datavorous/entrance-exam-dataset huggingface dataset:
What would be the effect on Kc if (i) More PCl5is added?

Options:
A Kc will increase
B Kc will decrease
C Kc will remain unchanged
D Kc first increase then decrease

The correct answer is:
Kc will remain unchanged

Solution:
Correct Option is : (C)
Kc will remain unchanged

Topic: Chemistry, Ionic Equilibrium, NEET
What would be the effect on Kc if Pressure is increased?

Options:
A Kc will increase
B Kc will decrease
C Kc will remain unchanged
D Kc first increase then decrease

The correct answer is:
Kc will remain unchanged

Solution:
Correct Option is : (C)
Kc will remain unchanged

Topic: Chemistry, Ionic Equilibrium, NEET
What would be the electrode potential for the given half-cell reaction at pH=5?_________.
2H2O→O2+4H⊕+4e-;Ered0=1.23V
R=8.314Jmol-1K-1;Temp=298K;oxygen under standard .atm.pressureof1bar

Options:

The correct answer is:
0.93

Solution:
On applyinf Nernst equation-

E=E°-0.0591nlogQE=-1.23-0.05914log[H+]4
=1.23+0.0591×pH=-1.23+0.0591×5
=-1.23+0.2955=-0.9345V=-0.93V

The Nernst equation defines the relationship between cell potential to standard potential and to the activities of the electrically active (electroactive) species. It relates the effective concentrations (activities) of the components of a cell reaction to the standard cell potential.

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2020 (08 Jan Shift 1)
What would be the heat released when an aqueous solution containing 0.5 mole of $\mathrm{HNO}_3$ is mixed with 0.3 mole of $\mathrm{OH}^{-}$?
(enthalpy of neutralisation is -57.1 kJ )

Options:
A 28.5 kJ
B 17.1 kJ
C 45.7 kJ
D 1.7 kJ

The correct answer is:
17.1 kJ

Solution:
0.5 mole of $\mathrm{HNO}_3$ equivalents to 0.5 mole of $\mathrm{H}^{+}$. But only 0.3 mole of $\mathrm{OH}^{-}$ions are available. Thus, 0.3 mole of $\mathrm{H}^{+}$ions will combine with 0.3 mole of $\mathrm{OH}^{-}$ions to form 0.3 mole of $\mathrm{H}_2 \mathrm{O}$ and 0.2 mole of $\mathrm{H}^{+}$ions will remain unreacted. Heat evolved when 1 mole of $\mathrm{H}^{+}$ions combine with 1 mole of $\mathrm{OH}^{-}$ions $=57.1 \mathrm{~kJ}$. Therefore, heat evolved wher 0.3 mole of $\mathrm{H}^{+}$ions combine with 0.3 mole of $\mathrm{OH}^{-}$ions $=0.3 \times 57.1=17.13 \mathrm{~kJ}$.

Topic: Chemistry, Thermodynamics (C), JIPMER, JIPMER 2015
What would be the molality of 20% (mass/mass) aqueous solution of KI ? (molar mass of KI=166 g mol-1 )

Options:
A 1.35
B 1.51
C 1.08
D 1.48

The correct answer is:
1.51

Solution:
Molality =Moles of soluteweight of solventg×1000
100 gm of solution contains 20 gm KI
So, the moles of KI =20166
Weight of solvent =100-20=80 gm
Molality =20×1000166×80
=1.51 Molal

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2019 (09 Apr Shift 2)
What would be the $\mathrm{pH}$ of a solution obtained by mixing $5 \mathrm{~g}$ of acetic acid and $7.5 \mathrm{~g}$ of sodium acetate and making the volume equal to $500 \mathrm{~mL}$ ?
$\left(\mathrm{K}_{\mathrm{a}}=1.75 \times 10^{-5}, \mathrm{pK}_{\mathrm{a}}=4.76\right)$

Options:
A
$\mathrm{pH}=4.70$

B
$\mathrm{pH} < 4.70$
C
$\mathrm{pH}$ of solution will be equal to $\mathrm{pH}$ of acetic acid

D
$4.76 < \mathrm{pH} < 5.0$

The correct answer is:
$4.76 < \mathrm{pH} < 5.0$

Solution:
$\begin{aligned}
\mathrm{pH} & =\mathrm{pK}_a+\log \frac{[\text { salt }]}{[\text { acid }]} \\
& =4.76+\log \frac{\frac{7.5}{500}}{\frac{5}{500}}=4.7+\log 1.5=4.87
\end{aligned}$
Hence correct answer is $4.76 < \mathrm{pH} < 5.0$

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2013 (25 Apr Online)
What would happen if oxygen availability to activated sludge flocs is reduced?

Options:
A
It will slow down the rate of degradation of organic matter

B
The center of flocs will become anoxic, which would cause death of bacteria and eventually breakage of flocs

C
Flocs would increase in size as anaerobic bacterial would grow around flocs

D
Protozoa would grow in large numbers

The correct answer is:
The center of flocs will become anoxic, which would cause death of bacteria and eventually breakage of flocs

Solution:
Activated sludge microorganisms require oxygen as they oxidise wastes to obtain energy for growth. Insufficient oxygen will slow down or kill off aerobic bacteria to the breakage of flocs.
It may also result in the production of the foul-smelling by-products of anaerobic decomposition, so, sufficient oxygen should always be maintained in the aeration tank to ensure complete waste stabilisation.

Topic: Biology, Microbes in Human Welfare
What would happen to a reversible reaction at equilibrium when an inert gas is added while the pressure remains unchanged

Options:
A More of the product will be formed
B Less of the product will be formed
C More of the reactants will be formed
D It remains unaffected

The correct answer is:
It remains unaffected

Solution:
When an inert gas is added to the system in equilibrium at constant pressure, then the total volume will increase. Hence, the number of moles per unit volume of various reactants and products will decrease. Hence, the equilibrium will shift towards the direction in which there is increase in number of moles of gases.

Topic: Chemistry, Chemical Equilibrium, JEE Main
What would happen when a solution of potassium chromate is treated with an excess of dilute nitric acid?

Options:
A
$\mathrm{Cr}_2 \mathrm{O}_7^{2-}$ and $\mathrm{H}_2 \mathrm{O}$ are formed

B
$\mathrm{CrO}^{2-}{ }_4$ is reduced to $+3$ state of $\mathrm{Cr}$

C
$\mathrm{CrO}_4^{2-}$ is oxidized to $+7$ state of $\mathrm{Cr}$

D
$\mathrm{Cr}^{3+}$ and $\mathrm{Cr}_2 \mathrm{O}^{2-}{ }_7$ are formed

The correct answer is:
$\mathrm{Cr}_2 \mathrm{O}_7^{2-}$ and $\mathrm{H}_2 \mathrm{O}$ are formed

Solution:
$\mathrm{Cr}_2 \mathrm{O}_7^{2-}+\mathrm{OH}^{-} \rightarrow 2 \mathrm{CrO}_4^{2-}+\mathrm{H}^{+}$. The above equilibrium shifts to L.H.S. on addition of acid.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2003
What would happen when one grows a recombinant bacterium in the bioreactor but forgets to add antibiotic tothe medium in which the recombinant is growing?

Options:

None

Solution:
In the absence of an antibiotic, there will be no stress on recombinants to retain the plasmid containing the gene of interest. Maintaining a high copy number of plasmids would be a metabolic burden to the microbial cells and they would thus tend to loose the plasmid.

Topic: Biology, Biotechnology - Principles and Processes
Wheatstone bridge principle is used to measure the specific resistance S1 of given wire, having length L, radius r. If X is the resistance of wire, then specific resistance is : S1=Xπr2L. If the length of the wire gets doubled then the value of specific resistance will be :

Options:
A S14
B 2 S1
C S12
D S1

The correct answer is:
S1

Solution:
The specific resistance of a material, also known as electrical resistivity, is a measure of how strongly a specific material opposes the flow of electric current.

Specific resistance is an intrinsic property of a material and is dependent on factors such as the material's composition, temperature, and crystal structure. It is the counterpart of conductivity, with which it is inversely proportional.

Hence, even if the length of the wire is doubled, the specific resistance remains the same, i.e., S1.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2024 (27 Jan Shift 2)
Wheezing occurs in

Options:
A Asthma
B Emphysema
C Silicosis
D Asbestosis

The correct answer is:
Asthma

Solution:
Asthma is a respiratory disorder with inflammation of bronchi and bronchioles. This results in spasmodic contractions. The main cause of asthma is allergic reactions or hypersensitivity. Airway narrowing is also seen due to which there is difficulty in breathing and this produces wheezing sound. This can be heard through stethoscope and is the characteristic feature of this condition.

Topic: Biology, Breathing and Exchange of Gases, NEET
When 0.01mol of an organic compound containing 60% carbon was burnt completely, 4.4 g of CO2 was produced. The molar mass of compound is _____ gmol-1 (Nearest integer)

Options:

The correct answer is:
200

Solution:
Let M is the molar mass of the compound g/mol
mass of compound =mole × molecular mass = 0.01M g

mass of carbon =0.01M×60100 ....(i)

moles of carbon =mass of compoundmolecular mass of carbon = 0.01M12×60100 ....(ii)

Thus, moles of CO2 from combustion =4.444=0.1 moles of carbon dioxide

Using POAC,
Thus, from (i) and (ii),
⇒0.01M12×60100=4.444

⇒M=4.444×10060×120.01=200 g/mol

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2023 (29 Jan Shift 2)
When 0.15 g of an organic compound was analyzed using Carius method for estimation of bromine, 0.2397 g of AgBr was obtained. The percentage of bromine in the organic compound is _________.(Nearest integer)

[Atomic mass : Silver =108, Bromine =80]

Options:

The correct answer is:
68

Solution:
Moles of Br= Moles of AgBr obtained

⇒ Mass of Br=0.2397188×80 g

therefore %Br in the organic compound

=WBrWT×100

=0.2397×80188×0.15×100=0.85×80

=68

⇒ Nearest integer is '68'

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2021 (20 Jul Shift 2)
When 0.1 mol MnO42- is oxidized the quantity of electricity required to completely MnO42- to MnO4- is

Options:
A 96500Â C
B 2Ã96500Â C
C 9650 C
D 96.50Â C

The correct answer is:
9650 C

Solution:
MnO42-âMnO4-
0.1 mole
v.f.=1
So, 0.1 mole=96500Ã0.1
=9650 C charge is required

Topic: Chemistry, Electrochemistry, JEE Main
When 1 mole of an ideal monoatomic gas is compressed adiabatically the internal energy change involved is 24 cals. The temperature rise(in kelvin) is

Options:

The correct answer is:
8

Solution:
⇒Δ U=nC v ΔT
⇒24=1× 3 2 R×ΔT ∴ for mono atomic gas  C v = 3 2 R
⇒ΔT= 24×2 3×2 =8 K

Topic: Chemistry, Thermodynamics (C), JEE Main
When 0.01 moles of the following acids are dissolved in 1 L of H2O , the [H+] will be greatest in: -

Options:
A HNO2;pka=3.0
B HCOOH ;pka=3.75
C HCN;pka=9.4
D CH3COOH;pka=4.75

The correct answer is:
HNO2;pka=3.0

Solution:
[H+]=kaÃc
kaâpka â[H+]â

Topic: Chemistry, Ionic Equilibrium, JEE Main
When 2.44 grams of benzoic acid C6H5COOH dissolved in 25 grams of benzene, it shows depression of freezing point equal to 2.2 K. Molal depression constant of benzene is 5.0 K kg mol-1. What is the percentage association of acid, if it forms dimer in solution?

Options:
A 50%
B 90%
C 95%
D 77%

The correct answer is:
90%

Solution:
Given ,

WB=2.44g , Kf=5 K kg mol-1WA =25g , ∆Tf =2.2K
WB=2.44g , Kf=5 K kg mol-1WA =25g , ∆Tf =2.2K Now ,∆Tf = Kf ×WBMB ×1000WAMB=5×2.44×10002.2×25=221.8g/mol2C6H5COOH ⇔C6H5COOH 2If x is the degree of association , (1-x) mole ofbenzoic acid left undissociated and corresponding x2as associated moles of C6H5COOH at equilibrium.Total number of moles at equilibrium 1-x+x2=1-x2i=Normal molecular massAbnormal molecular mass1-x2 =122221.8x=0.900=90%

Topic: Chemistry, Solutions, JEE Main
When 2 litre of ideal gas expands isothermally into vacuum to a total volume of 6 litre, the change in internal energy is _____ J. (Nearest integer)

Options:

The correct answer is:
0

Solution:
For ideal gas U=fT and for isothermal process, ΔU=0,
Also Pext = 0, thus work done in vacuum is also zero. Internal energy is a fuction of temperature.

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2023 (30 Jan Shift 1)
When 2-butyne is treated with $\mathrm{H}_2$ /Lindlar's catalyst, compound $\mathrm{X}$ is produced as the major product and when treated with $\mathrm{Na}$ /liq. $\mathrm{NH}_3$ it produces $\mathrm{Y}$ as the major product. Which of the following statements is correct?

Options:
A
Y will have higher dipole moment and higher boiling point than $\mathrm{X}$ Y

B
$\mathrm{Y}$ will have higher dipole moment and lower boiling point than $\mathrm{X}$

C
X will have lower dipole moment and lower boiling point than $\mathrm{Y}$

D
$\mathrm{X}$ will have higher dipole moment and higher boiling point than Y

The correct answer is:
$\mathrm{X}$ will have higher dipole moment and higher boiling point than Y

Solution:
When 2-butyne is treated with $\mathrm{H}_2$ /Lindlar's catalyst, compound $\mathrm{X}$ (cis-2-butene) is produced as the major product; and when treated with $\mathrm{Na} /$ liq $\mathrm{NH}_3$ it produces $\mathrm{Y}$ (trans -2- butene) as the major product. Cisisomer $(\mathrm{X})$ will have higher dipole moment and higher boiling point than $\operatorname{trans}(\mathrm{Y})$.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
When 3.00 g of a substance X' is dissolved in 100 g of CCl4, it raises the boiling point by 0.60 K. The molar mass of the substance ' X' is ___ g mol-1.
(Nearest integer).
[Given Kb for CCl4 is 5.0 K kg mol-1

Options:

The correct answer is:
250

Solution:
ΔTb=Kb× molality

0.60=5×3/M100/100

M=Molar mass

M=250.

Topic: Chemistry, Solutions, JEE Main, JEE Main 2021 (25 Jul Shift 2)
When $3^{2023}$ is divided by 16 , the remainder obtained is

Options:
A 15
B 11
C 9
D 7

The correct answer is:
11

Solution:
$\begin{aligned} & \text3^{2023}=3 \cdot\left(3^2\right)^{1011}=3(8+1)^{1011}=3(1+8)^{1011} \\ & =3\left[{ }^{1011} C_0+{ }^{1011} C_1 8+{ }^{1011} C_2 8^2+\ldots .+{ }^{1011} C_{1011} 8^{1011}\right] \\ & =3(1+1011 \times 8+16 k) \\ & \therefore \quad \text { Remainder when } 3^{2023} \text { is divided by } 16 \\ & =\text { remainder when } 3(1+1011 \times 8) \text { is divided by } 16 \\ & \Rightarrow 3(1+(1010+1) \times 8)=3\left(1+8+16 k_1\right) \\ & =3(1+8)=27 \\ & \therefore \text { Remainder }=11 .\end{aligned}$

Topic: Mathematics, Binomial Theorem, TS EAMCET, TS EAMCET 2023 (13 May Shift 2)
When ${ }_3 \mathrm{Li}^{\prime}$ nuclei are bombarded by protons, and the resultant nuclei are ${ }_4 \mathrm{Be}^8$, the emitted particles will be

Options:
A
neutrons

B
alpha particles

C
beta particles

D
gamma photons

The correct answer is:
gamma photons

Solution:
Gamma-photon

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2006
When 0.04 F of electricity is passed through a solution of CaSO4, then the weight of Ca2+ metal deposited at the cathode is

Options:
A 0.2 g
B 0.4 g
C 0.6 g
D 0.8 g

The correct answer is:
0.8 g

Solution:
Ca 2 + + 2 e - → Ca â
E Ca = 4 0 2 = 2 0
w Ca = E Ca × No. of faradays = 2 0 × 0.04 = 0.8 g .

Topic: Chemistry, Electrochemistry, JEE Main
When 5.1 g of solid NH4HS is introduced into a two litre evacuated flask at 27°C, 20% of the solid decomposes into gaseous ammonia and hydrogen sulphide. The Kp for the reaction at 27°C is x×10-2. The value of x is (Integer answer)
[Given 1R=0.082 L atm K-1 mol-1

Options:

The correct answer is:
6

Solution:
Moles of NH4HS initially taken =5.1 g51 g/mol=0.1 mol

volume of vessel =2 L

NH4HS(s)⇌NH3( g)+H2S(g)

t=0 0.1 mol

t=∞0.1(1-0.2)0.1×0.20.1×0.2

Partial pressure of each gaseous component

P=nRTV=0.1×0.2×0.082×3002

=0.246 atm

⇒kP=PNH3×PH2 S=(0.246)2=0.060516

=6.05×10-2

⇒6

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2021 (27 Aug Shift 2)
When 5 V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is 2.5×10-4 m s-1 . If the electron density in the wire is 8×1028 m-3 , the resistivity of the material is close to:

Options:
A 1.56×10-5 Ω m
B 1.6×10-8 Ω m
C 1.6×10-7 Ω m
D 1.6×10-6 Ω m

The correct answer is:
1.56×10-5 Ω m

Solution:
Potential difference= 5 V, length =0.1 m=l and the electron speed = drift velocity vd=2.5×10-4 m s-1, electron density n = 8 × 1 0 2 8 m - 3 and the charge on each electron(e) =1.6×10-19 C. We know i=nAevd ...(i). And V=iR ...(ii)
Resistance R is also equal to, ρlA⇒R=ρlA
ρ=ARl ρ = Resistivity
=Al×Vi [from (ii)]=Avl×nA evd & [from (i)]=Vl×n×e×vd
= 5 0.1 × 8 × 1 0 2 8 × 1.6 × 1 0 - 1 9 × 2.5 × 1 0 - 4
=0.156×10-4 Ω m=1.56×10-5 Ω m

Topic: Physics, Current Electricity, JEE Main, JEE Main 2015 (04 Apr)
When 9.45 g of ClCH2COOH is added to 500 mL of water, its freezing point drops by 0.5°C. The dissociation constant of ClCH2COOH is x×10-3. The value of x is off to the nearest integer)

KfH2O=1.86 K kg mol-1

Options:

The correct answer is:
36

Solution:
ClCH2COOH⇌ClCH2COO⊖+H+

i=1+(2-1)α

i=1+α

ΔTf=i.Kf.m

0.5=(1+α)(1.86)9.4594.55001000

53.72=1+α⇒α=1.283.72

α=3293

Ka=Cα2C−Cα=Cα21−αC=0.1500/1000=0.2

Ka=0.232/9321−32/93=0.2×32293×61

=0.036

Ka=36×10−3

Topic: Chemistry, Solutions, JEE Main, JEE Main 2021 (24 Feb Shift 1)
When 10.6 g of a non volatile substance is dissolved in 750 g of ether, its boiling point is raised 0 .266 o C . The molecular weight of the substance is
(Given: Molal boiling point constant for ether is 2 .0 o C kg/mol )
Report your answer by rounding it up to nearest whole number.

Options:

The correct answer is:
106

Solution:
ÎTb=iÃKbÃm
0.226=1Ã2.0Ã10.6M0.75
M= 2×10.6 0.266×0.75 =106.26

Topic: Chemistry, Solutions, JEE Main
When $10 \mathrm{~g}$ of a non-volatile solute is dissolved in $100 \mathrm{~g}$ of benzene, it raises boiling point by $1^{\circ} \mathrm{C}$ then molecular mass of the solute is ( $K_b$ for benzene $=2.53 \mathrm{k}-\mathrm{m}^{-1}$ )

Options:
A $223 \mathrm{~g}$
B $233 \mathrm{~g}$
C $243 \mathrm{~g}$
D $253 \mathrm{~g}$

The correct answer is:
$253 \mathrm{~g}$

Solution:
$\Delta T_b=\frac{K_b \times W \times 1000}{m \times W}$
$m=\frac{K_b \times W \times 1000}{\Delta T_b \times W}=\frac{2.53 \times 10 \times 1000}{1 \times 100}=253 \mathrm{~g}$

Topic: Chemistry, Solutions, JEE Main
When \( 10 \mathrm{~mL} \) of an aqueous solution of \( \mathrm{Fe}^{2+} \) ions was titrated in the presence of dil \( \mathrm{H}_{2} \mathrm{SO}_{4} \) using diphenylamine indicator, \( 15 \mathrm{~mL} \) of \( 0.02 \mathrm{M} \) solution of \( \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \) was required to get the end point. The molarity of the solution containing \( \mathrm{Fe}^{2+} \) ions is \( \mathrm{x} \times 10^{-2} \mathrm{M} \). The value of \( \mathrm{x} \) is . (Nearest integer)

Options:

The correct answer is:
18

Solution:
milli-equivalents of Fe2+= milli-equivalents of K2Cr2O7

M×10×1=0.02×15×6

M=0.18=18×10-2 M

Topic: Chemistry, Redox Reactions, JEE Main
When 10 mL of 0.1 M acetic acid (pK a  = 5) is titrated against 10 mL of 0.1 M ammonia solution (pK b =5), the equivalent point will occur at pH :

Options:
A 5
B 6
C 7
D 9

The correct answer is:
7

Solution:
CH 3 COOHâ(aq) + NH 3 â(aq) â CH 3 COONH 4 â(aq) 1âmâmol 1âmâmol 1âmâmol
CH 3 COONH 4 â=â 1âmM 20âmL â=â0.05 M
Salt of weak Acid + weak base.
pH = 1 2 p K w + p K a - p K b
= 1 2 1 4 + 5 - 5 = 7

Topic: Chemistry, Ionic Equilibrium, JEE Main
When 10 mL of an aqueous solution of KMnO4 was titrated in acidic medium, equal volume of 0.1M of an aqueous solution of ferrous sulphate was required for complete discharge of colour. The strength of KMnO4 in grams per litre is _______×10-2. (Nearest integer) [Atomic mass of K=39,Mn=55,O=16]

Options:

The correct answer is:
316

Solution:
MnO4-vf=5+Fe2+⟶vf=1Mn2++Fe3+

Mili equ. of MnO4=- mili equ. of Fe2+

5[M×10]=1[0.1×10]

M=0.15=0.02 mole Lit

Strength of KMnO4=0.02×158gramLit=3.16 g/L=316×10-2 g/L

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2021 (27 Aug Shift 1)
When 12.2 g of benzoic acid is dissolved in 100 g of water, the freezing point of solution was found to be -0.93 °CKfH2O=1.86 K kg mol-1. The number n of benzoic acid molecules associated (assuming 100% association) is _.

Options:

The correct answer is:
2

Solution:
ΔTf=i×kf×m

0--0.93=i×1.86×12.2122×100×1000

i=0.931.86=0.5

i=1+1n-1α

12=1+1n-1×1

n=2

Topic: Chemistry, Solutions, JEE Main, JEE Main 2021 (26 Feb Shift 2)
When $20 \mathrm{~g}$ of naphthoic acid $\left(\mathrm{C}_{11} \mathrm{H}_8 \mathrm{O}_2\right)$ is dissolved in $50 \mathrm{~g}$ of benzene $\left(K_f=1.72 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)$, a freezing point depression of $2 \mathrm{~K}$ is observed. The van't Hoff factor $(i)$ is

Options:
A
$0.5$

B
1

C
2

D
3

The correct answer is:
$0.5$

Solution:
Actual molecular weight of napthoic acid $\left(\mathrm{C}_{11} \mathrm{H}_8 \mathrm{O}_2\right)=172$
$$
\begin{aligned}
\text { Molecular mass (calculated ) } & =\frac{1000 . K_f \cdot w}{W . \Delta T_f} \\
& =\frac{1000 \times 1.72 \times 20}{50 \times 2} \\
& =344
\end{aligned}
$$

Topic: Chemistry, Solutions, JEE Main
When $25 \mathrm{~g}$ of a non-volatile solute is dissolved in $100 \mathrm{~g}$ of water, the vapour pressure is lowered by $2.25 \times 10^{-1} \mathrm{~mm}$. If the vapour pressure of water at $20^{\circ} \mathrm{C}$ is $17.5 \mathrm{~mm}$, what is the molecular weight of the solute?

Options:
A $206$
B $302$
C $350$
D $276$

The correct answer is:
$350$

Solution:
Given,
weight of non-volatile solute,
$$
w=25 \mathrm{~g}
$$
Weight of solvent, $W=100 \mathrm{~g}$
Lowering of vapour pressure,
$$
p^{\circ}-p_s=0.225 \mathrm{~mm}
$$
Vapour pressure of pure solvent,
$$
p^{\circ}=17.5 \mathrm{~mm}
$$
Molecular weight of solvent $\left(\mathrm{H}_2 \mathrm{O}\right), M=18 \mathrm{~g}$
Molecular weight of solute, $m=$ ?
According to Raoult's law
$$
\begin{aligned}
\frac{p^{\circ}-p_s}{p^{\circ}} & =\frac{w \times M}{m \times W} \\
\frac{0.225}{17.5} & =\frac{25 \times 18}{m \times 100} \\
m & =\frac{25 \times 18 \times 17.5}{22.5} \\
& =350 \mathrm{~g}
\end{aligned}
$$

Topic: Chemistry, Solutions, JEE Main
When \( 25 \mathrm{~g} \) of \( \mathrm{Na}_{2} \mathrm{SO}_{4} \) is dissolved in \( 10^{3} \mathrm{Kg} \) of solution, its concentration will be

Options:
A \( 2.5 \mathrm{ppm} \)
B \( 25 \mathrm{ppm} \)
C \( 250 \mathrm{ppm} \)
D \( 100 \mathrm{ppm} \)

The correct answer is:
\( 25 \mathrm{ppm} \)

Solution:
ppm: When one-millionth part of solute is present in a solution called part per million (ppm).

ppm= Mass of soluteMass of solution×106 =0.025103×106=25 ppm

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
When 35 mL of 0.15 M lead nitrate solution is mixed with 20 mL of 0.12 M chromic sulphate solution, ____________×10-5 moles of lead sulphate precipitate out. (Round off to the Nearest Integer).

Options:

The correct answer is:
525

Solution:
3 PbNO32+Cr2SO43→3PbSO4+2CrNO33

35 ml 20 ml0.15 M0.12 M

=5.25 m.mol=2.4 m.mol5.25 m.mol

=5.25×10-3 mol

Therefore moles of PbSO4 formed

=5.25×10-3 =525×10-5

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2021 (16 Mar Shift 2)
When $0.50 Ã$ X-rays strike a material, the photoelectrons from the $\mathrm{k}$ shell are observed to move in a circle of radius $23 \mathrm{~mm}$ in a magnetic field of $2 \times 10^{-2}$ tesla acting perpendicularly to the direction of emission of photoelectrons. What is the binding energy of k-shell electrons?

Options:
A $3.5 \mathrm{keV}$
B $6.2 \mathrm{keV}$
C $2.9 \mathrm{keV}$
D $5.5 \mathrm{keV}$

The correct answer is:
$6.2 \mathrm{keV}$

Solution:
As we know,

$$

\mathrm{F}=\mathrm{qvB}=\mathrm{m} \frac{\mathrm{v}^{2}}{\mathrm{R}} \Rightarrow \mathrm{v}=\frac{\mathrm{q}}{\mathrm{m}} \mathrm{BR}

$$

The kinetic energy of the photoelectron

$$

\begin{array}{l}

=\frac{1}{2} m v^{2}=\frac{1}{2} \frac{e^{2} B^{2} R^{2}}{m} \\

=\frac{1}{2} \frac{\left(1.6 \times 10^{-19}\right)^{2}\left(2 \times 10^{-2}\right)^{2}\left(23 \times 10^{-3}\right)^{2}}{\left(9.1 \times 10^{-31}\right)} \\

=2.97 \times 10^{-15} \mathrm{~J} \\

=\frac{2.97 \times 10^{-15}}{1.6 \times 10^{-19}}=18.36 \mathrm{keV}

\end{array}

$$

Fnergy of the incident $\mathrm{n}=\frac{\mathrm{hc}}{\lambda}$

$$

=\frac{12.4}{0.50}=24.8 \mathrm{keV}

$$

Therefore, Binding energy $=24.8-18.6=$ $6.2 \mathrm{keV}$

Topic: Physics, Dual Nature of Matter, JEE Main
When 90Th228 transforms to 83Bi212 , then the number to the emitted α and β -particles are, respectively

Options:
A 8Î±,Â 7Î²
B 4Î±,Â 7Î²
C 4Î±,Â 4Î²
D 4Î±, 1Î²

The correct answer is:
4Î±, 1Î²

Solution:
Z=90ThA=228âZ′=83BiA′=212
Number of Î± -particles emitted
nÎ±=A-A′4=228-2124=4
Number of Î² -particles emitted
nÎ²=2nÎ±-Z+Z′=2Ã4-90+83=1.

Topic: Physics, Nuclear Physics, JEE Main