datavorous/entrance-exam-dataset huggingface dataset:
60 g of ice at 0 Â°C is added to 200 g of water initially at 70 Â°C in a calorimeter of unknown water equivalent W. If the final temperature of the mixture is 40 Â°C, then the value of W is [Take latent heat of fusion of ice Lf=80 cal g-1 and specific heat capacity of water s=1 cal g-1 Â°C-1]

Options:
A 70 g
B 80 g
C 40 g
D 20 g

The correct answer is:
40 g

Solution:
The principle of calorimetry states that for an isolated system,

The heat gained by the cold body (Here it is ice) = The heat lost by the hot body (Here it is water + calorimeter)

miceLf+mices∆T=mwater+Ws∆T

⇒60×80+60×1×40=200+W×1×30

⇒480+240=600+3W

⇒W=40 g

Topic: Physics, Thermodynamics, JEE Main
\( 60 \mathrm{~g} \) of gaseous \( \mathrm{C}_{2} \mathrm{H}_{6} \) is mixed with \( 28 \mathrm{~g} \) of carbon monoxide. The pressure of the resulting gaseous mixture is \( 3 \) atm. The partial pressure in atm of \( \mathrm{C}_{2} \mathrm{H}_{6} \) in the mixture is

Options:

The correct answer is:
2

Solution:
Moles of C2H6=60g30g/mol=2mol
moles of CO=28g28g/mol=1mol
Mole fraction C2H6=22+1=23
Partial pressure of C2H6=Ptotal× Mole fraction of C2H6
=3atm×23=2atm .

Topic: Chemistry, States of Matter, JEE Main
60% of the angiosperms shed their pollens at the

Options:
A 2 celled stage
B 3 celled stage
C 4 celled stage
D 1 celled stage

The correct answer is:
2 celled stage

Solution:
Pollen grains represent male gametocyte. When pollen grains mature it consists of the vegetative cell which is bigger and has ample food reserve and generative cell which is relatively smaller. In over 60% of angiosperms pollen grains are shed at this 2 celled stage and in the remaining of angiosperms generative cell undergoes mitotic division resulting in the formation of two male gametes before pollen grains are shed (3 celled stage)

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
60 words can be made using all the letters of the word BHBJO, with or without meaning. If these words are written as in a dictionary, then the $50^{\text {th }}$ word is :

Options:
A JBBOH
B OBBJH
C OBBHJ
D HBBJO

The correct answer is:
OBBJH

Solution:
$\mathrm{B} \mathrm{B} \mathrm{H} \mathrm{J} \mathrm{O}$
B____$4 !=24$
H____$\frac{4 !}{2 !}=12$
j____$\frac{4 !}{2 !}=12$
O B B H J
O B B J H $\rightarrow 50^{\text {th }}$ rank

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2024 (05 Apr Shift 2)
64 identical drops each charged upto potential of 10 mV are combined to form a bigger drop. The potential of the bigger drop will be _____ mV.

Options:

The correct answer is:
160

Solution:
The formula to calculate the potential V of each tiny drop is given by

V=kqr ...1

When all the drops forms the bigger drop the radius of the new drop r' can be calculated as follows:

Volume of big drop=64×volume of each tiny drop⇒43πr'3=64×43πr3⇒r'=4r

So, the potential of the new big drop V' can be written as

V'=k64qr'=k64q4r=16kqr ...2

Divide equation (2) by equation (1) and simplify to obtain the potential of the new drop.

V'V=16kqrkqr⇒V'=16V ...3

Substitute the value of the known parameter into equation (3) to calculate the required value of the potential for the new drop.

V'=16×10 mV=160 mV

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2023 (12 Apr Shift 1)
\( 64 \mathrm{~g} \) non-volatile solute is added to \( 702 \mathrm{~g} \) benzene. The vapour pressure of benzene has decreased from \( 200 \mathrm{~mm} \) of \( \mathrm{Hg} \) to \( 180 \mathrm{~mm} \) of \( \mathrm{Hg} \). Molecular weight of the solute is

Options:
A \( 128 \)
B \( 64 \)
C \( 96 \)
D \( 256 \)

The correct answer is:
\( 64 \)

Solution:
V.P. has decreased by 10%
This happens only when 9 moles solvent & one mole solute are present
702 g benzene = 9 moles
64 g=1 mole

Topic: Chemistry, Solutions, JEE Main
\( 7.6 \mathrm{gm} \mathrm{KBr} \) in \( 1250 \mathrm{ml} \) solution was found to have an osmotic pressure of \( 1.804 \) atm at \( 27^{\circ} \mathrm{C} \). Calculate degree of
dissociation in \( \% \) (Atomic weight \( \mathrm{K}=39, \mathrm{Br}=80 \) )

Options:
A \( 43.4 \% \)
B \( 58.4 \% \)
C 4. \( 64 \% \)
D \( 80 \% \)

The correct answer is:
\( 43.4 \% \)

Solution:
KBr⇌K++Br-1 0 01-α α α (1+α)

π=iCRT

i=1.804CRT C=7.6119/12501000

=1.804×119×12507.6×1000×0.0821×300

i=1.434

i=1+α1

∴ α=i-1=1.434-1

α=0.434, %=43.4

Topic: Chemistry, Solutions, JEE Main
0.078 g of hydrocarbon occupy 22.4 mL of volume at 1 atm and 0Â°C. The empirical formula of the hydrocarbon is CH . The molecular formula is -

Options:
A C2H2
B C4H4
C C6H6
D C8H8

The correct answer is:
C6H6

Solution:
Molar mass of hydrocarbon can be determtextnd with the help of ideal gas equation:
PV=nRT
PV= w M RT
1× 22.4 1000 L= 0.078 M ×0.082×273 k
m=78 g
Molecular formula  =n×  Emperial formula
n= molecular formula empirial mass
Empirial mass =CH
So, Empirical mass =12 H=13
∴ n= 78 13 =6
Molecular formula = CH 6 = C 6 H 6

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
$80 \mathrm{~g}$ of oxygen contains as many atoms as in

Options:
A $80 \mathrm{~g}$ of hydrogen
B $1 \mathrm{~g}$ of hydrogen
C $10 \mathrm{~g}$ of hydrogen
D $5 \mathrm{~g}$ of hydrogen

The correct answer is:
$5 \mathrm{~g}$ of hydrogen

Solution:
Number of moles of oxygen $=\frac{80}{16}$
Number of atoms of oxygen $=\frac{80}{16} \times \mathrm{N}_{0} \times 2$
$$
=5 \times \mathrm{N}_{0} \times 2
$$
Number of moles of $5 \mathrm{~g}$ of hydrogen $=\frac{5}{1}$
Number of atoms in $5 \mathrm{~g}$ of hydrogen $=5 \times \mathrm{N}_{0} \times 2$
Hence, the number of atoms in $80 \mathrm{~g}$ of oxygen is equal to the number of atoms in $5 \mathrm{~g}$ of hydrogen.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
80 kg of a radioactive material reduces to 10 kg in 1 h. The decay constant of the material is

Options:
A 5.80 Ã 10-4 s-1
B 1.16Â ÃÂ 10-3Â s-1
C 2.32Â Ã10-3Â s-1
D 4.64Ã10-3Â s-1

The correct answer is:
5.80 Ã 10-4 s-1

Solution:
Fraction of material that remains undecayed
1080=121hT1/2
3=1hT1/2
or T1/2 = 603 min = 20 min =1200 s
Now, λ=0.693T1/2=0.6931200= 5.8 × 10-4 s-1

Topic: Physics, Nuclear Physics, JEE Main
80 mole percent of MgCl2 is dissociated in aqueous solution. The vapour pressure of 1.0 molal aqueous solution of MgCl2 at 38oC is _____ mm Hg. (Nearest integer)

Given: Vapour pressure of water at 38oC is 50 mm Hg

Options:

The correct answer is:
48

Solution:
One molal solution means one of solute present in one kg of solvent.

MgCl2⇌Mg2++2Cl- 1 0 01-0.8 0.8 1.6

Hence, overall moles after dissociation =2.6

The relative lowering of vapour pressure

po-pspo=moles of solutemoles of solute+moles of solvent

p°-psp°=2.6100018+2.6

For dilute solution

p°-psp°=2.6100018

p=47.66≃48 mm Hg

Topic: Chemistry, Solutions, JEE Main, JEE Main 2023 (12 Apr Shift 1)
$\longrightarrow{ }_{82} \mathrm{~Pb}^{208}$. The number of $\alpha$
and $\beta$ -particles emitted during the above reaction is

Options:
A $8 \alpha$ and $4 \beta$
B $8 \alpha$ and $16 \beta$
C $4 \alpha$ and $2 \beta$
D $6 \alpha$ and $4 \beta$

The correct answer is:
$6 \alpha$ and $4 \beta$

Solution:
No. of $\alpha$ particles $=\frac{232-208}{4}=\frac{24}{4}=6 \alpha$
No. of $\beta$ particles $=2 \times 6-(90-82)$
$$
\begin{array}{l}
=12-8 \\
=4 \beta
\end{array}
$$
Hence, $6 \alpha$ and $4 \beta$ particles are emitted.

Topic: Chemistry, Chemical Kinetics, JEE Main
84Po210 decays with a particle to 82Pb206 with a half-life of 138.4 days. If 1.0 g of 84Po210 is placed in a sealed tube, how much helium will accumulate in 69.2 days. Express the answer in cm3 at STP.

Options:
A 28.21 cm3
B 31.25 cm3
C 36.85 cm3
D 38.47 cm3

The correct answer is:
31.25 cm3

Solution:
t1/2=138.4 days, t = 69.2 day
Number of half-lifes n=tt1/2=69.2138.4=12
Amount of Po left 12 after half-life =121/2g=0.707 g
Amount of Po used in 12 half-life =1-0.707=0.293 g
Now 84Po210→82Pb206+2He4
210 g Po on decay will produce = 4g He
0.293 g Po on decay will produce =4 × 0.293 210=5.581×10-3g He
Volume of He at STP =5.581 × 10-3 × 224004=31.25 mL=31.25 cm3

Topic: Chemistry, Chemical Kinetics, JEE Main
92238A→ 90234B+D24+Q

In the given nuclear reaction, the approximate amount of energy released will be:

[Given, mass of 92238A=238.05079×931.5 MeV c-2, mass of B90234=234.04363×931.5 MeV c-2, mass of D24=4.00260×931.5MeV c-2 ]

Options:
A 3.82MeV
B 5.9MeV
C 2.12MeV
D 4.25MeV

The correct answer is:
4.25MeV

Solution:
The value of Q is

Q=(mA-mB-mD)×931.5 MeV⇒Q=(238.05079-234.04363-4.00260)×931.5 MeV⇒Q=4.25 MeV

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2023 (13 Apr Shift 1)
${ }_{92} \mathrm{U}^{235}$ nucleus absorbs a neutron and disintegrates in ${ }_{54} \mathrm{Xe}^{139},{ }_{38} \mathrm{Sr}^{94}$ and $x$. What will be the product $x$ ?

Options:
A 3 neutrons
B 2 neutrons
C $\alpha$-particle
D $\beta$-particle

The correct answer is:
2 neutrons

Solution:
${ }_{92} \mathrm{U}^{235}+{ }_0 n^1 \rightarrow{ }_{54} \mathrm{Xe}^{139}+{ }_{38} \mathrm{Sr}^{94}+2_0 n^1$
Related Theory
Moderators are used to slow down the neutrons produced from fission. This is necessary because many nuclear fuels (uranium-235, for instance) require neutrons to be slow-moving in order to absorb them. Nuclei with low mass numbers are most effective at doing so, therefore materials like water or graphite are often used.
Caution
In a nuclear fission, a heavy nucleus disintegrates into lighter nuclei. Large amount of energy is released during nuclear fission. Nuclear fission occurs by itself and no additional energy is required for a nuclear fission to occur.

Topic: Physics, Structure of Atom, JEE Main
\(9.3 \mathrm{~g}\) of pure aniline is treated with bromine water at room temperature to give a white precipitate of the product ' \(\mathrm{P}\) '. The mass of product ' \(\mathrm{P}\) ' obtaind is \(26.4 \mathrm{~g}\). The percentage yield is _______ \(\%\).

Options:

The correct answer is:
80

Solution:
$93 \mathrm{~g}$ of aniline produces $330 \mathrm{~g}$ of $2,4,6-$ tribromoaniline. Hence $9.3 \mathrm{~g}$ of aniline should produce $33 \mathrm{~g}$ of 2, 4, 6-tribromoaniline. Hence percentage yield $\frac{26.4 \times 100}{33}=80 \%$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2024 (05 Apr Shift 1)
\(9.3 \mathrm{~g}\) of pure aniline upon diazotisation followed by coupling with phenol gives an orange dye. The mass of orange dye produced (assume 100\% yield/conversion) is _______ g. (nearest integer)

Options:

The correct answer is:
20

Solution:
Reaction suggests that 1 mole of aniline give 1 mole of orange dye.
$\begin{aligned}
& \text { so }(\mathrm{mol})_{\text {aniline }}=(\mathrm{mole})_{\text {orange dye }} \\
& \frac{9.3 \mathrm{~g}}{93 \mathrm{~g} \mathrm{~mol}^{-1}}=\frac{\text { mass of orange dye }}{199 \mathrm{~g} \mathrm{~mol}^{-1}}
\end{aligned}$
mass of orange dye $=19.9 \mathrm{~g} \simeq 20 \mathrm{~g}$

Topic: Chemistry, Amines, JEE Main, JEE Main 2024 (06 Apr Shift 1)
96 cosπ33 cos2π33 cos4π33 cos8π33 cos16π33 is equal to

Options:
A 3
B 1
C 4
D 2

The correct answer is:
3

Solution:
Given,

Expression 96·cosπ33·cos2π33·cos4π33.........cos16π33

Now we know that,

cosA·cos2A·cos22A·cos23A.....·cos2n-1A=sin2nA2nsinA

Now using the above formula in given expression we get,

96·cosπ33·cos2π33·cos4π33.........cos16π33

=96×sin32π3325sinπ33

=96×sinπ-π3325sinπ33

=96×sinπ3325sinπ33 as sinπ-α=sinα

=96×132=3

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main, JEE Main 2023 (10 Apr Shift 1)
\( 99 \% \) of a first order reaction was completed in \( 32 \mathrm{~min} \). When will \( 99.9 \% \) of the reaction complete?

Options:
A \( 24 \mathrm{~min} \)
B \( 8 \mathrm{~min} \)
C \( 4 \mathrm{~min} \)
D \( 48 \mathrm{~min} \)

The correct answer is:
\( 48 \mathrm{~min} \)

Solution:
k=2.30332log⁡1001
∴t=2.3032.30332log⁡100log⁡1000.1
t=48 min

Topic: Chemistry, Chemical Kinetics, JEE Main
100 balls each of mass m moving with speed v simultaneously strike a wall normally and reflected back with same speed, in time t s. The total force exerted by the balls on the wall is

Options:
A 100mvt
B 200mvt
C 200 mvt
D mv100t

The correct answer is:
200mvt

Solution:
.

Force exerted by each ball is F=∆p∆t, here, p is momentum and t is time.

So, for one ball F=mv-m-vt=2mvt

Now, total force exerted by the balls on the wall is Fnet=100×F=200mvt.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2023 (31 Jan Shift 1)
100 cells each of e.m.f. $5 \mathrm{~V}$ and internal resistance $1 \mathrm{ohm}$ are to be arranged so as to produce maximum current in a $25 \mathrm{ohms}$ resistance. Each row is to contain equal number of cells. The number of rows should be

Options:
A 2
B 4
C 5
D 10

The correct answer is:
2

Solution:
If $m=$ Number of rows
and $n=$ Number of cells in a row
Then $m \times n= 100$ $\ldots . . .(i)$
Also condition of maximum current is $R=\frac{n r}{m}$
$\Rightarrow \quad 25=\frac{1 \times n}{m} \Rightarrow n=25 m$ $\ldots . . .(ii)$
On solving (i) and (ii) $m=2$

Topic: Physics, Current Electricity, JEE Main
100 g of an ideal gas is kept in a cylinder of 416 L volume at 27 °C under 1.5 bar pressure. The molar mass of the gas is g mol-1. (Nearest integer)
(Given : R=0.083 L bar K-1 mol-1)

Options:

The correct answer is:
4

Solution:
PV=nRT

1.5×416=100M.wt×0.083×300

M.Wt. =3.99=4 g/mole

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2022 (28 Jun Shift 2)
100âg of liquid A (molar mass 140âgâ mol â1 ) was dissolved in 1000âg of liquid B (molar mass 180âgâ mol â1 ). The vapour pressure of pure liquid B was found to be 500âtorr . Calculate vapour pressure of A (in torr) the solution if the total vapour pressure of the solution is 475âtorr.

Options:

The correct answer is:
32.0

Solution:
Step I : Calculation of vapour pressure of pure liquid A (pÂ°A)
âNumber of moles of liquid A
n A = W A M A = (100 g) 140 g mol - 1 = 0.7143 mol
Number of moles of liquid B
n B = W B M B = (1000 g) 180 g mol - 1 = 5.5556 mol
Mole fraction of A
x A = n A n A + n B
= (0.7143 mol) (0.7143 + 5.5556) mol = 0.7143 6.2699 = 0.1139
Mole fraction of B(x B )â=â1â0.1139=0.8861
Vapour pressure of pure liquid B, B(p B o )â=â500âtorr
Total vapour pressure of solution (p)=475âtorr
According to the Raoult's law
pâ=â p A o â x A â+â p B o â x B
475âtorrâ=â p A o âÃâ(0.1139)â+â500âtorrâÃâ(0.8861)
475âtorrâ=â p A o âÃâ(0.1139)â+â443.05âtorr
p A o = 475 - 443.05 torr (0.1139) = 31.95 0.1139 torr = 280.5 torr
Step II : Calculation of vapour pressure of A in the solution (pA)
According to Raoult's law,
p A â=â p A o x A â=â(280.5âtorr)âÃâ(0.1139)
p A â=â32.0âtorr

Topic: Chemistry, Solutions, JEE Main
100 g of propane is completely reacted with 1000 g of oxygen. The mole fraction of carbon dioxide in the resulting mixture is x×10-2. The value of x is (Nearest integer)

Options:

The correct answer is:
19

Solution:
C3H8(g) + 5O2(g)⟶3CO2( g)+4H2O(ℓ)

t=02.27 mole 31.25 mol

t=∞0 19.9 mol 6.81 mol 9.08 mol

mole fraction of CO2 in the final reaction mixture (heterogeneous)

XCO2=6.8119.9+6.81+9.08

=0.1902=19.02×10-2⇒19

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2021 (27 Aug Shift 2)
$100 \mathrm{~g}$ of water is heated from $30^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$. Ignoring the slight expansion of the water, the change in its internal energy is (specific heat of water is $4148 \mathrm{~J} / \mathrm{kg} / \mathrm{K}$ ):

Options:
A
$8.4 \mathrm{~kJ}$

B
$84 \mathrm{~kJ}$

C
$2.1 \mathrm{~kJ}$

D
$4.2 \mathrm{~kJ}$

The correct answer is:
$8.4 \mathrm{~kJ}$

Solution:
$\Delta \mathrm{Q}=\Delta \mathrm{U}+\Delta \mathrm{W}$ (ignoring expansion)
$\Delta \mathrm{U}=\mathrm{ms} \Delta \mathrm{T}=0.1 \times 4.184 \times 20=8.368 \mathrm{~kJ}$

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2011
$100 \mathrm{~g}$ of water is supercooled to $-10^{\circ} \mathrm{C}$. At this point, due to some disturbance mechanised or otherwise some of it suddenly freezes to ice. What will be the temperature of the resultant mixture and how much mass would freeze? $\left[S_w=1 \mathrm{cal} / \mathrm{g} /{ }^{\circ} \mathrm{C}\right.$ and $\left.L^{w_{\text {Fusion }}}=80 \mathrm{cal} / \mathrm{g}\right]$

Options:

None

Solution:
As given that, mass of water $(\mathrm{m})=100 \mathrm{~g}$
Change in temperature $\Delta T=0-(-10)=10^{\circ} \mathrm{C}$
Specific heat of water $\left(S_w\right)=1 \mathrm{cal} / \mathrm{g} /{ }^{\circ} \mathrm{C}$
Latent heat of fusion of water $L_{\text {fusion }}^w=80 \mathrm{cal} / \mathrm{g}$
Heat required to bring water in super cooling from $-10^{\circ} \mathrm{C}$ to $0^{\circ} \mathrm{C}$
$$
Q=m s_w \Delta T=100 \times 1 \times 10=1000 \mathrm{cal}
$$
Let $m$ gram of ice be melted, so, $Q=m L$
or $m=\frac{Q}{L}=\frac{1000}{80}=12.5 \mathrm{~g}$
So, there is $m=12.5 \mathrm{~g}$ water and $100-12.5 \mathrm{~g}$ ice in mixture. As small mass of ice is melted. Hence the temperature of the mixture will remain $0^{\circ} \mathrm{C}$.

Topic: Physics, Thermal Properties of Matter
100 ml of 5 mâ H 2 SO 4 of density 1 gm/ml is mixed with 100 ml of 8 mâ H 2 SO 4 of density 1.25 g/mL. If there is no change in volume of resulting solution due to mixing, the molarity of the resulting mixture is-

Options:
A 5.5âM
B 6.5âM
C 4.5 M
D 5.26âM

The correct answer is:
4.5 M

Solution:
On mixing total vol of solution=200 ml
Molarityâ= molalityÃdÃ1000 molalityâÃ M â² +1000
or
1 Molarity = 1 d 1 molality + M â² 1000
Let molarity of 5 mâ H 2 SO 4 = M 1
1M1=1d1m+Mâ²1000
Where Mâ²=mol wt. of H2SO4
1M1=1115+981000=0.298â0.3
M1=10/3
Let molarity of 8âmâ H 2 SO 4 = M 2
1M2=11.2518+Mâ²1000
=11.25Ã18+981000
or 1M2=11.25 Ã 0.223
or M 2 = 1.25 0.223 =5.6âM
M mix. = M 1 V 1 + M 2 V 2 V 1 +V 2 =4.5 M

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
100 mL of 0.1 M HCl is taken in a beaker and to it 100 mL 0.1 M NaOH of is added in steps of 2 mL and the pH is continuously measured. Which of the following graphs correctly depicts the change in pH ?

Options:
A
B
C
D

The correct answer is:

Solution:
At equivalence point pH is 7 and pH increases with addition of NaOH.

Topic: Chemistry, Redox Reactions, JEE Main
100 mL of 0.1 M HCl is taken in a beaker and to it 100 mL 0.1 M NaOH of is added in steps of 2 mL and the pH is continuously measured. Which of the following graphs correctly depicts the change in pH ?

Options:
A
B
C
D

The correct answer is:

Solution:
At equivalence point pH is 7 and pH increases with addition of NaOH.

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2020 (03 Sep Shift 2)
100 mL of 0.3 M acetic acid is shaken with 0.8 g wood charcoal. The final concentration of acetic acid in the solution after adsorption is 0.125 M . The mass of acetic acid adsorbed per gram of charcoal is

Options:
A 1.05 g
B 0.0131 g
C 1.31 g
D 0.131 g

The correct answer is:
1.31 g

Solution:
Mass adsorbed = 100 0.3−0.125 60 1000 =1.05 g
Mass adsorbed per gram of charcoal =0.3-0.12560.8=1.3125

Topic: Chemistry, Surface Chemistry, JEE Main
$100 \mathrm{~mL}$ of $\frac{N}{5} \mathrm{HCl}$ was added to $1 \mathrm{~g}$ of pure $\mathrm{CaCO}_{3}$. What would remain after the reaction?

Options:
A $0.5 \mathrm{~g}$ of $\mathrm{CaCO}_{3}$
B Neither $\mathrm{CaCO}_{3}$ nor $\mathrm{HCl}$
C $50 \mathrm{~mL}$ of $\mathrm{HCl}$
D $25 \mathrm{~mL}$ of $\mathrm{HCl}$

The correct answer is:
Neither $\mathrm{CaCO}_{3}$ nor $\mathrm{HCl}$

Solution:
Number of moles of $\mathrm{CaCO}_{3}=\frac{W}{M}=\frac{1}{100}$ $=0.01 \mathrm{~mol}$
$\therefore$ Molar mass of $\mathrm{CaCO}_{3}=100 \mathrm{~g} \mathrm{~mol}^{-1}$
$100 \mathrm{~mL}$ of $\frac{N}{5} \mathrm{HCl}=\frac{100}{1000} \times \frac{1}{5}=0.02 \mathrm{~mol}$
The reaction is
$\therefore 0.01$ mole of $\mathrm{CaCO}_{3}$ reacts with $0.02$ mole of $\mathrm{HCl}$.
$\therefore$ Both are completely consumed in the given reaction.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
$100 \mathrm{~mL}$ of $\frac{N}{5} \mathrm{HCl}$ was added to $1 \mathrm{~g}$ of pure $\mathrm{CaCO}_{3}$. What would remain after the reaction?

Options:
A $0.5 \mathrm{~g}$ of $\mathrm{CaCO}_{3}$
B Neither $\mathrm{CaCO}_{3}$ nor $\mathrm{HCl}$
C $50 \mathrm{~mL}$ of $\mathrm{HCl}$
D $25 \mathrm{~mL}$ of $\mathrm{HCl}$

The correct answer is:
Neither $\mathrm{CaCO}_{3}$ nor $\mathrm{HCl}$

Solution:
Number of moles of $\mathrm{CaCO}_{3}=\frac{W}{M}=\frac{1}{100}$ $=0.01 \mathrm{~mol}$
$\therefore$ Molar mass of $\mathrm{CaCO}_{3}=100 \mathrm{~g} \mathrm{~mol}^{-1}$
$100 \mathrm{~mL}$ of $\frac{N}{5} \mathrm{HCl}=\frac{100}{1000} \times \frac{1}{5}=0.02 \mathrm{~mol}$
The reaction is
$\therefore 0.01$ mole of $\mathrm{CaCO}_{3}$ reacts with $0.02$ mole of $\mathrm{HCl}$.
$\therefore$ Both are completely consumed in the given reaction.

Topic: Chemistry, Some Basic Concepts of Chemistry, COMEDK, COMEDK 2012
100 mL of Na3PO4 solution contains 3.45 g of sodium. The molarity of the solution is .....×10-2 mol L-1. (Nearest integer)
[Atomic Masses - Na:23.0u,O:16.0u,P:31.0u]

Options:

The correct answer is:
50

Solution:
Number of mole of Na+ ion =3.4523 mol

So number of mole of Na3PO4=3.453×23=0.050 mol

Molarity =0.050100×1000=0.50=50×10-2M

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2021 (26 Aug Shift 2)
100 ml of O2g effuse from a porous container in 100 sec. 75 ml of unknown gas effuses under the same conditions of temperature and pressure in 150 sec. Calculate vapour density of unknown gas?

Options:
A 16
B 32
C 64
D 128

The correct answer is:
64

Solution:
ro2runknown=v1t1v2t2=MunknownMo2
⇒  100 × 150 100 × 75  =  M unknown 32
⇒ M unknown =128
∴ V.D.= 128 2 =64

Topic: Chemistry, States of Matter, JEE Main
116 g of a substance upon dissociation reaction, yields 7.5 g of hydrogen, 60 g of oxygen and 48.5 g of carbon. Given that the atomic masses of H,O and C are 1,16 and 12, respectively. The data agrees with how many formulae of the following?

A. CH3COOH

B. HCHO

C. CH3OOCH3

D. CH3CHO

Options:

The correct answer is:
2

Solution:
Organic Compound containg C,H,O⟶dissociationH7.5gram+O60gram+C48.5gram% of H in the given organic compound =7.5116×100=6.465%

(A) % of H=460×100=6.6%

(B) % of H=230×100=6.6%

(c) % of H=662×100=7.677%

(D) % of H=444×100=9.09%

The percentage of Hydrogen in the given organic compound is equal to 6.465% which is approximately equal to the percentage of hydrogen in CH3COOH and HCHO.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2022 (27 Jun Shift 2)
120 g of an organic compound which contains only carbon and hydrogen on complete combustion gives 330 g of CO2 and 270 g of water. The percentage of carbon and hydrogen in the organic compound are respectively

Options:
A 25 and 75
B 40 and 60
C 60 and 40
D 75 and 25

The correct answer is:
75 and 25

Solution:
Weight of the organic compound, W=120g,

Weight of carbondioxide (x)=330g

Weight of water (y)=270g

Percentage of carbon=1244XxWX100 =1244X330120X100 =75%Percentage of Hydrogen =218XyWX100 =218X270120X100 =25%

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2022 (24 Jun Shift 2)
1.22 g C6H5COOH is added into the 100 g solvent CH3COCH3 where data of Î Tb and Kb are given as:
ÎTb = 0.17, solute by Kb = 1.7 K kg/mol
Find out the molecular weight (g/mol) of the solute in the solution by interpreting the result.

Options:

The correct answer is:
122

Solution:
Δ T b = 0.17 = K b · m
0.17 = 1.7 × 1.22/M 100/1000
M = 122 g/mole
Benzoic acid is not associated or dissociated.

Topic: Chemistry, Solutions, JEE Main