datavorous/entrance-exam-dataset huggingface dataset:
Calculate the cell potential of following cell

PtsH2g0.1 barBOH0.1M|HA0.1MH2g1 bar|Pt

Given KaHA=10-7,KbBOH=10-5

Options:
A 0.39 V
B 0.36 V
C 0.93 V
D 0.63 V

The correct answer is:
0.39 V

Solution:
Ecell=0.0592logâ¡[H+]cathode2Ã[PH2]anode[H+]anode2Ã[PH2]cathode
OH-=CKb=0.1Ã10-5=10-3M
[H+]anode=10-1410-3=10-11M
[H+]cathode=CKa=0.1Ã10-7=10-4M
So Ecell=0.0592logâ¡(10-4)2Ã0.1(10-11)2Ã1
=0.39 V.

Topic: Chemistry, Electrochemistry, JEE Main
Calculate the compressional force required to prevent the metallic rod of length l cm and cross-sectional area A cm2 when heated through t Â°C, from expanding lengthwise. Young's modulus of elasticity of the metal is E and mean coefficient of linear expansion is Î± per degree celsius.

Options:
A EAαt
B EAαt1+αt
C EAαt1-αt
D El⁡αt

The correct answer is:
EAαt1+αt

Solution:
The change in natural length = Î lâ¡ t = lâ¡ Î± t
The natural length of rod at temperature t â C = lâ¡ + lâ¡ Î± t
The decrease in natural length due to developed stress = Î lâ¡
But the length of rod remains constant.
â´ Î lâ¡ t - Î lâ¡ = 0
â´ Î lâ¡ = Î lâ¡ t = lâ¡ Î± t
â´ E = stress strain = F/A - Î lâ¡ lâ¡ + Î lâ¡ t
â´ E = - EA Î lâ¡ lâ¡ + Î lâ¡ t = - EA Î lâ¡ t lâ¡ + Î lâ¡ t
= - EA lâ¡ Î± t lâ¡ + lâ¡ Î± t = - EA Î± t 1 + Î± t

Topic: Physics, Thermal Properties of Matter, JEE Main
Calculate the concentration (in percentage by weight) of a solution obtained by mixing 300 g 25% by weight solution of NH4Cl and 150 g of 40% by weight solution of NH4Cl .

Options:

The correct answer is:
30

Solution:
Weight of solution I and II = 300 + 150 = 450 g
Weight of NH4Cl in I solution=25Ã300100= 75 g
Weight of NH4Cl in II solution =40Ã150100= 60 g
â´ Total weight of NH4Cl = 75 + 60 = 135 g
â´ % by wt. of mixed solution =135450Ã 100 = 30%

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
Calculate the couple (in Î¼N m) acting on a magnet of length 10 cm and pole strength 15 A m, kept in a field of B=2Ã10-5 T at an angle of 30Â°.

Options:

The correct answer is:
15

Solution:
C=MBsin⁡θ
=m×2l×2×10-5sin⁡30°
=15×10×10-2×2×10-5×12
=15×10-6 N m

Topic: Physics, Magnetic Properties of Matter, JEE Main
Calculate the density (in kg m-3 ) of potassium hving a bcc structure with nearest neighbour distance of 4.52 Å. (Given atomic weight of potassium is 39)

Options:

The correct answer is:
910

Solution:
For bcc, d=32a
Or a=2d3=2×4.521.732=5.219A∘=522pm
ρ=Z×Ma3×N0×10-30
=2×39(522)3×(6.023×1023)×10-30=0.91 g/cm3=910 kg m-3

Topic: Chemistry, Solid State, JEE Main
Calculate the depression in the freezing point of a solution containing $0.1 \mathrm{~g}$ of $\mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]$ in $100 \mathrm{~g}$ of $\mathrm{H}_2 \mathrm{O}$ ? (molecular weight of $\mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]=329 ; K_f=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^1$ )

Options:
A 0
B 1.223
C 0.0226
D 0.226

The correct answer is:
0.0226

Solution:
Given,
$$
\begin{aligned}
& \text { Mass of } \mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]=0 \mathrm{lg} \\
& \text { Mass of solvent }=100 \mathrm{~g} \\
& \text { Molar mass of } \mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]=329 \mathrm{~g} \mathrm{~mol}^{-1} \\
& k_f=1.86 \mathrm{~K} \mathrm{Kg} / \mathrm{mol} \\
& \mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right] \rightleftharpoons \\
& \left.i \mathrm{~K}^{+}+\mathrm{Fe}(\mathrm{CN})_6\right]^{-} \\
& i\left(\operatorname{van}^{\prime} \mathrm{t} \text { Hoff factor }\right)=3+1=4
\end{aligned}
$$
From, depression in freezing point,
$$
\begin{aligned}
& T_f=i \times K_f \times m \\
& T_f=4 \times 1.86 \mathrm{Kkg} / \mathrm{moI} \times \frac{0 \mathrm{gg}}{329 \mathrm{~g}} \times \frac{1000}{100 \mathrm{~kg}} \\
& \mathrm{~T}_f=0.0226 \mathrm{~K}
\end{aligned}
$$
Thus, option (3) is correct.

Topic: Chemistry, Solutions, JEE Main
Calculate the emf of the cell
$\mathrm{Cu}(s)\left|\mathrm{Cu}^{2+}(a q)\right|\left|\mathrm{Ag}^{+}(a q)\right| \mathrm{Ag}(s)$
Given
$E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^0=+0.34 \mathrm{~V}, E_{\mathrm{Ag}^{+} / \mathrm{Ag}}^0=0.80 \mathrm{~V}$

Options:
A $+0.46 \mathrm{~V}$
B $+1.14 \mathrm{~V}$
C $+0.57 \mathrm{~V}$
D $-0.46 \mathrm{~V}$

The correct answer is:
$+0.46 \mathrm{~V}$

Solution:
$\begin{aligned} E_{\text {cell }}^{\circ} & =E_{\text {red (cathode) }}^{\circ}-E_{\text {oxi (anode) }}^{\circ} \\ & =E_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\circ}-E_{\mathrm{Cu}^{+} / \mathrm{Cu}^{2+}}^0 \\ & =0.80-(+0.34)=+0.46 \mathrm{~V}\end{aligned}$

Topic: Chemistry, Electrochemistry, JEE Main
Calculate the emf of the cell in which the following reaction takes place :

Ni(s) + 2 Ag+ (0.002 M) â¶ Ni2+ (0.160 M) + 2 Ag(s)

Given that E cell o = 1 . 0 5 V

Options:

The correct answer is:
0.91

Solution:
From the given cell reaction and nernst equation.
E cell = E cell o - 0 . 0 5 9 1 n log Ni 2 + Ag + 2
= 1 . 0 5 V - 0 . 0 5 9 1 2 log 0 . 1 6 0 0 . 0 0 2 2
= 1 . 0 5 - 0 . 0 5 9 1 2 log 4 × 1 0 4
= 1 . 0 5 - 0 . 0 5 9 1 2 4 . 6 0 2 1
= 1 . 0 5 - 0 . 1 4 = 0 .9 1 V
E cell = 0 . 9 1 V

Topic: Chemistry, Electrochemistry, JEE Main
Calculate the enthalpy change for the reaction, \( \mathrm{Fe}_{2} \mathrm{O}_{3}+3 \mathrm{CO} \rightarrow 2 \mathrm{Fe}+3 \mathrm{CO}_{2} ; \) from the following data:
\( 2 \mathrm{Fe}+\frac{3}{2} \mathrm{O}_{2} \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3} ; \Delta \mathrm{H}=-177.1 \mathrm{kcal} \)
\( \mathrm{C}+\frac{1}{2} \mathrm{O}_{2} \rightarrow \mathrm{CO} ; \Delta \mathrm{H}=-32.8 \mathrm{kcal} \)
\( \mathrm{C}+\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} ; \Delta \mathrm{H}=-94.3 \mathrm{kcal} \)

Options:

The correct answer is:
-7.4

Solution:
Let the enthalpy of the given reaction be △Hr.

Fe2O3(s)+3CO(g)→2Fe(s)+3CO2 (g) ...(i)

Given equation:

2Fe(s)+32O2(g)→Fe2O3(s); ΔH=-177.1 kcal ....(ii)

C(s)+12O2(g)→CO(g); ΔH=-32.8kcal ....(iii)

C(s)+O2(g)→CO2(g); ΔH=-94.3kcal ....(iv)

Enthalpy is an extensive property.

Equation (i) can be obtained from equation (ii), (iii) and (iv) as given below:

Equation (iv)×3-Equation (iii)×3-Equation (ii)≡Equation (i)

Thus, according to Hess' law,[(-98.3)×3]-[(-32.8)×3]-[-177.1]=△Hr

△Hr=-282.9+98.4+177.1=-282.9+275.5

△Hr=-7.4 kcal

Thus, the enthalpy of equation (i) is -7.4 kcal.

Topic: Chemistry, Thermodynamics (C), JEE Main
Calculate the enthalpy change when \( 50 \mathrm{~mL} \) of \( 0.01 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2} \) reacts with \( 25 \mathrm{~mL} \) of \( 0.01 \mathrm{M} \mathrm{HCl} \). Given that, \( \Delta \mathrm{H}^{\circ} \) of neutralization of a strong acid and a strong base is \( 140 \mathrm{kcal} \mathrm{mol}^{-1} \).

Options:
A \( 14 \mathrm{kcal} \)
B \( 35 \mathrm{cal} \)
C \( 10 \mathrm{cal} \)
D \( 7.5 \mathrm{cal} \)

The correct answer is:
\( 35 \mathrm{cal} \)

Solution:
Number of moles of HCl =MV1000=0.01×251000=25×10-5
HCl→H++Cl-

Moles of H+=25×10-5
Number of moles of CaOH2=MV1000=0.01×501000=50×10-5
Moles of OH-=2×50×10-5=10-3
In the process of neutralization, 25×10-5 moles of H+ will be completely neutralized.
∴∆H=140×25×10-5 kcal=0.035 kcal=35 cal

Topic: Chemistry, Thermodynamics (C), JEE Main
Calculate the equilibrium constant for the reaction (in 1012)

Fe 2 + + Ce 4 + â Fe 3 + + Ce 3 +

Given, E o Ce 4 + / Ce 3 + = 1.44 V, E o Fe 3 + / Fe 2 + = 0.68 V RTF=.0591

Options:

The correct answer is:
7.07

Solution:
Fe 2 + + Ce 4 + â Fe 3 + + Ce 3 +
E o = E o Ce 4 + / Ce 3 + - E o Fe 3 + / Fe 2 +
= 1.44 - 0.68 = 0.76 V
âµ Eo = 0.059 log K
â log K = E o 0.0592
= 0.760.059=12.85
â K = 7.07Ã1012

Topic: Chemistry, Electrochemistry, JEE Main
Calculate the force acting between two small magnets, placed in end on position \( 0.1 \mathrm{~m} \) apart from their centres. Given magnetic moment of each magnets is \( 5 \mathrm{Am}^{2} \)

Options:
A \( 0.6 \mathrm{~N} \)
B \( 0.8 \mathrm{~N} \)
C \( 0.15 \mathrm{~N} \)
D \( 0.2 \mathrm{~N} \)

The correct answer is:
\( 0.15 \mathrm{~N} \)

Solution:
According to Question

Two small bar magnet is r=0.1 m apart from their centre

Magnetic moment=M1=M2=5 Am2

Now The relation of forceF and magnetic moment M in two small bar magnet

F=μ04π·6M1M2 r 4

Put the all known values in the above relation
F=10-7×6×5×50.14

F=0.15 N

Topic: Physics, Magnetic Effects of Current, JEE Main
Calculate the fractional error \( \left(\frac{\Delta x}{x}\right) \), if \( x=a^{n} \)

Options:
A \( \pm\left(\frac{\Delta a}{a}\right)^{n} \)
B \( \pm n\left(\frac{\Delta a}{a}\right) \)
C \( \pm n \log _{\mathrm{e}} \frac{\Delta a}{a} \)
D \( \pm \frac{n \log \Delta a}{a} \)

The correct answer is:
\( \pm n\left(\frac{\Delta a}{a}\right) \)

Solution:
x=an
∴ Δxx= ±nΔaa

Topic: Physics, Mathematics in Physics, JEE Main
Calculate the $\mathrm{H}^{+}$ion concentration in a $1.00(\mathrm{M}) \mathrm{HCN}$ litre solution $\left(K_a=4 \times 10^{-10}\right)$

Options:
A $4 \times 10^{-14}$ mole /litre
B $2 \times 10^{-5} \mathrm{~mole} / \mathrm{litre}$
C $2.5 \times 10^{-5} \mathrm{~mole} / \mathrm{litre}$
D None of these

The correct answer is:
$2 \times 10^{-5} \mathrm{~mole} / \mathrm{litre}$

Solution:
$\left[H^{+}\right]=\sqrt{K \times C}$
$\left[H^{+}\right]=\sqrt{4 \times 10^{-10} \times 1} ;\left[H^{+}\right]=2 \times 10^{-5}$ mole/litre

Topic: Chemistry, Ionic Equilibrium, JEE Main
Calculate the half-life of the first-order reaction
C2H4Og→CH4g+COg
if the initial pressure of C2H4Og is 80 mm and the total pressure at the end of 20 minutes is 120 mm.

Options:
A 40 min
B 120 min
C 20 min
D 80 min

The correct answer is:
20 min

Solution:
According to the question,
p0-p+p=120 mm
or p0+p=120 mm
or p=120-80=40 mm.
For a first order reaction,
k=1tln⁡p0p0-p=120ln⁡8080-40=120ln⁡2 …(i)
But we know that t12=1n2k …(ii)
From equations (i) and (ii)
k=120kt12
t12=20 min.

Topic: Chemistry, Chemical Kinetics, JEE Main
Calculate the heat produced (in kJ ) when 224 gm of CaO is completely converted to CaCO 3 by reaction with CO 2 at 27 o C in a container of fixed volume.

Given : ÎH f o [ CaCO 3 ( s ) ]=â1207 kJ/mol ; Î H f o [ CaO( s ) ]=â635 kJ/mol

Î H f â CO 2 , g = - 394 kJ/mol ; Use R = 8.3 K -1 mol -1

Options:
A -702.04 kJ
B 721.96 kJ
C 712 kJ
D 721 kJ

The correct answer is:
-702.04 kJ

Solution:
CaO s + CO 2 g â CaCO 3 s
Î r H â = Î H f â CaCO 3 - Î H f â CaO - Î H f â CO 2
= - 1 2 0 7 - - 6 3 5 - - 3 9 4
= - 178 kJ/mol
â´ Î E r â = Î H r â - Î n g RT
Î E r â = - 1 7 8 - - 1 Ã 8.3 Ã 3 0 0 1 0 0 0
= - 175.51 kJ
n CaO = 2 2 4 5 6 = 4
â´ q v = n Â· Î r E = 4 Ã - 175.51
= - 702.04 kJ

Topic: Chemistry, Thermodynamics (C), JEE Main
Calculate the increase in internal energy of system when 40 joule heat is supplied and the work done by a system is 8 joule

Options:
A 25 J
B 30 J
C 32 J
D 28 J

The correct answer is:
32 J

Solution:
q=40 J
w=-8 J (work done by the system)
ΔE=q+w=40-8=32 J .

Topic: Chemistry, Thermodynamics (C), JEE Main
Calculate the limit of resolution of a telescope objective having a diameter of 200 cm , if it has to detect light of wavelength 500 nm coming from a star.

Options:
A 305×10-9 radian
B 610×10-9 radian
C 152.5×10-9 radian
D 457.5×10-9 radian

The correct answer is:
305×10-9 radian

Solution:
Limit of resolution of the telescope =1.22λD
θ=1.22×500×10-9200×10-2=305×10-9 radian

Topic: Physics, Ray Optics, JEE Main, JEE Main 2019 (08 Apr Shift 2)
Calculate the maximum acceleration of a moving car so that a body lying on the floor of the car remains stationary. The coefficient of static friction between the body and the floor is 0.15 g=10 m s-2

Options:
A 50 m s-2
B 1.2 m s-2
C 150 m s-2
D 1.5 m s-2

The correct answer is:
1.5 m s-2

Solution:
When the body remains stationary on the floor of the car, let's consider the following free-body diagram of the body:

The formula to calculate the Newtonian force on the body is given by

F=mamax ...1

And, the frictional force on the body by the surface of the car is given by

f=μsmg ...2

Equate equations (1) and (2) and simplify to obtain the required maximum acceleration of the car.

mamax=μsmg⇒amax=μsg ...3

Substitute the values of the known parameters into equation (3) to calculate the required parameter.

amax=0.15×10 m s-2= 1.5 m s-2

Topic: Physics, Laws of Motion, JEE Main
Calculate the maximum acceleration of a moving car so that a body lying on the floor of the car remains stationary. The coefficient of static friction between the body and the floor is 0.15 g=10 m s-2

Options:
A 50 m s-2
B 1.2 m s-2
C 150 m s-2
D 1.5 m s-2

The correct answer is:
1.5 m s-2

Solution:
When the body remains stationary on the floor of the car, let's consider the following free-body diagram of the body:

The formula to calculate the Newtonian force on the body is given by

F=mamax ...1

And, the frictional force on the body by the surface of the car is given by

f=μsmg ...2

Equate equations (1) and (2) and simplify to obtain the required maximum acceleration of the car.

mamax=μsmg⇒amax=μsg ...3

Substitute the values of the known parameters into equation (3) to calculate the required parameter.

amax=0.15×10 m s-2= 1.5 m s-2

Topic: Physics, Laws of Motion, NEET, NEET 2023 (All India)
Calculate the minimum mass of Cl2 in grams required to produce 37.25 g KCl?

Cl2+2KOH→KCl+KClO+H2O

Options:

The correct answer is:
35.5

Solution:
Cl2+2KOHâKCl+KClO+H2O
37.25 g
37.2574.5mol
0.5 mol KCl
Mole of Cl2 required = 0.5 mol
Mass of Cl2 required =0.5Ã71=35.5 g

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
Calculate the molal depression constant of a solvent which has freezing point $16.6^{\circ} \mathrm{C}$ and latent heat of fusion $180.75 \mathrm{Jg}^{-1}$

Options:
A 2.68
B 3.86
C 4.68
D 2.86t6

The correct answer is:
3.86

Solution:
$\begin{gathered}K_f=\frac{R T_f^2}{1000 \times L_f}, R=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \\ T_f=273+16.6=289.6 \mathrm{~K} ; L_f=180.75 \mathrm{Jg}^{-1} \\ K_f=\frac{8.314 \times 289.6 \times 289.6}{1000 \times 180.75} = 3.86\end{gathered}$

Topic: Chemistry, Solutions, JEE Main
Calculate the molality of KI if the density of 20% (mass/mass) aqueous KI is 1.202 gmL-1. Report your final answer by rounding off the molality correct upto one place of decimal and then multiplying it by 10.

Options:

The correct answer is:
15.00

Solution:
Weight of KI in 100g of water = 20 g
Weight of water in the solution = 100 - 20 = 80 g= 0.08 kg
Molar mass of KI = 39 + 127 = 166 g mol-1
Molality of the solution (m) = Number of moles of KI Mass of water in kg
= 2 0 g / 1 6 6 g mol - 1 0 Â· 0 8 kg
= 1.506 mol kg-1 = 1.506 m
By rounding off to one place, we get the molality as 1.5 and by further multiplying it with 10, we get "15" as the final answer.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
Calculate the molecular weight of the given organic compound. If in Victor Meyer's method 0.2g of an organic substance displaced 56ml of air at STP.

Options:
A 56
B 112
C 80
D 28

The correct answer is:
80

Solution:
Molecular mass = Weight of organic substance/Air displaces at STP×22400
= 0.2/56 × 22400 = 80

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
Calculate the number of hours of service that can be derived at 1 atm, 300 K from an acetylene lamp containing 640 g calcium carbide. Given that the lamp requires 50 L acetylene gas at 1 atm 300 K for one hour. [Take 0.0821 Ã 300=25]

Options:

The correct answer is:
5

Solution:
CaC2+2H2O â CaOH2+C2H2
moles of Ca C 2 = 640 64 =10
â´ moles of C2H2 formed = 10
â´ V C 2 H 2 = 10 Ã 0.0821 Ã 300 = 250 Litres
â´ number of hours = 250 50 =5

Topic: Chemistry, States of Matter, JEE Main
Calculate the number of moles of NH4Cl that should be added to one litre of 1.0 M NH4OH  to prepare buffer solution with pH = 9  K b =2 ×  10 −5 , take log 2 = 0.3

Options:
A 3.4
B 2.6
C 1.5
D 2

The correct answer is:
2

Solution:
It is basic buffer
pOHâ=â pK b â+âlogâ [salt] [weakâbase]
pOHâ=â14â9=5
5â=â4.7+ logâ x 1
0.3â=âlogââxâxâ=â2

Topic: Chemistry, Ionic Equilibrium, JEE Main
Calculate the number of moles of NaOH required to completely neutralise 100 g of 118% oleum

Options:
A 2.4
B 1.2
C 4.8
D 8.4

The correct answer is:
2.4

Solution:
Percent labeling of oleum = 100 + Total mass of water required to completely convert oleum intoH2SO4

So, we need 18 g of water for the reaction SO3+H2O→H2SO4

one mole of water is required, which in turn means that one mole of SO3​ is present. Therefore, in a 100 g sample, 80 g is SO3â.

Hence, the mass of H2SO4​= 20 g.

Number of moles of SO3=8080=1

Number of moles of H2SO4=2098=0.20

one mole of H2SO4â requires two moles of NaOH for complete neutralization

one mole of SO3â requires two moles of NaOH for complete neutralization

SO3+2NaOH→Na2SO4+H2O

So, number of moles of NaOH required for neutralization of 100g of 118% oleum=2×Number of SO3=2×number of moles of H2SO4

=2×1+2×0.2=2.4 moles

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
Calculate the number of oxygen atoms required to combine with 7g of N2 to form N2O3 when 80% of N2 is converted to N2O3 .

Options:
A 2.3×1023
B 3.6×1023
C 1.8×1023
D 5.4×1021

The correct answer is:
3.6×1023

Solution:
for formation of N2O3
2N2+3O2→2N2O3
2 moles of N2 combines with 3mole of O2
Given, 7gm N2 (80% only reacts ) i.e.5.6 gm of N2
Mole of N2=5.628=0.2
Hence 20.2=3xmoleofO2
x=0.3 moles
Number of atoms =0.3×2×6.023×1023
=3.61×1023

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
Calculate the number of unit cells present in a cube-shaped ideal crystal of NaCl of mass 1.00 g [Atomic masses: Na=23, Cl=35.5]

Options:
A 2.57×1021 unit cells
B 5.14×1021 unit cells
C 1.28×1021 unit cells
D 1.71×1021 unit cells

The correct answer is:
2.57×1021 unit cells

Solution:
Number of NaCl unit in one gram =158.5×6.023×1023=1.029×1022
A unit cell contains 4 Na+ ion and 4 Cl- ions
Unit cell =1.029×10224=2.57×1021 unit cell

Topic: Chemistry, Solid State, JEE Main
Calculate the number of waves made by a Bohr electron in one complete revolution in nth orbit of He+ ion, if ratio of de-Broglie wavelength associated with electron moving in nth orbit and 2nd orbit is 2.0 :-

Options:

The correct answer is:
4

Solution:
nλ=2πr
nλ=2×π×0.529×n2Z
λn=2π×0.529×nZ ------ (i)
λ2=2π×0.529×2Z ------ (ii)
Eq (i) ÷ (ii)
λnλ2=2=n2
n=4

Topic: Chemistry, Structure of Atom, JEE Main
Calculate the octane number of gasoline fuel, which contains 25% n-heptane and 75% is iso-octane .

Options:
A 25
B 50
C 75
D 100

The correct answer is:
75

Solution:
Octane number is defined as the percentage by volume of iso-octane present in the mixture of iso-octan⁡e and n-heptan⁡e which has the same antiknocking properties as the fuel under examination.
Given fuel (25% n-heptane +75% iso-octane) Hence, octane number = 75 (because iso octane is 75%)

Topic: Chemistry, Hydrocarbons, JEE Main
Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185000 in 450 mL of water at 37o C.

Options:

The correct answer is:
30.95

Solution:
Mass of polymer (WB) = 1.0 g
Molar mass of polymer (MB) = 185000 g mol-1
Volume of solution (V) = 450 mL = 0.450 L
Temperature (T) = 37 + 273 = 310 K
Solution constant (R) = 8.314 x 103 Pa L K-1 mol-1
Osmotic pressure π = CRT
= W B × R × T M B × V
π = 1.0 g × 8.314 × 10 3 Pa L K - 1 mol - 1 × 310 K 185000 g mol - 1 × 0.450 L
= 30.95 Pa

Topic: Chemistry, Solutions, JEE Main
Calculate the pH of solution when 100 mL, 0.1M CH3COOH and 100 mL, 0.1M HCOOH are mixed together. (Given: KaCH3COOH=2Ã10â5,KaHCOOH=6Ã10â5 )

Options:

The correct answer is:
2.7

Solution:
As the volume is doubled, concentration would be half. The total H + is obtained by addition of H + from both acids.
[H+⁡]=Ka1C1+Ka2C2
=2×10-5×0.12+6×10-5×0.12
=2×10-3
pH=2.7

Topic: Chemistry, Ionic Equilibrium, JEE Main
Calculate the partial pressure of carbon monoxide from the following
$\mathrm{CaCO}_{3(s)} \xrightarrow{\Delta} \mathrm{CaO}_{(s)}+\mathrm{CO}_2 \uparrow: \mathrm{K}_p=8 \times 10^{-2}$
$\mathrm{CO}_{2(\mathrm{~g})}+\mathrm{C}_{(\mathrm{s})} \rightarrow 2 \mathrm{CO}_{(\mathrm{g})} ; \mathrm{K}_p=2$

Options:
A 0.2
B 0.4
C 1.6
D 4

The correct answer is:
0.4

Solution:
Given, $\mathrm{CaCO}_3(\mathrm{~s}) \xrightarrow{\Delta} \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g}) \uparrow$
$\mathrm{C}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})=2 \mathrm{CO}(\mathrm{g})$
$\mathrm{Kp}_2=\frac{[p C O]^2}{\left[p \mathrm{CO}_2\right]} ; p \mathrm{pO}=\sqrt{\left[\mathrm{Kp}_1 \times \mathrm{Kp}_2\right]}$
$p C O=\sqrt{\left[8 \times 10^{-2} \times 2\right]}=\sqrt{16 \times 10^{-2}}=4 \times 10^{-1}=0.4$

Topic: Chemistry, Chemical Equilibrium, JEE Main
Calculate the percentage of all the monochlorinated products obtained from the chlorination of 2 - methyl butane. The relative reactivity of 1 ∘ , 2 ∘ , and 3 ∘ hydrogen to chlorination is (1 : 3.8 : 5).

Options:
A 15%, 45%, 20%, 25%
B 14%, 35%, 23%, 28%
C 12%, 30%, 22%, 29%
D 17%, 36%, 20%, 23%

The correct answer is:
14%, 35%, 23%, 28%

Solution:
There are four different (A, B, C and D) monochloro products.

Relative amount

(A)

(A) is obtained from the reaction of three equivalent 1 â H atoms.	3 x 1 = 3.0

(B)

(B) is obtained from the reaction of two equivalent 2 â H atoms.	2 x 3.8 = 7.6

(C)

(C) is obtained from the reaction of one equivalent 3 â H atoms.	1 x 5 = 5.0

(D)	(D) is obtained from the reaction of six equivalent 1 â H atoms.
6 x 1 = 6.0

Total = 21.6

Therefore, percentages of the A, B, C and D monochloro products are as follows :
A = 3 2 1 . 6 × 1 0 0 = 1 3 . 9 ≈ 1 4 %
B = 7 . 6 2 1 . 6 × 1 0 0 = 3 5 . 1 8 ≈ 3 5 %
C = 5 . 0 2 1 . 6 × 1 0 0 = 2 3 . 1 5 ≈ 2 3 %
D = 6 . 0 2 1 . 6 × 1 0 0 = 2 7 . 8 ≈ 2 8 %

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Calculate the percentage of packing efficiency in simple cubic unit cell.

Options:

The correct answer is:
52.40

Solution:
In simple cubic, packing efficiency
= Volume of one atom × 1 0 0 Volume of unit cell
= 4 3 πr 3 8r 3  × 100 =  π 6  × 100 = 52.4%
where, r = radius of one sphere or atom.

Topic: Chemistry, Solid State, JEE Main
Calculate the $\mathrm{pH}$ of a buffer solution containing $0.01 \mathrm{M}$ salt and $0.004 \mathrm{M}$ weak acid.
$\left(\mathrm{pK}_{\mathrm{a}}=4.762\right)$

Options:
A $4.36$
B $4.76$
C $5.16$
D $5.36$

The correct answer is:
$5.16$

Solution:
$\begin{aligned} \mathrm{pH} & =\mathrm{pK}_{\mathrm{a}}+\log _{10} \frac{[\text { Salt }]}{[\text { Acid }]} \\ & =4.762+\log _{10} \frac{0.01}{0.004} \\ & =4.762+\left(\log _{10} 5-\log _{10} 2\right) \\ & =4.762+(0.699-0.3010) \\ & =4.762+0.398 \\ & =5.16\end{aligned}$

Topic: Chemistry, Ionic Equilibrium, JEE Main
Calculate the ratio of wavelength for an α particle and proton accelerated through same potential difference.

Options:
A 12
B 12
C 122
D 22

The correct answer is:
122

Solution:
Using de Broglie equation λ=h2qmV
λ= wavelength, q = charge, m = mass, V = Potential and h = Plank constant
λa=h2qamaV λp=h2qpmpV
ma=4mp,qa=2qp
λaλp=qpmpqama=12.14=18
λaλp=122

Topic: Chemistry, Structure of Atom, JEE Main
Calculate the resulting temperature when 20 g of boiling water is poured into an ice-cold brass vessel (specific heat = 0.1 cal g-1 Â°C-1 ) of mass 100 g.

Options:
A 66.66 Â°C
B 6.66 Â°C
C 0.66 Â°C
D 50 Â°C

The correct answer is:
66.66 Â°C

Solution:
The general formula for final temperature is
θ=m1s1θ1+m2s2θ2+…m1s1+m2s2+…
In this problem
θ=m1s1θ1+m2s2θ2m1s1+m2s2
Here, m1=20 g, m2=100 g
θ1=100 °C, θ2=0 °C
s1=1 cal g-1 °C-1
s2=0.1 cal g-1 °C-1
∴ θ=66.66 °C

Topic: Physics, Thermal Properties of Matter, JEE Main
Calculate the solubility product of Co2[Fe(CN)6] in water at 25° C.
Given, conductivity of saturated solutions of Co2[Fe(CN)6] is 2.06 ×  10 −6   Ω −1   cm −1 and that of water used is 4.1 ×  10 −7   Ω −1   cm −1 . The ionic molar conductivities of Co2+ and [Fe(CN)6]4- are 86.0 Ω  cm 2   mol −1 and 444.0  Ω −1   cm 2   mol −1 , respectively.

Options:
A 7.87 ×  10 −17
B 7.87Ã10-6
C 7.87Ã10-8
D 7.87Ã10-9

The correct answer is:
7.87 ×  10 −17

Solution:
ÎºCo2[Fe(CN)6]=2.06Ã10-6-0.41Ã10-6=1.65Ã10-6
ÎmCo2[Fe(CN)6]=2ÎmCo2++Îm[Fe(CN)6]4-
=2×86+444=616  Scm 2   mol −1
Îm=1000Îºc
c=2.7Ã10-6M
KspCo2[Fe(CN)6]=4c3=7.87Ã10-17M3.

Topic: Chemistry, Ionic Equilibrium, JEE Main