datavorous/entrance-exam-dataset huggingface dataset:
The molar heat capacity of rock salt at low temperatures varies with temperature according to Debye's law; C=kT3Î¸3 where k=1940 J mol-1 K-1 and Î¸=281 K. The heat required to raise the temperature of 2 moles of rock salt from 10 K to 50 K is

Options:
A 800 J
B 373 J
C 273 J
D 400 J

The correct answer is:
273 J

Solution:
dQ=nCdT ,
dQ=nkT3Î¸3 dT
Q=nkÎ¸3 â«T1T2T3dt=nkÎ¸3T24-T144
=2Ã1940504-1042813Ã4=272.79 J
Q=273 J

Topic: Physics, Thermodynamics, JEE Main
The molar heat capacity of water at constant pressure, CP is 75 J K-1mol-1 . When 10 kJ of heat is supplied to 1 kg water which is free to expand, the increase in temperature of water is

Options:
A 2.4 K
B 4.8 K
C 3.3 K
D 7.2 K

The correct answer is:
2.4 K

Solution:
q m â=âmcâÎT
qm=10 kJ or 10000 J, m=1000g
c=75 J K-1mol-1 . Number of moles =100018=55.5 mol
ÎTâ=â 10000 55.5âÃâ75 â=â2.4âK

Topic: Chemistry, Thermodynamics (C), JEE Main
The molar neutralization heat for $\mathrm{KOH}$ and $\mathrm{HNO}_3$ as compared to molar neutralization heat of $\mathrm{NaOH}$ and $\mathrm{HCl}$

Options:
A Less
B More
C Equal
D Depends on pressure

The correct answer is:
Equal

Solution:
Heat of neutralisation between strong acid and a strong base is about $-13.7 \mathrm{Kcal}$.

Topic: Chemistry, Thermodynamics (C), JEE Main
The molar neutralization heat for $\mathrm{KOH}$ and $\mathrm{HNO}_3$ as compared to molar neutralization heat of $\mathrm{NaOH}$ and $\mathrm{HCl}$

Options:
A Less
B More
C Equal
D Depends on pressure

The correct answer is:
Equal

Solution:
Heat of neutralisation between strong acid and a strong base is about $-13.7 \mathrm{Kcal}$

Topic: Chemistry, Thermodynamics (C), JEE Main
The molar solubility (in mol $\mathrm{L}^{-1}$ ) of a sparingly soluble salt $M X_{4}$ is ${ }^{\prime} S$. The corresponding solubility product is $K_{\mathrm{sp}} . S$ in terms of $'K_{\text {ip }}$' s given by the relation

Options:
A $S=\left(\frac{K_{\mathrm{s} \rho}}{128}\right)^{1 / 4}$
B $S=\left(\frac{K_{s p}}{256}\right)^{1 / 5}$
C $S=\left(256 K_{s p}\right)^{1 / 5}$
D $S=\left(128 K_{\mathrm{sp}}\right)^{1 / 4}$

The correct answer is:
$S=\left(\frac{K_{s p}}{256}\right)^{1 / 5}$

Solution:
$\because$ For a compound $A_{x}B_{y}$
$$
K_{s p}=x^{x} \cdot y^{y} \cdot S^{x+y} . \text { (where, } S=\text { solubility })
$$
For $M x_{4}$
$$
\begin{array}{l}
K_{\mathrm{sp}}=1 \times(4)^{4} \cdot \mathrm{S}^{1+4} \\
K_{\mathrm{sp}}=256 \times \mathrm{S}^{5}
\end{array}
$$
$\therefore S=\left(\frac{K_{s}}{256}\right)^{1 / 5}$

Topic: Chemistry, Ionic Equilibrium, JEE Main
The molar solubility of Zn(OH)2 in 0.1MNaOH solution is x×10-18M. The value of x is _____ . (Nearest integer)

Given : The solubility product of Zn(OH)2 is 2×10-20

Options:

The correct answer is:
2

Solution:
Zn(OH)2( s)⇌Zn+2(aq)+2OH-(aq) S (0.1+2s)≈0.1

Ksp=S(0.1)2

2×10-20=s×10-2⇒s=2×10-18

=x×10-18

x=2

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2021 (01 Sep Shift 2)
The molar solubility product is $\mathrm{K}_{\mathrm{sp}}$. ' $\mathrm{s}$ ' is given in terms of $\mathrm{K}_{\mathrm{sp}}$ by the relation

Options:
A
$\mathrm{s}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{128}\right)^{1 / 4}$

B
$\mathrm{s}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{256}\right)^{1 / 5}$

C
$\mathrm{S}=\left(256 \mathrm{~K}_{\mathrm{sp}}\right)^{1 / 5}$

D
$\mathrm{s}=\left(128 \mathrm{~K}_{\mathrm{sp}}\right)^{1 / 4}$

The correct answer is:
$\mathrm{s}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{256}\right)^{1 / 5}$

Solution:
$\mathrm{Ksp}=108 s^5$
$1 \times 4^4 \times s^{1+4}=256 s^5=\mathrm{Ksp}$

Topic: Chemistry, Ionic Equilibrium, JEE Main
The molar solubility product is $\mathrm{K}_{\mathrm{sp}}$. ' $\mathrm{s}$ ' is given in terms of $\mathrm{K}_{\mathrm{sp}}$ by the relation

Options:
A
$\mathrm{s}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{128}\right)^{1 / 4}$

B
$\mathrm{s}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{256}\right)^{1 / 5}$

C
$\mathrm{S}=\left(256 \mathrm{~K}_{\mathrm{sp}}\right)^{1 / 5}$

D
$\mathrm{s}=\left(128 \mathrm{~K}_{\mathrm{sp}}\right)^{1 / 4}$

The correct answer is:
$\mathrm{s}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{256}\right)^{1 / 5}$

Solution:
$\mathrm{Ksp}=108 s^5$
$1 \times 4^4 \times s^{1+4}=256 s^5=\mathrm{Ksp}$

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2004
The molarity (in mol/liter) of a 13% solution (by weight) of sulphuric acid with a density of 1.02 g/mL is
(Note: Report your answer in the required format by relevant rounding up.)

Options:

The correct answer is:
1.36

Solution:
13% solution of Sulfuric acid consists of 13 g of H 2 SO 4 and 87 g (100 g-13g) of water.
The molarity is,
M= number of moles volume = n V = ( given mass of H 2 SO 4 molar mass ) ( total mass of solution/ density ) M= ( 13âg 98âg/mol ) ( 100âg/1.02âg/mL ) = 0.133âmol 98âmL = 0.133âmol 0.098âL M=1.357âmol/L ≃1.36

Topic: Chemistry, Solutions, JEE Main
The molarity of a $\mathrm{NaOH}$ solution by dissolving $4 \mathrm{~g}$ of it in $250 \mathrm{ml}$ water is

Options:
A $0.4 \mathrm{M}$
B $0.8 \mathrm{M}$
C $0.2 \mathrm{M}$
D $0.1 \mathrm{M}$

The correct answer is:
$0.4 \mathrm{M}$

Solution:
Hints : Molarity $=\frac{4 / 40}{250 / 1000}=0.4$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
The molarity of a solution obtained by mixing 750 mL of 0.5(M) HCl with 250 mL of 2(M) HCl will be

Options:
A 1.75 M
B 0.975 M
C 0.875 M
D 1.00 M

The correct answer is:
0.875 M

Solution:
M = M 1 V 1 + M 2 V 2 V 1 + V 2
= 7 5 0 × 0 . 5 + 2 5 0 × 2 1 0 0 0
= 3 7 5 + 5 0 0 1 0 0 0 = 0 . 8 7 5 M

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2013 (07 Apr)
The molarity of HNO3 in a sample which has density 1.4 g/mL and mass percentage of 63% is ________ (Molecular Weight of HNO3=63 )

Options:

The correct answer is:
14

Solution:
63%ww⟶HNO3 solution
M=63×1.463×100×1000mole/L
M=14mole/L

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2020 (09 Jan Shift 1)
The molarity of $\mathrm{NO}_{3}{ }^{-}$in the solution after $2 \mathrm{~L}$ of $3 \mathrm{M} \mathrm{AgNO}_{3}$ is mixed with $3 \mathrm{~L}$ of $1 \mathrm{M} \mathrm{BaCl}_{2}$ is

Options:
A $1.2 \mathrm{M}$
B $1.8 \mathrm{M}$
C $0.5 \mathrm{M}$
D $0.4 \mathrm{M}$

The correct answer is:
$1.2 \mathrm{M}$

Solution:
$2 \mathrm{~L}$ of $3 \mathrm{M} \mathrm{AgNO}_{3}$ will contains 6 moles of $\mathrm{AgNO}_{3}$
$3 \mathrm{~L}$ of $1 \mathrm{M} \mathrm{BaCl}_{2}$ will contain 3 moles of $\mathrm{BaCl}_{2}$
$2 \mathrm{AgNO}_{3}+\mathrm{BaCl}_{2} \longrightarrow 2 \mathrm{AgCl}$
$+\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}$
So, 6 moles of $\mathrm{AgNO}_{3}$ will react with 3 moles on $\mathrm{BaCl}_{2}$ it means, two solution will react completely to form 3 moles of $\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2} \equiv 6$ moles of $\mathrm{NO}_{3}^{-}$ions in $2+$ $3=5 \mathrm{~L}$ solution
Hence, molarity of $\mathrm{NO}_{3}^{-}=\frac{6}{5}=1.2 \mathrm{M}$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
The molarity of pure water is

Options:
A 55.6
B 5.56
C 100
D 18

The correct answer is:
55.6

Solution:
Molarity of pure water$=\frac{1000}{18}=55.6 \mathrm{M}$

Topic: Chemistry, Solutions, JEE Main
The molarity of the solution prepared by dissolving 6.3 g of oxalic acid H2C2O4·2H2O in 250 mL of water in mol L-1 is x×10-2. The value of x is _________ . (Nearest integer)
[Atomic mass : H : 1.0, C : 12.0, 0 : 16.0 J]

Options:

The correct answer is:
20

Solution:
M=Wsolute×1000GMMsolute×Vml

=6.3×1000126×250=0.2 M
=20×10-2
So X=20

Topic: Chemistry, Solutions, JEE Main, JEE Main 2021 (31 Aug Shift 1)
The mole fraction of a solute in a 100 molal aqueous solution ________ ×10-2.

(Round off to the Nearest Integer).

[Given : Atomic masses: H :1.0 u, O :16.0 u]

Options:

The correct answer is:
64

Solution:
100 molal aqueous solution means there is 100 mole solute in 1 kg=1000 gm water.

Now,

mole-fraction of solute=nsolute nsolute +nsolvent

=100100+100018=18002800=0.6428

=64.28×10-2

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2021 (17 Mar Shift 1)
The mole fraction of a solute in a binary solution is $0.1 .$ At $298 \mathrm{~K}$, molarity of this solution is same as as its molality. Density of this solution at $298 \mathrm{~K}$ is $2.0 \mathrm{~g} \mathrm{~cm}^{-3}$. The ratio of molecular weights of the solute and the solvent $\left(\mathrm{M}_{\text {solute }} / \mathrm{M}_{\text {solvent }}\right)$ is

Options:
A 9
B $\frac{1}{9}$
C $4.5$
D $\frac{1}{4.5}$

The correct answer is:
9

Solution:
$\mathrm{X}_{2}=0.1 \Rightarrow \frac{\mathrm{n}_{2}}{\mathrm{n}_{2}+\mathrm{n}_{1}}=0.1$ $\ldots(1)$
$\mathrm{M}=\frac{\mathrm{W}_{2} \times 1000}{\mathrm{M}_{2} \times \mathrm{W}} \times \mathrm{d}$ $\ldots(2)$
$\mathrm{m}=\frac{\mathrm{W}_{2} \times 1000}{\mathrm{M}_{2} \times \mathrm{W}_{1}(\mathrm{~g})}$ $\ldots(3)$
$\mathrm{d}=2 \mathrm{~g} / \mathrm{cc}$ $\ldots(4)$
Given M = m (numerically)
$\begin{array}{l}
\Rightarrow \frac{\mathrm{W}_{2} \times 1000}{\mathrm{M}_{2} \times\left(\mathrm{W}_{2}+\mathrm{W}_{1}\right)} \times \mathrm{d}=\frac{\mathrm{W}_{2} \times 1000}{\mathrm{M}_{2} \times \mathrm{W}_{1}(\mathrm{~g})} \\
\Rightarrow \mathrm{W}_{2}+\mathrm{W}_{1}=\mathrm{W}_{1} \times \mathrm{d}=2 \mathrm{~W}_{1} \\
\Rightarrow \mathrm{W}_{2}=\mathrm{W}_{1}
\end{array}$
Now, $\frac{n_{2}}{n_{2}+n_{1}}=0.1 \Rightarrow \frac{\frac{w_{2}}{M_{2}}}{\frac{W_{2}}{M_{2}}+\frac{W_{1}}{M_{1}}}=0.1 \Rightarrow \frac{\frac{W_{2}}{M_{2}}}{W_{2}\left(\frac{1}{M_{2}}+\frac{1}{M_{1}}\right)}=0.1$
$\Rightarrow \frac{\frac{1}{\mathrm{M}_{2}}}{\frac{\mathrm{M}_{1}+\mathrm{M}_{2}}{\mathrm{M}_{1} \mathrm{M}_{2}}}=0.1 \Rightarrow \frac{\mathrm{M}_{1}}{\mathrm{M}_{1}+\mathrm{M}_{2}}=\frac{1}{10} \Rightarrow \frac{\mathrm{M}_{1}}{\mathrm{M}_{1}+\mathrm{M}_{2}-\mathrm{M}_{1}}=\frac{1}{10-1} \Rightarrow \frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}=\frac{1}{9} \Rightarrow \frac{\mathrm{M}_{2}}{\mathrm{M}_{1}}=9$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
The mole fraction of a solute in a solution is 0.1. At 298 K, molarity of this solution is the same as its molality. Density of this solution at 298 K is 2.0 g cm-3 . The ratio of the molecular weights of the solute and solvent, MWsoluteMWsolvent , is

Options:

The correct answer is:
9

Solution:
XsoluteXsolvent=0.10.9=19
WsoluteWsolvent×MsolventMsolute=19 ...(i)
Wsolute+Wsolvent=Wsolution=density×volume
Wsolute+Wsolvent=2×V ...(ii)
Molarity = molality
nsoluteVsolution=nsoluteWsolvent
Wsolvent=Vsolution=Wsolute+Wsolvent2
2Wsolvent=Wsolute+Wsolvent
Wsolute=Wsolvent ...(iii)
Using equation (i) and (iii), we get
MsoluteMsolvent=9

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
The mole fraction of a solvent in aqueous solution of a solute is 0.8. The molality (in mol kg-1 ) of the aqueous solution is:

Options:
A 13.88×10-3
B 13.88×10-1
C 13.88
D 13.88×10-2

The correct answer is:
13.88

Solution:
Since mole fraction of solvent is 0.8, it implies that 10 moles of aqueous solution will have 8 moles of water and 2 moles of solute.

∴2 moles of solute is present in →8 moles of water

Or, 2moles of solute is present in →8×18gm of water

144 gm of water has 2 moles of solute
1 gm of water will have 2144moles of solute
1 kg of water will have 2144×1000=13.88 moles

∴ Molality = 13.88 m

Topic: Chemistry, Solutions, JEE Main, JEE Main 2019 (12 Apr Shift 1)
The mole fraction of benzene in a solution containing $39 \%$ by mass in an organic solvent of molecular mass 122 is

Options:
A $0.5$
B $0.6$
C $0.4$
D $0.35$

The correct answer is:
$0.6$

Solution:
Given, $39 \%$ by mass of organic solvent in benzene $\therefore 39 \mathrm{~g}$ of organic solvent in $100 \mathrm{~g}$ solution.
or
$39 \mathrm{~g}$ of organic solvent in $100-39=61$ g benzene.
Now, molar mass of organic solvent $=122 \mathrm{~g} / \mathrm{mol}$
Molar mass of benzene $\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)=12 \times 6+6=78 \mathrm{~g} / \mathrm{mol}$
$\therefore$ Moles of benzene $\left(n_{B}\right)=\frac{61}{78}$
Moles of organic solvent $\left(n_{\mathrm{O}}\right)=\frac{39}{122}$
Mole fraction of benzene $=\frac{n_{B}}{n_{B}+n_{\mathrm{O}}}=\frac{\frac{61}{78}}{\frac{61}{78}+\frac{39}{122}}=0.6$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
The mole fraction of ethanol in water is $0.08$. Its molality is

Options:
A $6.32 \mathrm{~mol} \mathrm{~kg}^{-1}$
B $4.83 \mathrm{~mol} \mathrm{~kg}^{-1}$
C $3.82 \mathrm{~mol} \mathrm{~kg}^{-1}$
D $2.84 \mathrm{~mol} \mathrm{~kg}^{-1}$

The correct answer is:
$4.83 \mathrm{~mol} \mathrm{~kg}^{-1}$

Solution:
Hint:
$\begin{aligned} \text { Molality } &=\frac{\mathrm{x}_{\mathrm{EtOH}}}{\mathrm{x}_{\mathrm{H}_{2} \mathrm{O}}} \times \frac{1000}{\mathrm{MW}_{\text {solvent }}} \\ &=\frac{0.08}{0.92} \times \frac{1000}{18} \\ &=4.83 \mathrm{~mol} \mathrm{~kg}^{-1} \end{aligned}$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
The mole fraction of glucose C6H12O6 in an aqueous binary solution is 0.1. The mass percentage of water in it, to the nearest integer, is ………

Options:

The correct answer is:
47

Solution:
Let total mole of solution =1

so mole of glucose =0.1

mole of H2O=0.9

% w/w of H2O=0.9×180.9×18+0.1×180×100

=47.368

=47.37

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2020 (03 Sep Shift 1)
The molecular electronic configuration of $\mathrm{H}_2^{-}$ion is

Options:
A ${ }^{(\sigma 1 s)^2}$
B $(\sigma 1 s)^2\left(\sigma^{\times} 1 s\right)^2$
C $(\sigma 1 s)^2\left(\sigma^x 1 s\right)^1$
D $(\sigma 1 s)^3$

The correct answer is:
$(\sigma 1 s)^2\left(\sigma^x 1 s\right)^1$

Solution:
$(\sigma 1 s)^2\left(\sigma^x 1 s\right)^1$

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na molecular weight=206. What would be the maximum uptake of Ca2+ ions by the resin if expressed in mol per gm?

Options:
A 1412
B 1103
C 1206
D 2309

The correct answer is:
1412

Solution:
2C8H7SO3Na aq+Ca2+ aq→C8H7SO32Ca aq+2Na+aq

1 mol calcium ion reacts with 2 moles of resin

2 mol resin=412 gresin≡40 gCa2+ions
=1 mol of Ca2+ions
∴ Moles of Ca2+ per g of resin=1412

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2015 (04 Apr)
The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Molecular weight 206). What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin?

Options:
A 2309
B 1412
C 1103
D 1206

The correct answer is:
1412

Solution:
2C8H7SO3-Na+2 mol(2×206=412 g)+Ca2+1 mol→C8H7SO3-2Ca2++2Na+
1 mol of Ca2+≡412 g of resin
Maximum uptake of Ca2+ ions by the resin
=1 mol412 g=1412mol /g

Topic: Chemistry, s Block Elements, JEE Main
The molecular formula of second homologue in the homologous series of mono carboxylic acids is _________.

Options:
A C3H6O2
B C2H4O2
C CH2O
D C2H2O2

The correct answer is:
C2H4O2

Solution:
The general molecular formula for carboxylic acid is CnH2n+1COOH.

In organic chemistry, a homologous series is a sequence of compounds with the same functional group and similar chemical properties in which the members of the series can be branched or unbranched, or differ by molecular formula of CH2 and molecular mass of 14u.

Second homologue : CH3COOH
=C2H4O2

Hence, the answer is B.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (27 Jan Shift 2)
The molecular geometry of SF6 is octahedral. What is the geometry of SF4 (including lone pair(s) of electrons, if (any)?

Options:
A Tetrahedral
B Trigonal bipyramidal
C Pyramidal
D Square planar

The correct answer is:
Trigonal bipyramidal

Solution:
SF4⇒Steric No=5 so hybridisation is sp3d

Geometry is trigonal bipyramidal but shape is “See Saw”.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2020 (02 Sep Shift 2)
The molecular shapes of $\mathrm{SF}_4, \mathrm{CF}_4$ and $\mathrm{XeF}_4$ are

Options:
A
the same with 2,0 and 1 lone pairs of electrons on the central atom, respectively

B
the same with 1, 1 and 1 lone pair of electrons on the central atoms, respectively

C
different with 0, 1 and 2 lone pair of electrons on the central atoms, respectively

D
different with 1, 0 and 2 lone pairs of electron on the central atoms respectively

The correct answer is:
different with 1, 0 and 2 lone pairs of electron on the central atoms respectively

Solution:
No Solution Available

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2005
The molecular weight of protein is

Options:
A $ \lt 10000$
B $\gt10000$
C $\gt1000$
D $\gt1000$ and $ \lt 10000$

The correct answer is:
$\gt10000$

Solution:
Proteins are composed of two or more polypeptide chains. Relatively shorter peptides are known as oligopeptides whereas longer polymers are called polypeptides. Polypeptides containing more than 100 amino acids having molecular mass higher than 10,000 are generally called proteins.

Topic: Chemistry, Biomolecules, JEE Main
The molecule having smallest bond angle is :

Options:
A
$\mathrm{NCl}_3$

B
$\mathrm{AsCl}_3$

C
$\mathrm{SbCl}_3$

D
$\mathrm{PCl}_3$

The correct answer is:
$\mathrm{SbCl}_3$

Solution:
As the size of central atom increases lone pair bond pair repulsions increases so, bond angle decreases.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2012 (Offline)
The molecule in which hybrid MOs involve only one d-robital of the central atom is :

Options:
A Ni(CN)42-
B BrF5
C XeF4
D CrF63-

The correct answer is:
Ni(CN)42-

Solution:
Complex	Hybridisation

(1) Ni(CN)42-	dsp2

(2) XeF4	sp3d2

(3) CrF63	sp3d2

(4) BrF5	sp3d2

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2020 (04 Sep Shift 2)
The molecule that has minimum or no role in the formation of photochemical smog, is:

Options:
A N2
B O3
C H2C=O
D NO

The correct answer is:
N2

Solution:
N2 is an inert gas and hence, it does not contribute to photochemical smog.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2019 (12 Jan Shift 1)
The molecule which contains maximum number of lone pairs is

Options:
A IF7
B XeF6
C XeF4
D XeF2

The correct answer is:
IF7

Solution:
IF7= lone pairs of I=0
Lone pairs on F=3Ã7=21
Total = 21
XeF6= lone pairs on Xe=1
Lone pairs on F=18
Total = 19
XeF4= lone pairs on Xe=2
Lone pairs on F=12
Total = 14
XeF2= Lone pairs on Xe=3
Lone pairs on F=6
Total = 9

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The molecule/ion with square pyramidal shape is:

Options:
A Ni(CN)42-
B PCl5
C BrF5
D PF5

The correct answer is:
BrF5

Solution:
BrF5 is an interhalogen compound, meaning that the molecules of this compound contain two or more different halogen atoms but no atoms of the elements belonging to any other group.

The hybridisation of bromine pentafluoride forms an sp3d2 hybridisation, resulting in a square pyramidal molecular structure. The hybrid molecule has five sigma bonds due to the covalent electron sharing.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2024 (30 Jan Shift 2)
The molecules of a given mass of a gas have

r.m.s. velocity of $200 \mathrm{~ms}^{-1}$ at $27^{\circ} \mathrm{C}$ and $1.0 \times 10^{5}$ $\mathrm{Nm}^{-2}$ pressure. When the temperature and pressure of the gas are respectively, $127^{\circ} \mathrm{C}$ and $0.05 \times 10^{5} \mathrm{Nm}^{-2},$ the r.m.s. velocity of its molecules in $\mathrm{ms}^{-1}$ is :

Options:
A $100 \sqrt{2}$
B $\frac{400}{\sqrt{3}}$
C $\frac{100 \sqrt{2}}{3}$
D $\frac{100}{3}$

The correct answer is:
$\frac{400}{\sqrt{3}}$

Solution:
Here $v_{1}=200 \mathrm{~m} / \mathrm{s}$; temperature $\mathrm{T}_{1}=27^{\circ} \mathrm{C}=27+273=300 \mathrm{k}$

temperature $\mathrm{T}_{2}=127^{\circ} \mathrm{C}=127+273=400 \mathrm{k}$

$\mathrm{V}=?$

R.M.S. Velocity, $\mathrm{V} \propto \sqrt{\mathrm{T}}$

$$

\begin{array}{l}

\Rightarrow \quad \frac{v}{200}=\sqrt{\frac{400}{300}} \\

\Rightarrow \quad v=\frac{200 \times 2}{\sqrt{3}} \mathrm{~m} / \mathrm{s} \\

\Rightarrow \quad \mathrm{v}=\frac{400}{\sqrt{3}} \mathrm{~m} / \mathrm{s}

\end{array}

$$

Topic: Physics, Kinetic Theory of Gases, JEE Main
The moles of hydrocarbon produced by hydrolysis of \( 1 \) mole \( \mathrm{Mg}_{2} \mathrm{C}_{3} \) (magnesium carbide) is

Options:

The correct answer is:
1

Solution:
MgC2 on heating gives Mg2C3. This carbide contains C34- units and reacts with water to form propyne (methyl acetylene).

Mg2C3+4H2O⟶2MgOH2+CH3-C≡CHPropyne 1 mole

Topic: Chemistry, s Block Elements, JEE Main
The moles of methane required to produce 81 g of water after complete combustion is____×10-2 mol. [nearest integer]

Options:

The correct answer is:
225

Solution:
CH4+2O2⟶CO2+2H2O

POAC on H atom

nCH4×4=nH2O×2

nCH4=8118×2×14=8136

nCH4=2.25

=225×10-2

Hence, the number of moles of methane is 225×10-2

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2022 (26 Jun Shift 2)
The moment of inertia of a circular ring of mass 1 kg about an axis passing through its centre and perpendicular to its plane is 4 kg m2. The diameter of the ring is

Options:
A 2 m

B 4 m

C 5 m

D 6 m

The correct answer is:
4 m

Solution:
Moment of inertia of circular ring about an axis passing through its centre of mass and perpendicular to its plane
I=MR2
Here, I=4 kg m2, m=1 kg
R2=41=4
or R=2 m
Therefore, diameter of ring = 4 m.

Topic: Physics, Rotational Motion, JEE Main
The moment of inertia of a door of mass m, length 2 â and width â about its longer side is

Options:
A 1 1 m ℓ 2 2 4
B 5 m ℓ 2 2 4
C m ℓ 2 3
D None of these

The correct answer is:
m ℓ 2 3

Solution:
I about YY ′ = m â 2 1 2
Using parallel axis theorem :
I (about AD) = m â 2 1 2 + m â 2 4 = m â 2 3

Topic: Physics, Rotational Motion, JEE Main
The moment of inertia of a hollow cubical box of mass M and side length a, about an axis passing through centres of two opposite faces, is equal to xMa218. The value of x is

Options:

The correct answer is:
5

Solution:
Let m be the mass of each face of the cubical box.
I=2Ãma26+4Ãma212+ma22
=5ma23
But, m=M6
â´I=5Ma218

âx=5

Topic: Physics, Rotational Motion, JEE Main
The moment of inertia of a rigid body in terms of its angular momentum L and kinetic energy K is

Options:
A L2K
B L22K
C L2K2
D L2K

The correct answer is:
L22K

Solution:
Angular momentum of a rigid body about a fixed axis is given by
L=IÏ
Where I is moment of inertia and Ï is angular velocity about that axis.
Kinetic energy of body is given by
K=12 IÏ2
â´ K=12IIÏ2=L22I
â I=L22K

Topic: Physics, Rotational Motion, JEE Main
The moment of inertia of a semicircular ring about an axis, passing through the center and perpendicular to the plane of ring, is 1x MR2, where R is the radius and M is the mass of the semicircular ring. The value of x will be _______.

Options:

The correct answer is:
1

Solution:
The moment of inertia of a semicircular ring about a line perpendicular to the plane of the ring and passing through its centre is given as I=MR2, where m and R are the mass and radius of the ring. Comparing the given value

.I= MR2=1xMR2

⇒x=1

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2023 (08 Apr Shift 1)
The moment of inertia of a thin rectangular plate \( A B C D \) of uniform thickness about an axis passing through the centre \( O \) and perpendicular to the plane of the plate is

Options:
A \( I_{1}+I_{2} \)
B \( I_{2}+I_{4} \)
C \( I_{1}+I_{3} \)
D \( I_{1}+I_{2}+I_{3}+I_{4} \)

The correct answer is:
\( I_{2}+I_{4} \)

Solution:
According to the theorem of perpendicular axis, the moment of inertia for any plane body for either of its axes direction normal to the plane is equivalent to the total of the moment of inertia along the two normal axis in the plane of the body that intersect the first axis in the plane.

From the figure moment of inertia about an axis passing through the centre O and perpendicular to the plane of the plate is I=I2+I4.

Since the plate is rectangular of uniform thickness the x-axis and y-axis will not be symmetrical, i.e, AC and BD are not perpendicular to each other and this gives I1≠I3.

Topic: Physics, Rotational Motion, JEE Main