 1.12 ~Å, 1.21 ~Å$

B
$1.49 ~Å, 1.21 ~Å, 1.12 ~Å, 1.30 ~Å$

C
$1.21 ~Å, 1.12 ~Å, 1.49 ~Å, 1.30 ~Å$

D
$1.12 ~Å, 1.21 ~Å, 1.30 ~Å, 1.49 ~Å$

The correct answer is:
$1.12 ~Å, 1.21 ~Å, 1.30 ~Å, 1.49 ~Å$

Solution:
The bond length follows the order
$\mathrm{O}_2^{+} < \mathrm{O}_2 < \mathrm{O}_2^{-} < \mathrm{O}_2^{2-}$
According to this the possible values are
$1.12 ~Å, 1.21 ~Å, 1.30 ~Å, 1.49 ~Å$

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2013 (25 Apr Online)
The intersection of the spheres $x^2+y^2+z^2+7 x-2 y-z=13$ and $x^2+y^2+z^2-3 x+3 y+4 z=8$ is the same as the intersection of one of the sphere and the plane

Options:
A
$x-y-z=1$

B
$x-2 y-z=1$

C
$x-y-2 z=1$

D
$2 x-y-z=1$

The correct answer is:
$2 x-y-z=1$

Solution:
Required plane is $S_1-S_2=0$ where $S_1=x^2+y^2+z^2+7 x-2 y-z-13=0$ and
$$
\begin{aligned}
& S_2=x^2+y^2+z^2-3 x+3 y+4 z-8=0 \\
& \Rightarrow 2 x-y-z=1
\end{aligned}
$$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2004
The intersection of three lines x-y=0, x+2y=3 and 2x+y=6 is a/an

Options:
A Isosceles triangle
B Equilateral triangle
C Right angled triangle
D None of the above

The correct answer is:
Isosceles triangle

Solution:
.

L1:x-y=0

L2:x+2y=3

L3:2x+y=6

On solving L1 and L2:y=1 and x=1

On solving L1 and L3, we get

x=2

y=2

L2 and L3:

x+2y=3

2x+y=6

On solving L2 & L3, we get

x=3

y=0

AC=4+1=5

BC=4+1=5

AB=1+1=2

so it is an isosceles triangle

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2021 (26 Feb Shift 1)
The inverse function of fx=82x-8-2x82x+8-2x,x∈-1,1, is __________.

Options:
A 14loge1+x1-x
B 14loge1-x1+x
C 14logeloge1-x1+x
D 14log81+x1-x

The correct answer is:
14log81+x1-x

Solution:
y=82x-8-2x82x+8-2x

⇒ 1+y1-y=82x8-2x (Applying componendo-dividendo)

⇒ 84x=1+y1-y

⇒ 4x=log81+y1-y

f-1x=14log81+x1-xs

Topic: Mathematics, Functions, JEE Main, JEE Main 2020 (08 Jan Shift 1)
The inverse matrix of $\left[\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right]$ is

Options:
A $\left[\begin{array}{ccc}\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -4 & 3 & -1 \\ \frac{5}{2} & \frac{-3}{2} & \frac{1}{2}\end{array}\right]$
B $\left[\begin{array}{ccc}\frac{1}{2} & -4 & \frac{5}{2} \\ 1 & -6 & 3 \\ 1 & 2 & -1\end{array}\right]$
C $\frac{1}{2}\left[\begin{array}{lll}1 & 2 & 3 \\ 3 & 2 & 1 \\ 4 & 2 & 3\end{array}\right]$
D $\frac{1}{2}\left[\begin{array}{ccc}1 & -1 & -1 \\ -8 & 6 & -2 \\ 5 & -3 & 1\end{array}\right]$

The correct answer is:
$\left[\begin{array}{ccc}\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -4 & 3 & -1 \\ \frac{5}{2} & \frac{-3}{2} & \frac{1}{2}\end{array}\right]$

Solution:
$A^{-1}=\frac{\operatorname{adj}(A)}{|A|}=\frac{1}{|A|} \cdot \operatorname{adj}(A)$
$A=\left[\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right] ;|A|=0-1(1-9)+2(1-6)=8-10$
$|A|=-2 \neq 0$
$\operatorname{Adj} A=\left[\begin{array}{lll}A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33}\end{array}\right]$
$A_{11}=(-1)^{1+1}[(2)(1)-(3)(1)]=-1$
$A_{12}=8, \quad A_{13}=-5, A_{21}=1, A_{22}=-6$
$A_{23}=3, A_{31}=-1, A_{32}=2, A_{33}=-1$
$\therefore A^{-1}=\frac{1}{-2}\left[\begin{array}{ccc}-1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1\end{array}\right]=\left[\begin{array}{ccc}1 / 2 & -1 / 2 & 1 / 2 \\ -4 & 3 & -1 \\ 5 / 2 & -3 / 2 & 1 / 2\end{array}\right]$.

Topic: Mathematics, Matrices, JEE Main
The inverse of $\left[\begin{array}{cc}2 & -3 \\ -4 & 2\end{array}\right]$ is

Options:
A $\frac{-1}{8}\left[\begin{array}{ll}2 & 3 \\ 4 & 2\end{array}\right]$
B $\frac{-1}{8}\left[\begin{array}{ll}3 & 2 \\ 2 & 4\end{array}\right]$
C $\frac{1}{8}\left[\begin{array}{ll}2 & 3 \\ 4 & 2\end{array}\right]$
D $\frac{1}{8}\left[\begin{array}{ll}3 & 2 \\ 2 & 4\end{array}\right]$

The correct answer is:
$\frac{-1}{8}\left[\begin{array}{ll}2 & 3 \\ 4 & 2\end{array}\right]$

Solution:
Let $A=\left[\begin{array}{cc}2 & -3 \\ -4 & 2\end{array}\right], \left.\quad \therefore A|=| \begin{array}{cc}2 & -3 \\ -4 & 2\end{array} \right\rvert\,=4-12=-8$
The matrix of cofactors of the elements of $A$ viz.
$\left[\begin{array}{ll}c_{11} & c_{12} \\ c_{21} & c_{22}\end{array}\right]=\left[\begin{array}{cc}2 & -(-4) \\ -(-3) & 2\end{array}\right]=\left[\begin{array}{ll}2 & 4 \\ 3 & 2\end{array}\right]$
$\therefore \operatorname{adj} A=$ transpose of the matrix of cofacters
of elements of $A=\left[\begin{array}{ll}2 & 3 \\ 4 & 2\end{array}\right]$

Topic: Mathematics, Matrices, JEE Main
The inverse of a symmetric matrix is

Options:
A Symmetric
B Skew symmetric
C Diagonal matrix
D None of these

The correct answer is:
Symmetric

Solution:
Let $A$ be a symmetric matrix.
Then $A A^{-1}=I \Rightarrow\left(A A^{-1}\right)^T=I$
$\Rightarrow\left(A^{-1}\right)^T A^T=I \Rightarrow\left(A^{-1}\right)^T=\left(A^T\right)^{-1}$
$\Rightarrow\left(A^{-1}\right)^T=(A)^{-1},\left(\because A^T=A\right)$
$\Rightarrow A^{-1}$ is a symmetric matrix.

Topic: Mathematics, Matrices, JEE Main
The inverse of \( f(x)=\frac{2}{3} \frac{10^{x}-10^{-x}}{10^{x}+10^{-x}} \) is

Options:
A \( \frac{1}{3} \log _{10} \frac{1+x}{1-x} \)
B \( \frac{1}{2} \log _{10} \frac{2+3 x}{2-3 x} \)
C \( \frac{1}{3} \log _{10} \frac{2+3 x}{2-3 x} \)
D \( \frac{1}{6} \log _{10} \frac{2-3 x}{2+3 x} \)

The correct answer is:
\( \frac{1}{2} \log _{10} \frac{2+3 x}{2-3 x} \)

Solution:
Given,

y=2310x-10-x10x+10-x

y=23102x-1102x+1

⇒102x=3y+22-3y

Taking log both sides, we get

2xlog1010=log103y+22-3y

⇒x=12log102+3y2-3y

∴ f-1x=12log102+3x2-3x.

Topic: Mathematics, Functions, JEE Main
The inverse of the function $\frac{10^x-10^{-x}}{10^x+10^{-x}}$ is

Options:
A $\frac{1}{2} \log _{10}\left(\frac{1+x}{1-x}\right)$
B $\frac{1}{2} \log _{10}\left(\frac{1-x}{1+x}\right)$
C $\frac{1}{4} \log _{10}\left(\frac{2 x}{2-x}\right)$
D None of these

The correct answer is:
$\frac{1}{2} \log _{10}\left(\frac{1+x}{1-x}\right)$

Solution:
$y=\frac{10^x-10^{-x}}{10^x+10^{-x}} \Rightarrow x=\frac{1}{2} \log _{10}\left(\frac{1+y}{1-y}\right)$
Let $y=f(x) \Rightarrow x=f^{-1}(y)$
$\Rightarrow f^{-1}(y)=\frac{1}{2} \log _{10}\left(\frac{1+y}{1-y}\right) \Rightarrow f^{-1}(x)=\frac{1}{2} \log _{10}\left(\frac{1+x}{1-x}\right) .$

Topic: Mathematics, Functions, JEE Main
The inverse of the function $f(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}}+2$ is given by

Options:
A $\log _e\left(\frac{x-2}{x-1}\right)^{1 / 2}$
B $\log _e\left(\frac{x-1}{3-x}\right)^{1 / 2}$
C $\log _e\left(\frac{x}{2-x}\right)^{1 / 2}$
D $\log _e\left(\frac{x-1}{x+1}\right)^{-2}$

The correct answer is:
$\log _e\left(\frac{x-1}{3-x}\right)^{1 / 2}$

Solution:
$y=\frac{e^x-e^{-x}}{e^x+e^{-x}}+2 \Rightarrow y=\frac{e^{2 x}-1}{e^{2 x}+1}+2$
$\begin{aligned} & \Rightarrow e^{2 x}=\frac{1-y}{y-3}=\frac{y-1}{3-y} \Rightarrow x=\frac{1}{2} \log _e\left(\frac{y-1}{3-y}\right) \\ & \Rightarrow f^{-1}(y)=\log _e\left(\frac{y-1}{3-y}\right)^{1 / 2} \Rightarrow f^{-1}(x)=\log _e\left(\frac{x-1}{3-x}\right)^{1 / 2}\end{aligned}$

Topic: Mathematics, Functions, JEE Main
The inverse of y=5logx is:

Options:
A x=5logy
B x=ylog5
C y=x1log5
D x=51logy

The correct answer is:
y=x1log5

Solution:
Given, y=5logx

y=xlog5 ∴ bloga=alogb

y1log5=x

Replacing x→y and y→x

x1log5=y

Topic: Mathematics, Functions, JEE Main, JEE Main 2021 (17 Mar Shift 1)
The inversion of cane sugar is represented by
$\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6+\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6$
It is a reaction of

Options:
A Second order
B Unimolecular
C Pseudo unimolecular
D None of the three

The correct answer is:
Pseudo unimolecular

Solution:
Inversion of cane sugar is a Pseudo unimolecular reaction.

Topic: Chemistry, Chemical Kinetics, JEE Main
The inward and outward electric flux for a closed surface in units of $N-m^2 / C$ are respectively $8 \times 10^3$ and $4 \times 10^3$. Then the total charge inside the surface is [where $\varepsilon_0=$ permittivity constant]

Options:
A $4 \times 10^3 \quad C$
B $-4 \times 10^3 \quad C$
C $\frac{\left(-4 \times 10^3\right)}{\varepsilon}$ C
D $-4 \times 10^3 \varepsilon_0 \mathrm{C}$

The correct answer is:
$-4 \times 10^3 \varepsilon_0 \mathrm{C}$

Solution:
By Gauss's law $\phi=\frac{1}{\varepsilon_0}$ (Geaclosed)
$\Rightarrow Q_{\text {enclosed }}=\phi \varepsilon_0=\left(-8 \times 10^3+4 \times 10^3\right) \varepsilon_0$
$=-4 \times 10^3 \varepsilon_0$ Coulomb.

Topic: Physics, Electrostatics, JEE Main
The ion that has sp3d2 hybridization for the central atom is:

Options:
A BrF2-
B ICl4-
C IF6-
D ICl2-

The correct answer is:
ICl4-

Solution:
Steric number =σ+lp=4+2; sp3d2 hybridized

Steric number =σ+lp=2+3=5; sp3d hybridized

Steric number =σ+lp=6+1=7; sp3d3 hybridized

Steric number =σ+lp=2+3=5; sp3d hybridized

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2019 (08 Apr Shift 2)
The ionic mobility of alkali metal ions in aqueous solution is maximum for

Options:
A
${K}^+$

B
${Rb}^+$

C
${Li}^+$

D
${Na}^+$

The correct answer is:
${Rb}^+$

Solution:
No Solution Available

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2006
The ionic product of water with increase in temperature.

Options:
A remains constant
B increases
C decreases
D may increase or decrease

The correct answer is:
increases

Solution:
With an increase in temperature, there is an increase in dissociation of \(\mathrm{H}_2 \mathrm{O}\) which gives rise to an increase ion the concentration of \(\mathrm{H}^{+}\)and \(\mathrm{OH}^{-}\) ions. Hence, the ionic product of water increases with an increase in temperature.

Topic: Chemistry, Ionic Equilibrium, AP EAMCET, AP EAMCET 2020 (21 Sep Shift 1)
The ionic radii (in â«) of N3-, O2- and F- are respectively:

Options:
A 1.71, 1.36 and 1.40
B 1.36, 1.40 and 1.71
C 1.36, 1.71 and 1.40
D 1.71, 1.40 and 1.36

The correct answer is:
1.71, 1.40 and 1.36

Solution:
As Z e ↑ ionic radius decreases for isoelectronic species.

Z=number of protonse=number of electrons
N 3 - Z e = 7 1 0
O 2 - Z e = 8 1 0
F - Z e = 9 1 0
Trend of ionic radius: N 3 - > O 2 - > F -

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2015 (04 Apr)
The ionic radii of K+,Na+,Al3+ and Mg2+ are in the order :

Options:
A Na+<K+<Mg2+<Al3+
B Al3+<Mg2+<K+<Na+
C Al3+<Mg2+<Na+<K+
D K+<Al3+<Mg2+<Na+

The correct answer is:
Al3+<Mg2+<Na+<K+

Solution:
Al3+,Mg2+ and Na+ are isoelectronic ionic species. For monatomic ionic isoelectronic species as positive charge increases ionic size decreases.

The order of size of Na+& K+ is Na+<K+,

∴ Increasing order of ionic radii: Al3+<Mg2+<Na+<K+

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2021 (25 Jul Shift 1)
The ionic radii of O2–, F–, Na+ and Mg2+ are in the order:

Options:
A F– > O2– > Na+ > Mg2+
B O2– > F– > Na+ > Mg2+
C Mg2+ > Na+ > F– > O2–
D O2– > F– > Mg2+ > Na+

The correct answer is:
O2– > F– > Na+ > Mg2+

Solution:
O2-	F-	Na+	Mg2+

Z	8	9	11	12

No. of e-	10	10	10	10

In isoelectronic species greater is Z smaller is radius so order is O−2>F−>Na+>Mg2+.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2020 (04 Sep Shift 1)
The ionic radius of Na+ ions is 1.02A°. The ionic radii (in A°) of Mg2+ and Al3+, respectively, are

Options:
A 1.05 and 0.99
B 0.72 and 0.54
C 0.85 and 0.99
D 0.68 and 0.72

The correct answer is:
0.72 and 0.54

Solution:
As we know nuclear charge increases, attraction of nucleus on electron increases because of that size decreases.

Here all ions have same number of electrons so as the number of protons increases nuclear charge increases and size decreases.

The ionic radii order is Na+>Mg2+>Al3+

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2021 (18 Mar Shift 1)
The ionisation constant for NH4+ in water is 5.0Ã10-10 at 25oC .The rate constant for the reaction of NH4+ and OH- to form NH3 and H2O at 25oC is 3.0× 10 10   Lmol −1 s −1 . The rate constant for proton transfer from water to NH3 is xÃ105. The value of 'x' is

Options:

The correct answer is:
6

Solution:
NH4+âNH3+H+ K=5Ã10-10
Now,
H++OH-âH2O Kw=1014
now,
K c  =  k f k b
5× 10 4 = 3× 10 10 k b
k b =6× 10 5

Topic: Chemistry, Chemical Equilibrium, JEE Main
The ionisation constant of NH4+ in water is 5.6 ×10-10 at 25 °C . The rate constant for the reaction of NH4+ and OH- to form NH3 and H2O at 25°C is 3.4×1010/mol-1S-1 . The rate constant for proton transfer from water to NH3 is

Options:
A 6.07×105
B 0.607×105
C 60.7×105
D 6.07×1010

The correct answer is:
6.07×105

Solution:
NH4+⇌NH3+H+;k1=5.6×1010 …(i)
H2O⇌H++OH;k2=1×1014 …(ii)
K=k1k2=5.6×10-101×10-14
=5.6×104
By subtracting equation (ii) from equation (i) we get
NH4++OH⇌NH3+H2O
From equation
kf=3.4×1010/mil-1S-1
NH3+H2O⇌NH4++OH-
kr=?
K=kfkr
So kr=kfK=3.4×10105.6×104
=6.07×105

Topic: Chemistry, Chemical Equilibrium, JEE Main
The ionization constant of a weak monobasic acid is 25Ã10â6 while the equivalent conductance of its 0.01 M solution is 19.6 Scm2eqâ1 . The equivalent conductance of the electrolyte at infinite dilution (in Scm2eqâ1) will be

Options:
A 250
B 196
C 392
D 384

The correct answer is:
392

Solution:
K a = Cα 2
25Ã10â6=0.01 Î±2
Î±=5Ã10â2
Î±=ÎeqÎeqo
5Ã10â2=19.6Îeqo
Îeqo=19.65Ã10â2= 392

Topic: Chemistry, Ionic Equilibrium, JEE Main
The ionization constant of ammonium hydroxide is $1.77 \times 10^{-5}$ at $298 \mathrm{~K}$. Hydrolysis constant of ammonium chloride -

Options:
A $5.65 \times 10^{-12}$
B $5.65 \times 10^{-10}$
C $6.50 \times 10^{-12}$
D $5.65 \times 10^{-13}$

The correct answer is:
$5.65 \times 10^{-10}$

Solution:
$\mathrm{K}_{\mathrm{h}}=\frac{\mathrm{K}_{\mathrm{W}}}{\mathrm{K}_{\mathrm{b}}}=\frac{10^{-14}}{1.77 \times 10^{-5}}=5.65 \times 10^{-10}$

Topic: Chemistry, Electrochemistry, NEET, NEET 2009 (Mains)
The ionization constant of benzoic acid is 6.46×10-5 and Ksp for silver benzoic is 2.5×10-13 . How many times is silver benzoate more soluble in a buffer of pH=3.19 compared to its solubility in pure water?

Options:
A 4
B 3.32
C 3.01
D 2.5

The correct answer is:
3.32

Solution:
Suppose S is the molar solubility of silver benzoate in water, then
C6H5COOAgsâC6H5COOaq-+Agaq+
Ksp=S2
So S=2.5Ã10-13=5.0Ã10-7 M
If the solubility of salt of weak acid of ionization constant Ka is S, then Ksp,Ka and S' are related to each other at pH=3.19.
So [H+]=6.46Ã10-4M âµ pH=3.19
Ksp=Sâ²2KaKa+H+
Sâ²=2.5Ã10-136.46Ã10-56.46Ã10-5+6.46Ã10-412
Sâ²=2.5Ã10-13Ã7.106Ã10-46.46Ã10-512
=2.75Ã10-1212
=1.658Ã10-6 M
So the ratio of Sâ²S=1.658Ã10-65.0Ã10-7=3.32
Silver benzoate is 3.32 times more soluble in buffer of pH = 3.19 than in pure water.

Topic: Chemistry, Ionic Equilibrium, JEE Main
The ionization constant of benzoic acid is \(6.46 \times 10^{-5}\) and \(K_{s p}\) for silver benzoate is \(2.5 \times 10^{-13}\). How many times is silver benzoate more soluble in a buffer of \(\mathbf{p H}\) 3.19 compared to its solubility in pure water ?

Options:

None

Solution:
\(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOAg}^2 \rightleftharpoons \mathrm{C}_6 \mathrm{H}_5 \mathrm{COO}^{-}+\mathrm{Ag}^{+}\)
Solubility in water. Suppose solubility in water \(=\mathrm{xmol} \mathrm{L}^{-1}\).
Then \(\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{COO}^{-}\right]=\left[\mathrm{Ag}^{+}\right]=\mathrm{x} \mathrm{mol} \mathrm{L}{ }^{-1}\).
\(\therefore x^2=K_{s p}\) or \(x=\sqrt{K_{s p}}=\sqrt{2.5 \times 10^{-13}}\)
\(=5 \times 10^{-7} \mathrm{~mol} \mathrm{~L}^{-1}\)
Solubility in buffer of \(\mathbf{p H}=\mathbf{3 . 1 9}\)
\(\mathrm{pH}=3.19\) means \(-\log \left[\mathrm{H}^{+}\right]=3.19\)
or \(\log \left[\mathrm{H}^{+}\right]=-3.19=-4.81\)
or \(\log \left[\mathrm{H}^{+}\right]=6.457 \times 10^{-4}\)
\(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COO}^{-}\)ions now combine with the \(\mathrm{H}^{+}\)ions to form benzoic acid but \(\left[\mathrm{H}^{+}\right]\)remains almost constant because we have buffer solution. Now
\(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH} \rightleftharpoons \mathrm{C}_6 \mathrm{H}_5 \mathrm{COO}^{-}+\mathrm{H}^{+}\)
\(\mathrm{K}_{\mathrm{a}}=\frac{\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}\right]}\)
or \(\frac{\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}\right]}{\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{COO}^{-}\right]}\)
\(=\frac{\left[\mathrm{H}^{+}\right]}{\mathrm{K}_{\mathrm{a}}}=\frac{6.457 \times 10^{-4}}{6.46 \times 10^{-5}}=10\) ...(i)
Suppose solubility in the buffer solution is \(\mathrm{y} \mathrm{mol} \mathrm{L}^{-1}\). Then as most of the benzoate ions are converted into benzoic acid molecules (which remain almost ionized), we have \(\mathrm{y}=\left[\mathrm{Ag}^{+}\right]=\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{COO}^{-}\right]+\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}\right]\)
\(\quad=11\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{COO}^{-}\right]\)
\({\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{COO}^{-}\right]=\mathrm{y} / 11 }\)
\(\therefore \quad \mathrm{K}_{\text {sp }}=\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{COO}^{-}\right]\left[\mathrm{Ag}^{+}\right]\)
i. e., \(2.5 \times 10^{-3}=y / 11 \times \mathrm{y}\)
or \(y^2=2.75 \times 10^{-12}\)
or \(y=1.66 \times 10^{-6}\)
\(\therefore y / x=\frac{1.66 \times 10^{-6}}{5 \times 10^{-7}}=3.32\)
Silver benzoate is \(3.32\) times more soluble in buffer of \(\mathrm{pH} 3.19\) then in pure water.

Topic: Chemistry, Equilibrium
The ionization energies of Li and Na are 520 kJ mol-1 and 495 kJ  mol −1 respectively. The energy required to convert all the atoms present in 7 mg of Li vapours and 23 mg of sodium vapours to their respective gaseous cations respectively are:

Options:
A 52 J, 49.5 J
B 520 J, 495 J
C 49.5 J, 52 J
D 495 J, 520 J

The correct answer is:
520 J, 495 J

Solution:
For 1 mole Li i.e. 7 g will absorb 520 kJ heat
So for 7 mg lithium heat absorbed will be 520 J
Similarly for sodium
1 mole Na i.e. 23 g will absorb 495 kJ Heat
So for 23 mg sodium heat absorbed will be 520 J

Topic: Chemistry, s Block Elements, JEE Main
The ionization energy of an element is

Options:
A The same as the electron affinity of the element
B Equal in magnitude but of opposite sign to the electron affinity of the element
C The energy released when an electron is added to an atom of the element
D The energy required to remove the outermost electron of an atom of the element

The correct answer is:
The energy required to remove the outermost electron of an atom of the element

Solution:
The energy required to remove the outermost electron of an atom of the element

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main