datavorous/entrance-exam-dataset huggingface dataset:
The number of incorrect statement/s from the following is ______________

A. The successive half lives of zero order reactions decreases with time.

B. A substance appearing as reactant in the chemical equation may not affect the rate of reaction

C. Order and molecularity of a chemical reaction can be a fractional number

D. The rate constant units of zero and second order reaction are molL -1s-1 and mol-1 L s-1 respectively

Options:

The correct answer is:
1

Solution:
A. C0⟶t=C02kC02⟶t=C04kC04⟶t=C08kC08

Successive half lives of zero order reactions decreases with time

Note: successive half lives of first order reactions remains same

B. For zero order reactions, the rate of reaction is independent on the concentration of reactants.

C. In some reaction molecularity (theoretical value) may be equal to the order of a reaction (experimental value). The order of a reaction may be fractional, zero, positive or negative. Molecularity can never be fractional.

(D) Unit of k = Units of rateunits of concentrationorder

Zero order =mollit-1sec-1

Second order =mol-1lit-1

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2023 (10 Apr Shift 2)
The number of incorrect statement/s from the following is/are

A. Water vapours are adsorbed by anhydrous calcium chloride.

B. There is a decrease in surface energy during adsorption.

C. As the adsorption proceeds, ΔH becomes more and more negative.

D. Adsorption is accompanied by decrease in entropy of the system.

Options:

The correct answer is:
2

Solution:
Water vapours are absorbed by anhydrous calcium chloride. There is a decrease in surface energy during adsorption. As the adsorption proceeds, ΔH becomes more and more negative. Adsorption is accompanied by decrease in entropy of the system. Thus, option A and C are incorrect.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2023 (25 Jan Shift 2)
The number of integers a in the interval [1,2014] for which the system of equations
$$
x+y=a \frac{x^{2}}{x-1}+\frac{y^{2}}{y-1}=4
$$
has finitely many solutions is

Options:
A 0
B 1007
C 2013
D 2014

The correct answer is:
2013

Solution:
$\frac{x^{2}-1+1}{x-1}+\frac{y^{2}-1+1}{y-1}=4$
$x+1+\frac{1}{x-1}+y+1+\frac{1}{y-1}=4$
$a+2+\frac{1}{x-1}+\frac{1}{(a-1)-x}=4$ $\frac{(a-1)-x+x-1}{(x-1)[(a-1)-x]}=2-a$
$\because a \neq 2[$ for $a=2$ equation have infinitely many solution] $\therefore(x-1)[(a-1)-x]=-1$ $(x-1)[x-(a-1)]=1$ $x^{2}-a x+(a-2)=0$ $D>0$
$\therefore$ equation have 2 real roots so an be $1,3,4 \ldots \ldots .2014$ a can be $1,3,4 \ldots \ldots .2014$
ans 2013

Topic: Mathematics, Determinants, JEE Main
The number of integers for which the equation sin-1⁡x+cos-1⁡x+tan-1⁡x=n has real solution(s) is

Options:
A 0
B 1
C 2
D 3

The correct answer is:
2

Solution:
Given equation is Ï2+tan-1â¡x=n, âxâ[-1,1]
Now LHS âÏ4,3Ï4
Integers in this interval are 1 & 2

Hence, there are 2 integers for which the equation has real solutions

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The number of integers greater than 6000 that can be formed, using the digits 3, 5, 6, 7 and 8, without repetition is

Options:
A 72
B 216
C 192
D 120

The correct answer is:
192

Solution:
4 digit and 5 digit numbers are possible.

4 digit numbers:
6 / 7 / 8 ↑ ¯ ↑ ¯ ↑ ¯ ↑ ¯ 3 4 3 2 Total numbers possible = 3 . 4 . 3 . 2 = 72
5 digit numbers
↑ ¯ ↑ ¯ ↑ ¯ ↑ ¯ ↑ ¯ 5 4 3 2 1 Total numbers possible = 5 . 4 . 3 . 2 . 1 = 120
∴Total numbers=72+120=192

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2015 (04 Apr)
The number of integers, greater than 7000 that can be formed, using the digits 3,5,6,7,8 without repetition is

Options:
A 120
B 168
C 220
D 48

The correct answer is:
168

Solution:
Case 1

forming 4 digit number,

When 7 is at 1000's place

7 _ _ _,

So total number of ways will be 4×3×2=24 ways,

Case 2

forming 4 digit number when 8 is at 1000's place

8 _ _ _,

So total number of ways will be 4×3×2=24 ways,

Case 3

forming 5 digit number,

So total number of ways will be 5!=120 ways,

Now adding all the cases we get, 120+24+24=168 ways.

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2023 (24 Jan Shift 2)
The number of integers x satisfying -3x4+det1xx21x2x41x3x6=0 is equal to

Options:
A 1
B 2
C 5
D 8

The correct answer is:
2

Solution:
Given, -3x4+det1xx21x2x41x3x6=0

⇒ x8+x5+x5-x4-x7-x7=3x4

⇒ x8-2x7+2x5-4x4=0

⇒ x4x4-2x3+2x-4=0

⇒ x4x3x-2+2x-2=0

⇒ x4x3+2x-2=0

∵ x is an integer, so x=0, 2.

Topic: Mathematics, Determinants, JEE Main
The number of integral points on the circle, touching the parabola y2=8x at 2,4 and passing through 0,4, are equal to

Options:
A 1
B 2
C 4
D 3

The correct answer is:
4

Solution:
Equation of tangent at 2,4 on the parabola y2=8x is

y4=8x+22⇒y=x+2
Let the equation of the circle touching line y=x+2 at 2,4 is

x-22+y-42+λx-y+2=0 which passes through 0,4

⇒4+0+λ0-4+2⇒λ=2
⇒ Required circle is x2+y2-2x-10y+24=0
⇒x-12+y-52=2
If x and y are integers, then

x-12=1=y-52
⇒x=0,2 and y=4,6
⇒4 integral points lie on the circle

Topic: Mathematics, Parabola, JEE Main
The number of integral terms in the expansion of 312+514680 is equal to

Options:

The correct answer is:
171

Solution:
Given expansion is 312+514680.

The general term in the expansion x+an is Tr+1=Crn·xn-r·ar

⇒Tr+1=Cr680312680-r514r, 0≤r≤680

=Cr6803680-r25r4

For 3680-r25r4 to be rational r should be a multiple of 4.

⇒r=0,4,8,12,.....680

⇒an=a+n-1d

⇒680=0+n-14

⇒n=171

That means r can take 171 values.

Hence, the required answer is 171.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2023 (11 Apr Shift 1)
The number of integral terms in the expansion of $(\sqrt{3}+8 \sqrt{5})^{256}$ is

Options:
A
35

B
32

C
33

D
34

The correct answer is:
33

Solution:
$\mathrm{T}_{\mathrm{r}+1}={ }^{256} \mathrm{C}_{\mathrm{r}}(\sqrt{3})^{256-\mathrm{r}}(\sqrt[8]{5})^{\mathrm{r}}={ }^{256} \mathrm{C}_{\mathrm{r}}(3)^{\frac{256-\mathrm{r}}{2}}(5)^{\mathrm{r} / 8}$
Terms will be integral if $\frac{256-\mathrm{r}}{2} \& \frac{\mathrm{r}}{8}$ both are +ve integer. As $0 \leq \mathrm{r} \leq 256 \therefore \mathrm{r}=0,8,16,24, \ldots \ldots \ldots \ldots 256$
For above values of $\mathrm{r},\left(\frac{256-\mathrm{r}}{2}\right)$ is also an integer.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2003
The number of integral terms in the expansion of 516+7191824 is

Options:
A 84
B 96
C 91
D 102

The correct answer is:
102

Solution:
Tr+1=1824Cr5161824-r719r
=1824Cr51824-r67r9
=1824Cr5304-r67r9
Both r6 and r9 are integers, if r is a multiple of 18
⇒r=0,18,36,......,1818
Let, the number of terms be n
⇒1818=0+n-118⇒n=102.

Topic: Mathematics, Binomial Theorem, JEE Main
The number of integral values of c for which both the roots of the equation x2-2cx+c2=1 are greater than -2 but less than 4 are

Options:
A 0
B 1
C 2
D more than 2

The correct answer is:
more than 2

Solution:
x-c2=1⇒x-c=±1
⇒x=c+1,c-1
c-1>-2 and c+1<4
⇒c>-1 and c<3
⇒-1<c<3
⇒c=0,1,2
Hence, the number of integral values of c are 3.

Topic: Mathematics, Quadratic Equation, JEE Main
The number of integral values of k, for which one root of the equation 2x2-8x+k=0 lies in the interval 1,2 and its other root lies in the interval 2,3, is :

Options:
A 2
B 0
C 1
D 3

The correct answer is:
1

Solution:
Let f(x)=2x2-8x+k=0

f(1)>0

f(2)<0

f(3)>0

Now, f(1)>0⇒2-8+k>0⇒k>6 ......(1)

f(2)<0⇒8-16+k<0⇒k<8 ......(2)

f(3)>0⇒18-24+k>0⇒k>6 ......(3)

From equations 1, 2 & 3,

k∈(6,8)

∴ Integral value of k=7

Therefore number of integral value of k is one

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2023 (01 Feb Shift 2)
The number of integral value(s) of k such that the system of equations kx-2y-z=x, ky-z=z+3x and 2x+kz=2y-z has non-trivial solution, is/are

Options:
A 0
B 1
C 2
D 3

The correct answer is:
3

Solution:
Rearranging all the equations, we get,
k-1x-2y-z=0
3x-ky+2z=0
2x-2y+k+1z=0
For a non-trivial solution,
k-1-2-13-k22-2k+1=0âk-1-k2-k+4+23k-1-1-6+2k=0
â-k3-k2+4k+k2+k-4+6k-2+6-2k=0
â-k3+9k=0âk=0,-3,3

Topic: Mathematics, Determinants, JEE Main
The number of integral values of m for which the equation, 1+m2x2-21+3mx+1+8m=0 has no real root, is

Options:
A 2
B 3
C Infinitely many
D 1

The correct answer is:
Infinitely many

Solution:
Since the given equation has no real solution.
⇒D<0
⇒41+3m2-4×1+m21+8m<0
⇒1+9m2+6m-1+8m+m2+8m3<0
⇒-2m2m-12<0
Clearly, for infinite values of m given condition is true.

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2019 (08 Apr Shift 2)
The number of integral values of m for which the quadratic expression 1+2m x2-21+3mx+41+m, x∈R is always positive, is

Options:
A 7
B 3
C 6
D 8

The correct answer is:
7

Solution:
∵ Quadratic expression is positive, hence 1+2m>0 and D<0

⇒m>-12 and

41+3m2-16 1+m1+2m<0

⇒9m2+1+6m-42m2+3m+1<0

⇒m2-6m-3<0

⇒m-3+23m-3-23<0

⇒m∈3-23, 3+23

∴m=0, 1, 2, 3, 4, 5, 6

∴ Number of integral values =7

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2019 (12 Jan Shift 2)
The number of integral values of m so that the abscissa of point of intersection of lines 3x+4y=9 and y=mx+1 is also an integer, is:

Options:
A 1
B 2
C 3
D 0

The correct answer is:
2

Solution:
3x+4y=9

y=mx+1

⇒3x+4mx+4=9

⇒(3+4m)x=5

⇒x will be an integer when

3+4m=5, -5, 1,-1

⇒m=12,-2,-12,-1

so, number of integral values of m is 2

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2021 (18 Mar Shift 1)
The number of integral values of $p$ in the domain $[-5,5]$, such that the equation $2 x^2+4 x y-\mathrm{p} y^2+4 x+\mathrm{q} y+1=0$ represents pair of lines, are

Options:
A 3
B 4
C 7
D 8

The correct answer is:
8

Solution:
Given equation of pair of lines is
$2 x^2+4 x y-p y^2+4 x+q y+1=0$
Comparing with $\mathrm{a} x^2+2 \mathrm{~h} x y+\mathrm{b} y^2+2 \mathrm{~g} x+2 \mathrm{f} y+\mathrm{c}=0$, we get $\mathrm{a}=2, \mathrm{~h}=2, \mathrm{~b}=-\mathrm{p}$
If the given equation represents a pair of straight lines, then
$\begin{aligned}
& \mathrm{h}^2 \geq \mathrm{ab} \\
& \Rightarrow 4 \geq-2 \mathrm{p} \\
& \Rightarrow 2 \geq-\mathrm{p} \\
& \Rightarrow \mathrm{p} \geq-2
\end{aligned}$
$\therefore \quad$ Possible values of $\mathrm{p}$ from domain $[-5,5]$ are $-2,-1,0,1,2,3,4,5$.
$\therefore \quad$ Number of integral values of $p=8$

Topic: Mathematics, Straight Lines, MHT CET, MHT CET 2023 (10 May Shift 1)
The number of inter halogens from the following having square pyramidal structure is

ClF3, IF7, BrF5, BrF3, I2Cl6, IF5, ClF, ClF5

Options:

The correct answer is:
3

Solution:
Square pyramidal structures are BrF5,IF5 and ClF5.

These molecules are made up of 6 equally spaced sp3d2 hybrid orbitals arranged at 90o angles. The shape of the orbitals is octahedral. One orbital contains a lone pair of electrons, so the remaining five atoms connected to the central atom gives the molecule a square pyramidal shape. Among the given options the structure with five bond pairs and one lone pair is these three structures only.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2022 (28 Jul Shift 1)
The number of ionic, covalent and coordinate bonds in $\mathrm{NH}_4 \mathrm{Cl}$ are respectively

Options:
A 1,3 and 1
B 1,3 and 2
C 1, 2 and 3
D 1,1 and 3

The correct answer is:
1,3 and 1

Solution:
Ionic bond $=1$, Covalent bond $=3$, Co-ordinate bond $=1$.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The number of ionisable hydrogens present in the product obtained from a reaction of phosphorus trichloride and phosphonic acid is:

Options:
A 3
B 0
C 2
D 1

The correct answer is:
2

Solution:
PCl3+H3PO3+ 2H2O→H4P2O5+3HCl

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2021 (18 Mar Shift 1)
The number of ions from the following that have the ability to liberate hydrogen from a dilute acid is _____.
\(\mathrm{Ti}^{2+}, \mathrm{Cr}^{2+}\) and \(\mathrm{V}^{2+}\)

Options:
A 2
B 3
C 1
D 0

The correct answer is:
3

Solution:
The ions $\mathrm{Ti}^{+2}, \mathrm{~V}^{+2} \mathrm{Cr}^{+2}$ are strong reducing agents and will liberate hydrogen from a dilute acid, eg.
$2 \mathrm{Cr}_{\text {(aq.) }}^{+2}+2 \mathrm{H}_{(\text {aq.) }}^{+} \longrightarrow 2 \mathrm{Cr}_{(\text {aq.) }}^{+3}+\mathrm{H}_2(\mathrm{~g})$

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2024 (05 Apr Shift 2)
The number of irrational terms in the binomial expansion of $\left(3^{1 / 5}+7^{1 / 3}\right)^{100}$ is

Options:
A 90
B 88
C 93
D 94

The correct answer is:
94

Solution:
General term of $\left(3^{1 / 5}+7^{1 / 3}\right)^{100}$ is given by $T_{\gamma+1}={ }^{100} C_{r}\left(3^{1 / 5}\right)^{100-r}\left(7^{1 / 3}\right)^{r}$
$={ }^{100} C_{1} \cdot 3^{\frac{100-r}{5}} \cdot 7^{\frac{r}{3}}$
For a rational term. $\frac{100-r}{5}$ and $\frac{r}{3}$ must be integer.
Hence, $r=0,15,30,45,60,75,90$
$\therefore$ There are seven rational terms. Hence, number of irrational terms
$$
=101-7=94
$$

Topic: Mathematics, Binomial Theorem, JEE Main
The number of isomers (including stereoisomers) of C5H10 are

Options:
A 10
B 11
C 12
D 13

The correct answer is:
13

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main
The number of isomers possible for the octahedral complex \( \left[\mathrm{CoCl}_{2}(e n)\left(\mathrm{NH}_{3}\right)_{2}\right]^{+} \)is,

Options:
A Two
B Three
C No isomer
D Four isomers

The correct answer is:
Two

Solution:
Two geometrical isomers are possible for the octahedral complex \( \left[\mathrm{CoCl}_{2}(e n)\left(\mathrm{NH}_{3}\right)_{2}\right]^{+} \), that is, cis and trans.
Trans
Cis

Topic: Chemistry, Coordination Compounds, JEE Main
The number of lattice points in an end-centred unit cell are:

Options:
A \( 8 \)
B \( 9 \)
C \( 10 \)
D \( 12 \)

The correct answer is:
\( 10 \)

Solution:
A total of ten lattice points are there in an end-centred unit cell:
(a) Eight corners, and
(b) Centres of any two opposite faces.

The shematic representation of lattice points in end-centred unit cell is shown below:

Topic: Chemistry, Solid State, JEE Main
The number of local maxima/maximum for the function y=x2-2sin⁡x, ∀x∈0,π2 is

Options:
A 0
B 1
C 2
D 3

The correct answer is:
0

Solution:
As, dydx=2x-cosâ¡x
Hence, the function shows a local minimum at x=Î± but no maximum

Topic: Mathematics, Application of Derivatives, JEE Main
The number of lone pair of electrons present in the valence shell of Xenon $(\mathrm{z}=54)$ in $\mathrm{XeOF}_4, \mathrm{XeF}_4, \mathrm{XeF}_2$ and $\mathrm{XeF}_6$ are respectively

Options:
A $1,2,3,1$
B $2,1,2,2$
C $3,1,2,1$
D $1,3,2,0$

The correct answer is:
$1,2,3,1$

Solution:
$\mathrm{Xe}$ in $\mathrm{XeOF}_4$ forms one $\mathrm{Xe}=0$ double bond and four $\mathrm{Xe}-\mathrm{F}$ single bonds using six of its eight valence electrons. The remaining two valence electrons form one lone pair. Similarly, there are two, three and one lone pairs in $\mathrm{XeF}_4$, $\mathrm{XeF}_2$ and $\mathrm{XeF}_6$ respectively.

Topic: Chemistry, Chemical Bonding and Molecular Structure, TS EAMCET, TS EAMCET 2023 (12 May Shift 2)
The number of lone pairs of electrons on the central I atom in I3- is

Options:

The correct answer is:
3

Solution:
The number of lone pairs of electron on the central atom is 3.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2021 (20 Jul Shift 1)
The number of matrices X with entries 0,2,3 for which the sum of all the principal diagonal elements of X·XT is 28 (where XT is the transpose matrix of X), is

Options:
A 12
B 18
C 36
D 44

The correct answer is:
36

Solution:
Let X=a1a2a3b1b2b3c1c2c3
XXT=a1a2a3b1b2b3c1c2c3a1b1c1a2b2c2a1b2c3
=a12+a22+a32a1b1+a2b2+a3b3a1c1+a2c2+a3c3a1b1+a2b2+a3b3b12+b22+b32b1c1+b2c2+b3c3a1c1+a2c2+a3c3b1c1+b2c2+b3c3c12+c22+c32
TrXXT=a12+a22+a32+b12+b22+b32+c12+c22+c32=28
There is only one possibility in which 7 entries are 2 and 2 entries are 0
Hence, the number of ways =C97=36

Topic: Mathematics, Matrices, JEE Main
The number of metals that show passivity with concentrate HNO3 among
Cr, Fe, Ni, Cu, Zn, Al, Ag, Sn

Options:

The correct answer is:
4.00

Solution:
Cr,Fe,Ni and Al react with conc. HNO3 to form a protective oxide layer which protest the metal surface from further reaction.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The number of mitotic division that must occur in a cell of the root tip to form 256 cells is

Options:
A 256
B 128
C 7
D 8

The correct answer is:
8

Solution:
Mitosis is a type of cell division where two identical daughter cells are formed. This cell division process occurs in certain steps. The chromosome number between parent cell and daughter cell remains the same. It is also called as equational division.
Each division of a single cell forms two new cell. Thus, if 'n' number of mitotic divisions are taking place then the total number of cells formed will be 2n
Root tip cells divide mitotically to give 256, thus, number of divisions required will be 2n = 256=28. n = 8 .
Number of division is equal to the number of generations.

Topic: Biology, Cell Cycle and Cell Division, NEET
The number of molecules and moles in 2.8375 litres of O2 at STP are respectively

Options:
A 7.527×1023 and 0.125 mol
B 7.527×1022 and 0.250 mol
C 1.505×1023 and 0.250 mol
D 7.527×1022 and 0.125 mol

The correct answer is:
7.527×1022 and 0.125 mol

Solution:
One mole of any ideal gas occupies 22.7L at standard temperature and pressure.

The number of moles =Given volumeVolume at STP

Moles =2.837522.7=0.125

The number of molecules =moles×NA

Molecules =0.125×6.022×1023

=7.527×1022

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2023 (10 Apr Shift 1)
The number of molecules from the following which contain only two lone pair of electrons is _____________

H2O,N2,CO,XeF4,NH3,NO,CO2, F2

Options:

The correct answer is:
3

Solution:
The lone pair of electrons can be identified by drawing structure of the given moleules as follows,

The number of molecules having only 2 lone pair of electrons = 3

Which are H2O,N2 and CO. XeF4 have 2 lone pairs on central atom, but we are asked about lone pair in molecule

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2023 (10 Apr Shift 2)
The number of molecules in one litre of an ideal gas at 300 K and 2 atmospheric pressure with mean kinetic energy 2×10-9 J per molecule is:

Options:
A 0.75×1011
B 3×1011
C 1.5×1011
D 6×1011

The correct answer is:
1.5×1011

Solution:
KE=32kT
PV=NNART
N=PVkT
=N=1.5×1011

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2021 (27 Jul Shift 1)
The number of molecule(s) or ion(s) from the following having non-planar structure is______.

NO3-,H2O2,BF3,PCl3,XeF4,

SF4,XeO3,PH4+,SO3,AlOH4-

Options:

The correct answer is:
6

Solution:
The species which have sp2 hybridisation with no lone pairs, sp3d2 hybridisation with two lone pairs are planar molecules.

The number of hybrid orbitals can be calculated as follows,

No.of hybrid orbitals = 1/2(V+M-C+A)

V = number of valence electrons

M = number of monovalent atoms attached to central atom.

C = Number of positive charges on the molecule

A = Number of negative charges on the molecule

SO3	sp2	Planar

BF3	sp2	Planar

NO3-	sp2	Planar

SF4	sp3d	Non-Planar

H2O2	sp3	Non-Planar

PCl3	sp3	Non-Planar

AlOH4-	sp3	Non-Planar

XeF4	sp3d2	Planar

XeO3	sp3	Non-Planar

PH4+	sp3	Non-Planar

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2022 (27 Jul Shift 2)
The number of molecules or ions from the following, which do not have odd number of electrons are _____ .
(A) NO2
(B) ICl4-
(C) BrF3
(D) ClO2
(E) NO2+
(F) NO

Options:

The correct answer is:
3

Solution:
ICl4-,BrF3 and NO2+ do not have odd number of e- as we can see from their lewis structures respectively:
, and

The number of electrons in NO223e-,NO15e- and ClO233e- are odd.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2023 (29 Jan Shift 1)
The number of molecules with energy greater than the threshold energy for a reaction increases five fold by a rise of temperature from 27oC to 42oC. Its energy of activation in J/mol is _________ (Take ln 5=1.6094; R=8.314 J mol-1)

Options:

The correct answer is:
84297

Solution:
k=Ae-EaRT

lnK2K1=EaR1T1-1T2

ln(5)=Ea8.3141300-1315

1.6094=Ea8.31415300×315

Ea=84297 J

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2020 (04 Sep Shift 2)
The number of moles $\mathrm{KMnO}_4$ that will be needed to react with one mole of sulphite ion in acidic solution is:

Options:
A $4 / 5$
B $2 / 5$
C 1
D $3 / 5$

The correct answer is:
$2 / 5$

Solution:
The balanced chemical equation is:
$2 \mathrm{MnO}_4^{-}+ 6 \mathrm{H}^{+}+5 \mathrm{SO}_3^{--} \longrightarrow 2 \mathrm{Mn}^{++}+5 \mathrm{SO}_4^{--}+3 \mathrm{H}_2 \mathrm{O}$
From the equation it is clear that
Moles of $\mathrm{MnO}_4^{-}$ require to oxidise 5 moles of $\mathrm{SO}_3^{--}$are 2
Moles of $\mathrm{MnO}_4^{-}$ require to oxidise 1 mole of $\mathrm{SO}_3^{--}$are $2 / 5$.

Topic: Chemistry, Redox Reactions, JEE Main
The number of moles of CuO, that will be utilized in Dumas method for estimating nitrogen in a sample of 57.5 g of N, N-dimethylaminopentane is ________ ×10-2. (Nearest integer)

Options:

The correct answer is:
1125

Solution:
Moles of NN in N, N-dimethylaminopentane =57.5115=0.5mol

C7H17 N+452CuO⟶7CO2+172H2O+12 N2+452Cu

no. molesof CuO reacted 45/2= no. molesof C7H17 N reacted 1

∴ no. of moles of CuO reacted =452×0.5=11.25=1125×10-2

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2021 (27 Aug Shift 1)
The number of moles of hydrogen that can be added to 1 mole of an oil is the highest in

Options:
A groundnut oil
B sunflower seed oil
C mustard oil
D linseed oil

The correct answer is:
linseed oil

Solution:
1 mole of linseed oil has maximum number of moles of hydrogen added. This oil is a typical drying oil. It consists of linolenic acid (80\%), linolenic acid ester $(15 \%)$ and triolein (5\%).

Topic: Chemistry, Hydrocarbons, JEE Main
The number of moles of $\mathrm{KMnO}_4$ reduced by one mole of $K I$ in alkaline medium is:

Options:
A One fifth
B five
C One
D Two

The correct answer is:
Two

Solution:
In alkaline medium
$2 \mathrm{KMnO}_4+\mathrm{Kl}+\mathrm{H}_2 \mathrm{O}-\cdots-\mathrm{MnO}_2+2 \mathrm{KOH}+\mathrm{KlO}_3$

Topic: Chemistry, Redox Reactions, JEE Main
The number of moles of \( \mathrm{KMnO}_{4} \) that will be needed to react with one mole of sulphite ion in acidic solution is

Options:

The correct answer is:
.4

Solution:
In acidic medium, every molecule of KMnO 4 accepts 5 - electron and SO 2 is oxidized to SO 4 2- by releasing two electron per SO 2 molecular.
In acidic medium, KMn O 4 oxidises sulphite into sulphate
S O 3 + 4 2 - â S O + 6 4 2 - + 2 e - Ã 5
MnO 4 - + 8 H + + 5 e - â Mn 2 + + 4 H 2 O Ã 2
5 moles of SO 3 2 - â¡ 2 moles of KMnO 4
1 mole of SO 3 2 - â¡ 2 5 moles of KMnO 4

Topic: Chemistry, Redox Reactions, JEE Main
The number of moles of methane required to produce \(11 \mathrm{~g} \mathrm{CO}_2(\mathrm{~g})\) after complete combustion is :
(Given molar mass of methane in \(\mathrm{g} \mathrm{mol}^{-1}: 16\))

Options:
A 0.35
B 0.5
C 0.75
D 0.25

The correct answer is:
0.25

Solution:
$\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}+2}+\frac{3 \mathrm{n}+1}{2} \mathrm{O}_2 \longrightarrow \mathrm{nCO}_2+(\mathrm{n}+1) \mathrm{H}_2 \mathrm{O}$
$0.25 \mathrm{~mol~} \mathrm{CH}_4$ gives $0.25 \mathrm{~mole}$ (or $11 \mathrm{gm}$ ) $\mathrm{CO}_2$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2024 (05 Apr Shift 2)
The number of moles of NH3, that must be added to 2 L of 0.80MAgNO3 in order to reduce the concentration of Ag+ ions to 5.0×10-8MKformation for AgNH32+=1.0×108 is __________. (Nearest integer)
[Assume no volume change on adding NH3]

Options:

The correct answer is:
4

Solution:
Let no. moles of NH3 added =x

Ag+(aq)+2NH3(aq)⇌AgNH32+(aq)

t=0 0.8 x2

t=∞ 5×10-8 x2-1.6 0.8

Keq=AgNH32+Ag+NH32

x2-1.6=0.4

x=4

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2021 (27 Aug Shift 1)
The number of moles present in 2 litre of $0.5 \mathrm{M} \mathrm{NaOH}$ is

Options:
A 0.5
B 0.1
C 1
D 2

The correct answer is:
1

Solution:
$M=\frac{n}{V(l)} ; \quad 0.5=\frac{n}{2} ; n=1$

Topic: Chemistry, Solutions, JEE Main