datavorous/entrance-exam-dataset huggingface dataset:
A mixture consists of two radioactive materials $A_1$ and $A_2$ with half lives of $20 \mathrm{~s}$ and $10 \mathrm{~s}$ respectively. Initially the mixture has $40 \mathrm{~g}$ of $A_1$ and $160 \mathrm{~g}$ of $A_2$. The amount of the two in the mixture will become equal after

Options:
A $60 \mathrm{~s}$
B $80 \mathrm{~s}$
C $20 \mathrm{~s}$
D $40 \mathrm{~s}$

The correct answer is:
$40 \mathrm{~s}$

Solution:
For $40 \mathrm{~g}$ amount
$40 \mathrm{~g} \underset{\text { half -life }}{\stackrel{20 \mathrm{~s}}{\longrightarrow}} 20 \mathrm{~g} \stackrel{20 \mathrm{~s}}{\longrightarrow} 10 \mathrm{~g}$
For $160 \mathrm{~g}$ amount
$\begin{array}{l}
160 \mathrm{~g} \stackrel{10 \mathrm{~s}}{\longrightarrow} 80 \mathrm{~g} \stackrel{10 \mathrm{~s}}{\longrightarrow} 40 \mathrm{~g} \\
40 \mathrm{~g} \stackrel{10 \mathrm{~s}}{\longrightarrow} 20 \mathrm{~g} \stackrel{10 \mathrm{~s}}{\longrightarrow} 10 \mathrm{~g}
\end{array}$
So, after $40 \mathrm{~s} A_1$ and $A_2$ remains same.

Topic: Physics, Nuclear Physics, NEET, NEET 2012 (Screening)
A mixture consists of two radioactive materials A1 and A2 with half-lives of 20 s and 10 s, respectively. Initially, the mixture has 40 g of A1 and 160 g of A2. The amount of the two in the mixture will become equal after

Options:
A 60 s
B 80 s
C 20 s
D 40 s

The correct answer is:
40 s

Solution:
Let after time t s, A1 and A2 will become equal in the mixture.
As N⁡ = N⁡ 0 1 2 n
where n is the number of half-lives
For A⁡ 1, N⁡ 1 = N⁡ 0 1 1 2 t / 2 0
For A⁡ 2, N⁡ 2 = N⁡ 0 2 1 2 t / 1 0
According to question, N1 = N2
4 0 2 t / 2 0 = 1 6 0 2 t / 1 0
2 t / 1 0 = 4 2 t / 2 0 or 2 t / 1 0 = 2 2 2 t / 2 0
2 t / 1 0 = 2 t 2 0 + 2
t 1 0 = t 2 0 + 2 or t 1 0 - t 2 0 = 2
ort20=2ort=40 s

Topic: Physics, Nuclear Physics, JEE Main
A mixture containing $\mathrm{Cu}^{2+}$ and $\mathrm{Ni}^{2+}$ can be separated for identification by

Options:
A Passing $\mathrm{H}_2 \mathrm{~S}$ in acid medium
B Passing $H_2 S$ in alkaline medium
C Passing $\mathrm{H}_2 \mathrm{~S}$ in neutral medium
D Passing $H_2 S$ in dry mixture

The correct answer is:
Passing $\mathrm{H}_2 \mathrm{~S}$ in acid medium

Solution:
$\mathrm{Cu}^{2+}$ will be precipitated as $\mathrm{CuS}$ in acidic medium where as $\mathrm{Ni}^{2+}$ is not precipitated.

Topic: Chemistry, Practical Chemistry, JEE Main
A mixture containing the following four compounds is extracted with 1 M HCl. The compound that goes to aqueous layer is:

Options:
A IV
B II
C I
D III

The correct answer is:
II

Solution:
When the given mixture is shaken with HCl amine get protonated and becomes a cation NH2+
which does not dissolve in an organic solvent
since,

Topic: Chemistry, Amines, JEE Main, JEE Main 2017 (08 Apr Online)
A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with concentrate H2SO4. The evolved gaseous mixture is passed through KOH pellets. What will be the weight in grams of the remaining product at STP?

Options:
A 2.8
B 3.0
C 1.4
D 4.4

The correct answer is:
2.8

Solution:
For gases n∝V:
CO2 is absorbed by KOH. So the remaning product is only CO. moles of CO formed from both reactions =120+120=110
Left mass of CO = moles×molar mass
=110×28
=2.8 g

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with concentrate H2SO4. The evolved gaseous mixture is passed through KOH pellets. What will be the weight in grams of the remaining product at STP?

Options:
A 2.8
B 3.0
C 1.4
D 4.4

The correct answer is:
2.8

Solution:
For gases n∝V:
CO2 is absorbed by KOH. So the remaning product is only CO. moles of CO formed from both reactions =120+120=110
Left mass of CO = moles×molar mass
=110×28
=2.8 g

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET, NEET 2018
A mixture of 2 moles of helium gas (atomic mass = 4u) , and 1 mole of argon gas (atomic mass = 40 u) is kept at 300 K in a container. The ratio of their rms speeds VrmsheliumVrms(argon), is close to:

Options:
A 0.45
B 2.24
C 0.32
D 3.16

The correct answer is:
3.16

Solution:
VRMS=3RTM
∴ (VRMS)He(VRMS)Ar=MArMHe=404=10= 3.16

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2019 (09 Jan Shift 1)
A mixture of 1.0 mole of Al and 3.0 mole of Cl 2 are alloved to react as: 2Al(s)+3 Cl 2 (g)→2 AlCl 3 (g). Then moles of excess reagent left unreacted is:

Options:
A 3.5
B 1.0
C 1.5
D 2.5

The correct answer is:
1.5

Solution:
2 Al + 3 Cl 2 â 2 AlCl 3 s Moles before reaction 1.0 3.0 0ââ Moles after reaction 0 1.5 1.0
Mole ratio for reactants and products are 2 : 3 : 2
Thus limiting reagent is Al .
Moles of AlCl3 formed = 1.0
Moles of excess reagent Cl2 left unreacted =1.5

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. H2SO4 . The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining product at STP will be

Options:
A 2.8
B 3.0
C 1.4
D 4.4

The correct answer is:
2.8

Solution:
Here gaseous mixture formed is CO and CO2 when it is passed through KOH, only CO2 is absorbed, so the remaining gas is CO.
Hence, weight of remaining gaseous product CO is 220×28=2.8g

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. H2SO4 . The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining product at STP will be

Options:
A 2.8
B 3.0
C 1.4
D 4.4

The correct answer is:
2.8

Solution:
Here gaseous mixture formed is CO and CO2 when it is passed through KOH, only CO2 is absorbed, so the remaining gas is CO.
Hence, weight of remaining gaseous product CO is 220×28=2.8g

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
A mixture of 3.0 moles of $\mathrm{Na}_2 \mathrm{O}$ and $1.5 \mathrm{~mol}$ of $\mathrm{KO}_2$ is dissolved in $1000 \mathrm{~mL}$ of water. The vapour pressure of the solution in Torr, at $100{ }^{\circ} \mathrm{C}$ is

Options:
A 740
B 760
C 580
D 608

The correct answer is:
608

Solution:
$\mathrm{Na}_2 \mathrm{O}$ and $\mathrm{KO}_2$, both are ionic compounds and are completely ionised as follows :
(i) $\mathrm{Na}_2 \mathrm{O} \longrightarrow 2 \mathrm{Na}^{+}+\mathrm{O}^{2-}(i=3)$
(ii) $\mathrm{KO}_2 \longrightarrow \mathrm{K}^{+}+2 \mathrm{O}^{2-}(i=3)$
Also,
$\because \quad \frac{p^{\circ}-p_s}{p^{\circ}}=i \cdot \chi_{\mathrm{Na}_2 \mathrm{O}}+i \cdot \chi_{\mathrm{KO}_2}$
$\text { and moles of } \mathrm{H}_2 \mathrm{O}=\frac{1000}{18}=55.5$
where, $p^{\circ}=$ atmospheric pressure
$p_s=\text { vapour pressure of solution }$
$ p^{\circ}=760$
$\therefore \frac{p^{\circ}-p_s}{p^{\circ}}=3 \times \frac{3}{3+3+55.5}+3 \times \frac{1.5}{1.5+1.5+55.5}$
$=\frac{9}{61.5}+\frac{1.5 \times 3}{58.5}=\frac{9}{61.5}+\frac{4.5}{58.5}$
$\frac{760-p_s}{760}=0.14+0.07=0.21 \approx 0.20$
$p_s=608$

Topic: Chemistry, Solutions, JEE Main
A mixture of 100 m mol of Ca(OH)2 and 2 g of sodium sulphate was dissolved in water and the volume was made up to 100 mL. What is the mass of calcium sulphate formed and the concentration of OH- in resulting solution, respectively? (Molar mass of CaOH2, Na2SO4 and CaSO4 are 74, 143 and 136 g mol-1, respectively; Ksp of CaOH2 is 5.5×10-6)

Options:
A 1.9 g, 0.14 mol L-1
B 13.6 g, 0.28 mol L-1
C 1.9 g, 0.28 mol L-1
D 13.6 g, 0.14 mol L-1

The correct answer is:
1.9 g, 0.28 mol L-1

Solution:
Mass of CaSO4=14×1361000=1.9 g
CaSO4=14×1361000=1.9 g
OH-=28100=0.28 M

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2019 (10 Jan Shift 1)
A mixture of benzaldehyde and formaldehyde on heating with aqueous NaOH solution gives main product has

Options:
A Benzyl alcohol and sodium formate
B Sodium benzoate and methyl alcohol
C Sodium benzoate and sodium formate
D Benzyl alcohol and methyl alcohol

The correct answer is:
Benzyl alcohol and sodium formate

Solution:
In Cannizaro reaction when formaldehyde reacts with other aldehydes lacking α-hydrogen, it is always oxidized and other aldehyde is reduced
HCHO+C6H5CHO→NaOH∆HCOO-Na++C6H5CH2OH

Topic: Chemistry, Aldehydes and Ketones, NEET
A mixture of benzaldehyde and formaldehyde on heating with aqueous NaOH solution gives

Options:
A Benzyl alcohol and sodium formate
B Sodium benzoate and methyl alcohol
C Sodium benzoate and sodium formate
D Benzyl alcohol and methyl alcohol

The correct answer is:
Benzyl alcohol and sodium formate

Solution:
In Cannizaro reaction when formaldehyde reacts with other aldehydes lacking α-hydrogen, it is always oxidized and other aldehyde is reduced
HCHO+C6H5CHO→NaOH∆HCOO-Na++C6H5CH2OH

Topic: Chemistry, Aldehydes and Ketones, JEE Main
A mixture of CH4 and C2H2 occupied a certain volume at a total pressure equal to 63 torr. The same gas mixture was burnt to CO2 and H2O( l). CO2(g) alone was collected in the same volume and at the same temperature, the pressure was found to be 69 torr.
What was the mole fraction of CH4 in the original gas mixture?

Options:
A 1921
B 1920
C 1718
D 1516

The correct answer is:
1921

Solution:
Let no. of moles of CH4 present = n1 mole
Let no. of moles of C 2 H 2 present = n 2 mole
CH 4 n 1 +2O 2 â CO 2 n 1 +2H 2 O
C 2 H 2 n 2 + 5 2 O 2 â2 CO 2 2n 2 +H 2 O
Total no. of moles at initial = n1+ n2
Total no. of moles at final = n1 + 2n2
At constant V & T
P1P2=No. of moles at initialNo. of moles at final
6369=n1+n2n1+2n2â 2123=n1+n2n1+2n2
ân2n1= 219ân1+n2n1=2119
â´n1n1+n2=1921

Alternative solutions

CH 4 g + 2 O 2 g â CO 2 g + 2 H 2 O l 6 3 - x Torr 6 3 - x Torr
C 2 H 2 g + 5 2 O 2 g â 2 CO 2 g + H 2 O l x Torr 2 x Torr
â P CO 2 = 6 3 - x + 2 x = 6 3 + x = 69 Torr
â´ x = 6 Torr
X CH 4 = 6 3 - 6 6 3 = 5 7 6 3 = 1 9 2 1

Topic: Chemistry, States of Matter, JEE Main
A mixture of $\mathrm{CaCl}_{2}$ and $\mathrm{NaCl}$ weighing $4.44 \mathrm{~g}$ is treated with sodium carbonate solution to precipitate all the $\mathrm{Ca}^{2+}$ ions as calcium carbonate. The calcium carbonate so obtained is heated strongly to get $0.56 \mathrm{~g}$ of $\mathrm{CaO}$. The percentage of $\mathrm{NaCl}$ in the mixture (atomic mass of $\mathrm{Ca}=40$ ) is

Options:
A 75
B $30.6$
C 25
D $69.4$

The correct answer is:
75

Solution:
$\underset{1 \mathrm{~mol}}{\mathrm{CaCO}_{3}} \stackrel{\Delta}{\longrightarrow} \underset{1 \mathrm{~mol}}{\mathrm{CaO}}+\mathrm{CO}_{2}$
$\mathrm{CaCl}_{2}+\mathrm{Na}_{2} \mathrm{CO}_{3} \longrightarrow \underset{1 \mathrm{~mol}}{\mathrm{CaCO}_{3}}+2 \mathrm{Na}$
$1 \mathrm{~mol} \mathrm{} \mathrm{CaO} \cong 1 \mathrm{~mol} \mathrm{} \mathrm{CaCl}_{2}$
$\frac{0.56}{56} \mathrm{~mol} \mathrm{CaO} \cong 0.01 \mathrm{~mol} \mathrm{CaCl}_{2}$
$=0.01 \times 111 \mathrm{~g} \mathrm{CaCl}_{2}$
$=1.11 \mathrm{~g} \mathrm{CaCl}_{2}$
Thus, in the mixture, weight of
$\mathrm{NaCl}=4.44-1.11=3.33 \mathrm{~g}$
$\therefore$ Percentage of $\mathrm{NaCl}=\frac{3.33}{4.44} \times 100$
$=75 \%$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
A mixture of ethane, ethene and ethyne is passed through ammoniacal $\mathrm{AgNO}_3$ solution. The gases which remain unreacted are

Options:
A Ethane and ethene
B Ethane and ethyne
C Ethene and ethyne
D Ethane only

The correct answer is:
Ethane and ethene

Solution:
Ethyne reacts with ammonical $\mathrm{AgNO}_3$ to give white ppt of silver acetylide while ethane and ethene do not react because acidic hydrogen is absent.

Topic: Chemistry, Hydrocarbons, JEE Main
A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of $290 \mathrm{~mm}$ at $300 \mathrm{~K}$.
The vapour pressure of propyl alcohol is $200 \mathrm{~mm}$. If the mole fraction of ethyl alcohol is $0.6$, its vapour pressure (in $\mathrm{mm}$) at the same temperature will be

Options:
A
$350$

B
$300$

C
$700$

D
$360$

The correct answer is:
$350$

Solution:
Let the vapour pressure of pure ethyl alcohol be $\mathrm{P}$,
According to Raoult's law
$\begin{aligned}
& 290=200 \times 0.4+\mathrm{P} \times 0.6 \\
& \mathrm{P}=\frac{290-80}{0.6}=350 \mathrm{~mm} \mathrm{~Hg}
\end{aligned}$
Hence, (A) is correct.

Topic: Chemistry, Solutions, JEE Main, JEE Main 2007
A mixture of gases O2,H2 and CO are taken in a closed vessel containing charcoal The graph that represents the correct behaviour of pressure with time is:

Options:
A
B
C
D

The correct answer is:

Solution:
A mixture of gases O2,H2 and CO are taken in a closed vessel containing charcoal The graph that represents the correct behaviour of pressure with time is given below:

Due to adsorption pressure of gas will decrease with time.

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2020 (09 Jan Shift 2)
A mixture of hydrogen and oxygen contains 40% hydrogen by mass when the pressure is 2.2 bar. The partial pressure of hydrogen is bar.

Options:

The correct answer is:
2

Solution:
According to Dalton's law of partial pressures, the total pressure by a mixture of gases is equal to the sum of the partial pressures of each of the constituent gases.

Pi=xiPPi= partial pressure of ith component.xi=mole fraction of the ith componentP = total pressure

Let WH2=40 g⇒nH2=402=20

WO2=60 g⇒nO2=6032=158

PH2=2020+158×2.2

=2020+1.875×2.2

=2021.875×2.2

=2.0114

≃2.01bar

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2022 (26 Jul Shift 1)
A mixture of hydrogen and oxygen has volume 500 cm3, temperature 300 K, pressure 400 kPa and mass 0.76 g. The ratio of masses of oxygen to hydrogen will be:

Options:
A 16 : 3
B 3 : 8
C 8 : 3
D 3 : 16

The correct answer is:
16 : 3

Solution:
A number of moles of, O2=m132, the number of moles of, H2=m22. Using the ideal gas equation, PV=m132+m22253×300,
400×103×500×10-6=m132+m2225×100⇒2×102=m1m2×132+12m2×25×100
⇒225=m1m2×132+12m2 ...(1)
⇒m1+m2=0.76
⇒m21+m1m2=76100 ...(2)
(1) divided by (2) ; m1m2×132+12m1m2+1=225×10076
Let assume,
⇒m1m2=x⇒x+1632x+1=876⇒ 76x+76×16=8×32x+8×32⇒76x+1216=256x+256
960=180x⇒x=9618=163

Topic: Physics, Kinetic Theory of Gases, JEE Main
A mixture of hydrogen and oxygen has volume 500 cm3, temperature 300 K, pressure 400 kPa and mass 0.76 g. The ratio of masses of oxygen to hydrogen will be:

Options:
A 16 : 3
B 3 : 8
C 8 : 3
D 3 : 16

The correct answer is:
16 : 3

Solution:
A number of moles of, O2=m132, the number of moles of, H2=m22. Using the ideal gas equation, PV=m132+m22253×300,
400×103×500×10-6=m132+m2225×100⇒2×102=m1m2×132+12m2×25×100
⇒225=m1m2×132+12m2 ...(1)
⇒m1+m2=0.76
⇒m21+m1m2=76100 ...(2)
(1) divided by (2) ; m1m2×132+12m1m2+1=225×10076
Let assume,
⇒m1m2=x⇒x+1632x+1=876⇒ 76x+76×16=8×32x+8×32⇒76x+1216=256x+256
960=180x⇒x=9618=163

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2021 (31 Aug Shift 2)
A mixture of light, consisting of wavelength $590 \mathrm{~nm}$ and an unknown wavelength, illuminates Young's double slit and gives rise to two overlapping interference patterns on the screen. The central maximum of both lights coincide. Further, it is observed that the third bright fringe of known light coincides with the $4^{\text {th }}$ bright fringe of the unknown light. From this data, the wavelength of the unknown light is

Options:
A
$393.4 \mathrm{~nm}$

B
$885.0 \mathrm{~nm}$

C
$442.5 \mathrm{~nm}$

D
$776.8 \mathrm{~nm}$

The correct answer is:
$442.5 \mathrm{~nm}$

Solution:
$$
\begin{aligned}
& 3 \lambda_1=4 \lambda_2 \\
& \Rightarrow \lambda_2=\frac{3}{4} \lambda_1=\frac{3}{4} \times 590=\frac{1770}{4}=442.5 \mathrm{~nm}
\end{aligned}
$$

Topic: Physics, Wave Optics, JEE Main, JEE Main 2009
A mixture of light, consisting of wavelength 590 nm and an unknown wavelength, illuminates Young's double-slit setup and gives rise to two overlapping interference patterns on the screen. The central maximum of both patterns coincide. Further, it is observed that the third bright fringe of known light coincides with the 4th bright fringe of the unknown light. From this data, the wavelength of the unknown light is

Options:
A 393.4 nm
B 885.0 nm
C 442.5 nm
D 776.8 nm

The correct answer is:
442.5 nm

Solution:
For interference, by Young's double slits, the path difference is xdD=nÎ» for bright fringes and xdD=2n+1Î»2 for dark fringes.
The central fringes when x=0, coincide for all wavelengths.
It is known that the third fringe of Î»1=590 nm coincides with the fourth bright fringe of unknown wavelength Î».
â´ xd D = 3 Ã 5 9 0 nm = 4 Ã Î» nm
â´ Î» = 3 Ã 5 9 0 4 = 4 4 2 . 5 nm

Topic: Physics, Wave Optics, JEE Main
A mixture of methane and ethene in a molar ratio of x:y has an average molecular mass of 20 u. The mean molar mass when they are mixed in the molar ratio of y:x will be -

Options:
A 20 u
B 25 u
C 24 u
D 15 u

The correct answer is:
24 u

Solution:
Mol. wt. of CH4=16
Mol. wt. of C2H4=28
â´20=16x+28yx+y
or 16x+28 y=20 x+20 y
or 4x=8y
or x=2y
In the gaseous mixture when the mole ratio of CH4 and C2H4 is y : x
then avg. mol. wt=16y+28xx+y=16y+56y3y
=72y3y=24 u

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
A mixture of one mole each of H2 , He and O2 each are enclosed in a cylinder of volume V at temperature T. If the partial pressure of H2 is 2atm, the total pressure of the gases in the cylinder is:

Options:
A 6 atm
B 38 atm
C 14 atm
D 22 atm

The correct answer is:
6 atm

Solution:
Pgas=ngasRTV

as n, T &V constant so

PH2=PO2=PHe=2atm

So, PTotal=PH2+PO2+PHe=6 atm

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2020 (03 Sep Shift 2)
A mixture of one mole of H2O and 1 mole of CO is taken in a 10 litre container and heated to 725 K. At equilibrium 40% of water by mass reacts with carbon monoxide according to the equation: COg + H2Og ⇌ CO2g + H2g. The equilibrium constant KC × 102 for the reaction is ________ (Nearest integer)

Options:

The correct answer is:
44

Solution:
CO(g)+H2O(g)⇌CO2( g)+H2( g)initial moles 1 1moles at equilibrium 1-0.4 1-0.4 0.4 0.4Molarity 0.6/10 0.6/10 0.4/10 0.4/10

Now, the equilibrium constant KC=CO2H2COH2O

⇒KC=0.04×0.040.06×0.06=0.44

∴Kc×102=44

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2023 (11 Apr Shift 1)
A mixture of one mole of monoatomic gas and one mole of a diatomic gas (rigid) are kept at room temperature $\left(27^{\circ} \mathrm{C}\right)$. The ratio of specific heat of gases at constant volume respectively is:

Options:
A $\frac{7}{5}$
B $\frac{3}{5}$
C $\frac{5}{3}$
D $\frac{3}{2}$

The correct answer is:
$\frac{3}{5}$

Solution:
$\frac{\left(\mathrm{C}_{\mathrm{v}}\right)_{\text {mono }}}{\left(\mathrm{C}_{\mathrm{v}}\right)_{\mathrm{dia}}}=\frac{\frac{3}{2} \mathrm{R}}{\frac{5}{2} \mathrm{R}}=\frac{3}{5}$

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2024 (08 Apr Shift 1)
A mixture of three alkyl chloride CH3Cl,CH3CH2Cl,CH3CH2CH2Cl under goes Wurtz coupling reaction. The product contains:

Options:
A Ethane, pentane, hexane only
B Propane, pentane, butane only
C Ethane, Propane, pentane, butane
D Ethane, Propane, butane, pentane, hexane

The correct answer is:
Ethane, Propane, butane, pentane, hexane

Solution:
Five products Ethane, Propane, butane, pentane, hexane will be formed
The possible cases will be:
I 2CH3ClâNaCH3-CH3âethane
II 2CH3CH2ClâNaCH3CH2-CH2CH3âbutanâ¡e
III 2CH3CH2CH2ClâNaCH3CH2CH2-CH2CH2CH3âhexane
IV CH3Cl+CH3CH2ClâNaCH3-CH2CH3âpropane
V CH3CH2Cl+CH3CH2CH2ClâNaCH3CH2-CH2CH2CH3âpentanâ¡e
VI CH3CH2CH2Cl+CH3ClâNaCH3CH2CH2-CH3âbutanâ¡e
In II and VI products are same so total number of products are 5

Topic: Chemistry, Hydrocarbons, JEE Main
A mixture of two gasses A and B is kept in a vessel which is maintained at a temperature T. In the vessel there are N molecules each of mass m, of the gas A and 2N molecules each of mass 2m of the gas B. The mean square velocity of the molecules of the gas B is v2 and the mean square velocity of the molecules of the gas A along the x direction is w2. The value of w2v2 is

Options:
A 2
B 1
C 13
D 23

The correct answer is:
23

Solution:
Mean kinetic energy of the two types of molecules should be equal. The mean square velocity of A type molecules =ω2+ω2+ω2=3ω2
Therefore, 12m3ω2=122mv2
This gives ω2v2=23

Topic: Physics, Kinetic Theory of Gases, JEE Main
A modulated signal Cmt has the form Cmt=30sin⁡300πt+10 cos⁡200πt-cos⁡400πt. The carrier frequency fc , the modulating frequency (message frequency) fm , and the modulation index μ are respectively given by :

Options:
A fc=200 Hz; fm=50 Hz; Î¼=12
B fc=150 Hz;fm=50 Hz; Î¼=23
C fc=150 Hz;fm=30 Hz; Î¼=13
D fc=200 Hz;fm=30 Hz; Î¼=12

The correct answer is:
fc=150 Hz;fm=50 Hz; Î¼=23

Solution:
Compare with
C m (t)= A c sin( w c t)+ μ A c 2 [ cos( w c − w m )t−cos( w c + w m ) ]
μ A c 2 =10
A c =30
μ= 2 3
w c =2π f c =300π ⇒ f c =150Hz
w c − w m =200π
w m =100π
f m =50Hz

Topic: Physics, Communication System, JEE Main, JEE Main 2016 (10 Apr Online)
A molecule $M$ associates in a given solvent according to the equation $\mathrm{M} \rightleftharpoons(\mathrm{M})_n$. For a certain concentration of M, the van't Hoff factor was found to be $0.9$ and the fraction of associated molecules was $0.2$. The value of $n$ is:

Options:
A
3

B
5

C
2

D
4

The correct answer is:
2

Solution:
van't Hoff factor (i) and the degree of association are related as below :
$$
\begin{aligned}
& i=1-\alpha\left(1-\frac{1}{n}\right) \\
& 0.9=1-0.2\left(1-\frac{1}{n}\right)
\end{aligned}
$$
On solving,
$$
\begin{aligned}
& \left(1-\frac{1}{n}\right)=\frac{1}{2} \\
& \frac{1}{n}=1-\frac{1}{2}=\frac{1}{2} \\
& \therefore n=2
\end{aligned}
$$

Topic: Chemistry, Solutions, JEE Main, JEE Main 2013 (22 Apr Online)
A molecule undergoes two independent first order reactions whose respective half lives are 12 min and 3 min. If both the reactions are occurring then the time taken for the 50% consumption of the reactant is ______ min. (Nearest integer)

Options:

The correct answer is:
2

Solution:
For parallel reaction

When adding the rate constants of the two reactions, we need to use the formula for combining rate constants of two reactions in parallel, which is:

k=k1+k2

where k1 and k2 are the rate constants of the individual reactions.

In this case, the half-lives of the two reactions are 12 min and 3 min, respectively. The rate constants of the two reactions can be calculated using the formula for half-life of a first-order reaction:

1t1/2=112+13=512

Net t1/2=125=2.4 min≈2 min

Hence time taken for 50% consumption of reactant will be close to 2 min.

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2023 (10 Apr Shift 1)
A monatomic gas expands at constant pressure on heating. The percentage of heat supplied that increases the internal energy of the gas and that is involved in the expansion is

Options:
A 75%, 25%

B 25%, 75%

C 60%, 40%

D 40%, 60%

The correct answer is:
60%, 40%

Solution:
ΔU= f 2 nRdT
ΔQ=( f 2 +1 )nRdT
ΔU ΔQ = f 2 1+ f 2
ΔU ΔQ = 3 5

∆U∆Q×100=35×100=60%

Remaining is used in expansion of the gas = 40%

Topic: Physics, Thermodynamics, NEET
A monatomic ideal gas sample is given heat Q. One half of this heat is used as work done by the gas and rest is used for increasing its internal energy. The equation of process in terms of volume and temperature is

Options:
A V2T3= constant
B V2T= constant
C VT3= constant
D V2T= constant

The correct answer is:
V2T3= constant

Solution:
PdV=nCvdT

â nRTVdV=nCvdT

âdVV=31dTT
â´ V2=CT3, Where C is a constant.

Topic: Physics, Thermodynamics, JEE Main
A monkey of mass $20 \mathrm{~kg}$ is holding a vertical rope. The rope will not break when a mass of $25 \mathrm{~kg}$ is suspended from it but will break if the mass exceeds $25 \mathrm{~kg}$. What is the maximum acceleration with which monkey can climb up along the rope?

Options:
A $5 \mathrm{~m} / \mathrm{s}^2$
B $10 \mathrm{~m} / \mathrm{s}^2$
C $25 \mathrm{~m} / \mathrm{s}^2$
D $2.5 \mathrm{~m} / \mathrm{s}^2$

The correct answer is:
$2.5 \mathrm{~m} / \mathrm{s}^2$

Solution:
Let $T=$ tension in the rope when monkey climb up with an acceleration $a$.
Then $T-m g=m a$
$\begin{aligned}
& 25 g-20 g=20 a \\
& 5 g=20 a \\
& \Rightarrow a=\frac{5 \times 10}{20}=2.5 \mathrm{~m} / \mathrm{s}^2
\end{aligned}$

Topic: Physics, Laws of Motion, JEE Main