datavorous/entrance-exam-dataset huggingface dataset:
Calculate the overall order of a reaction which has the rate expression: (i) Rate = k [A]1/2[B]3/2and (ii) Rate = k [A]3/2[B]–1

Options:
A (i) 2 and (ii) 1/2
B (i) 1/2 and (ii) 2
C (i) 1 and (ii) 1/2
D (i) 1/2 and (ii) 1/2

The correct answer is:
(i) 2 and (ii) 1/2

Solution:
Correct Option is : (A)
(i) 2 and (ii) 1/2

Topic: Chemistry, Chemical Kinetics, NEET
Calculate the pH of CH3OONa at 25o if the concentration of CH3COONa is 0.01M . (Given pKa of acetic acid is 4.74 ).

Options:
A 3.37
B 4.37
C 4.74
D 0.474

The correct answer is:
3.37

Solution:
pH=7+12pka+12logc
=7+12×4.74+12log⁡10-2
∴ pH=8.37

Topic: Chemistry, Ionic Equilibrium, NEET
Calculate the partial pressure of carbon monoxide from the following
CaCO 3 (s) → Δ CaO(s) +  CO 2 ↑;   K p  = 8 ×  10 −2
CO2(g)+C(s)→2CO(g) ; Kp=2

Options:
A 0.2
B 0.4
C 1.6
D 4

The correct answer is:
0.4

Solution:
Given, CaCO 3 (s) → Δ CaO(s) +  CO 2 (g)
C(s) +  CO 2 (g) → 2CO(g)
K P 2  =  [P CO ] 2 P CO 2 ;   P CO  =  K P 1  ×  K P 2
P CO = 8× 10 −2 ×2 = 16× 10 −2

Topic: Chemistry, Chemical Equilibrium, NEET
Calculate the pH of 0.08M solution of hypochlorous acid, HOCl. The ionization constant of the acid is 2.5 × 10⁻⁵.

Options:
A pH = 2.85
B pH = 4.60
C pH = 3.89
D pH = 3.55

The correct answer is:
pH = 2.85

Solution:
Correct Option is : (A)
pH = 2.85

Topic: Chemistry, Ionic Equilibrium, NEET
Calculate the pH of a 1.0 × 10-8M solution of HCl.

Options:
A pH = 2
B pH = 4
C pH = 7
D pH = 8

The correct answer is:
pH = 7

Solution:
Correct Option is : (C)
pH = 7

Topic: Chemistry, Ionic Equilibrium, NEET
Calculate the pH of a 0.10M ammonia solution.(The dissociation constant of ammonia, Kb = 1.77 × 10-5)

Options:
A pH = 11.2
B pH = 10.8
C pH = 11.0
D pH = 10.5

The correct answer is:
pH = 11.2

Solution:
Correct Option is : (A)
pH = 11.2

Topic: Chemistry, Ionic Equilibrium, NEET
Calculate the pH of a buffer when the reduction potential of hydrogen electrode placed in the buffer solution is found to be -0.413 V.

Options:
A 10
B 4
C 7
D 12

The correct answer is:
7

Solution:
Apply Nernst Equation on the given reduction reaction of hydrogen.

H+aq + e-→12H2g

Ecell= Ecell∘ -0.059nlog1H+

Where EH+/H2∘=0 V and n=1

-0.413=-0.059log1H+ or 0.4140.059=-logH+=pH

pH=7

Topic: Chemistry, Electrochemistry, NEET
Calculate the pH of a solution formed by mixing equal volumes of two solutions A and B of a strong acid having pH = 6 and pH = 4 respectively.

Options:
A 4.9
B 2.1
C 4.3
D 6.2

The correct answer is:
4.3

Solution:
Correct Option is : (C)
4.3

Topic: Chemistry, Ionic Equilibrium, NEET
Calculate the pH of a solution obtain by mixing 10 ml of 1 N sodium acetate and 50 ml of 2 N acetic acid (Ka=1.8Ã10-5),

Options:
A 3.74
B 5
C 6
D 7

The correct answer is:
3.74

Solution:
pH=pKa+log[Salt][Acid]
pH=-log(1.8×10-5)+log[10][100]
=-log1.8+5+log10-1
=-0.2553+5-1=3.7447

Topic: Chemistry, Ionic Equilibrium, NEET
Calculate the pH of a solution obtained by mixing 50 ml of 0.01 M Ba(OH)2 solution with 50 ml water.

Options:
A 6
B 10
C 2
D 12

The correct answer is:
12

Solution:
N1V1Ba(OH)2=N2V2mixture
[Normality of solution of Ba(OH)2=2× molarity, as its mol. wt = 2 x eq. wt as n- factor is 2 for it. ]
0.01×2×50=N2×(50+50) (as it is assumed as 100% ionised).
N2=0.01×100100=0.01 N
pOH=-log[OH-]=-log10-2=2log10
pOH=2
pH+pOH=14
pH+2=14
pH=12

Topic: Chemistry, Ionic Equilibrium, NEET
Calculate the pH of buffer solution prepared by dissolving 0.20 mol of sodium cyanate (NaCNO) and 1.0 mol of cynic acid (HCNO) in enough water to make 1.0 litre of slution. Ka(HCNO)=2.0×10-4.

Options:
A 0
B 3.0
C 4.4
D 5.0

The correct answer is:
3.0

Solution:
pH =-logKa(HCNO)+log[NaCNO][HCNO]
=−log(2.0× 10 −4 )+log 0.20 mol/L 1.0 mol/L
=3.7+(-0.67)
=3.03â3.0

Topic: Chemistry, Ionic Equilibrium, NEET
Calculate the pH of the following solutions a) 2 g of TlOH dissolved in water to give 2 litre of solution. b) 0.3 g of Ca(OH)2dissolved in water to give 500 mL of solution. c) 0.3 g of NaOH dissolved in water to give 200 mL of solution. d) 1mL of 13.6 M HCl is diluted with water to give 1 litre of solution.

Options:
A (a) pH = 12.29 (b) pH = 12.40 (c) pH = 13.04 (d) pH = 11.31
B (a) pH = 12.29 (b) pH = 13.40 (c) pH = 13.04 (d) pH = 0.58
C (a) pH = 12.61 (b) pH = 12.40 (c) pH = 13.04 (d) pH = 0.58
D (a) pH = 12.61 (b) pH = 13.40 (c) pH = 13.04 (d) pH = 0.58

The correct answer is:
(a) pH = 12.29 (b) pH = 12.40 (c) pH = 13.04 (d) pH = 11.31

Solution:
Correct Option is : (A)
(a) pH = 12.29 (b) pH = 12.40 (c) pH = 13.04 (d) pH = 11.31

Topic: Chemistry, Ionic Equilibrium, NEET
Calculate the pH of the resultant mixtures a) 10 mL of 0.2M Ca(OH)2+ 25 mL of 0.1M HCl b) 10 mL of 0.01M H2SO4+ 10 mL of 0.01M Ca(OH)2c) 10 mL of 0.1M H2SO4+ 10 mL of 0.1M KOH

Options:
A a) pH = 11.91, b) pH = 7.08, c) pH = 1.30
B a) pH = 9.08, b) pH = 11.21, c) pH = 7.00
C a) pH = 8.00, b) pH = 9.08, c) pH = 11.21
D a) pH = 7.00, b) pH = 8.00, c) pH = 9.08

The correct answer is:
a) pH = 11.91, b) pH = 7.08, c) pH = 1.30

Solution:
Correct Option is : (A)
a) pH = 11.91, b) pH = 7.08, c) pH = 1.30

Topic: Chemistry, Ionic Equilibrium, NEET
Calculate the pH of the solution in which 0.2M NH4Cl and 0.1M NH3are present. The pKb of ammonia solution is 4.75.

Options:
A pH = 9.25
B pH = 9.50
C pH = 9
D pH = 10.00

The correct answer is:
pH = 9

Solution:
Correct Option is : (C)
pH = 9

Topic: Chemistry, Ionic Equilibrium, NEET
Calculate the $\mathrm{pOH}$ of a solution $25^{\circ} \mathrm{C}$ that contains $1 \times 10^{-10} \mathrm{M}$ of hydronium ions i.e. $\mathrm{H}_3 \mathrm{O}^{+}$:

Options:
A 4.000
B 9.000
C 1.000
D 7.000

The correct answer is:
4.000

Solution:
Given $\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=1 \times 10^{-10} \mathrm{M}$ at $25^{\circ}\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]=10^{-14}$
$\therefore \quad\left[\mathrm{OH}^{-}\right]=\frac{10^{-14}}{10^{-10}}=10^{-4}$
Now, $\quad\left[\mathrm{OH}^{-}\right]=10^{-p^{\text {OH }}}$
$=10^{-4}=10^{-P^{\mathrm{OH}}}$
$\therefore \quad p^{\mathrm{OH}}=4$.

Topic: Chemistry, Ionic Equilibrium, NEET, NEET 2007
Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.

Options:
A -0.40 V
B 0.40 V
C -0.59 V
D 0.59 V

The correct answer is:
0.40 V

Solution:
Correct Option is : (B)
0.40 V

Topic: Chemistry, Electrochemistry, NEET
Calculate the power factor of the circuit shown in figure

Options:
A 0.2
B 0.4
C 0.8
D 0.6

The correct answer is:
0.8

Solution:
Resistance of the circuit,
R=R1+R2=40 Î©+40 Î©=80 Î©
Impedance of the circuit,
Z=R2+XL-XC2=802+100-402
=802+602=100 Î©
Power factor, cosâ¡Ï=RZ=80100=0.8

Topic: Physics, Alternating Current, NEET
Calculate the pressure of gas at constant volume, if a 10g of a gas at one atmospheric pressure is cooled from 273oC to 0oC .

Options:
A 1/2 atm
B 1/273 atm
C 2 atm
D 273 atm

The correct answer is:
1/2 atm

Solution:
T 1 â=â 273 o Câ=â273â+â273âKâ=â546âK
T 2  =  0 o C = 273 +  0 o C = 273 K
P1=1 ; P2=?
According to Gay-Lussacâs law
P1T1=P2T2
â´ P 2 = P 1 T 2 T 1 = 1Ã273âK 546âK âatm= 1 2 âatm
` Use relation P1T1=P2T2

Topic: Chemistry, States of Matter, NEET
Calculate the radius of Bohr’s fifth orbit for hydrogen atom.

Options:
A 1.3225 nm
B 1.5423 nm
C 1.6241 nm
D 1.8113 nm

The correct answer is:
1.3225 nm

Solution:
Correct Option is : (A)
1.3225 nm

Topic: Chemistry, Structure of Atom, NEET
Calculate the radius of Xe atom, If the edge of the unit cell FCC is 620 pm.

Options:
A 189.37 pm
B 209.87 pm
C 219.25 pm
D 235.16 pm

The correct answer is:
219.25 pm

Solution:
In FCC arrangement surface diagonal is given by 2a
Hence, 4r=2a
r=a22 ; r=62022=219.25 pm

Topic: Chemistry, Solid State, NEET
Calculate the ratio of current flowing through the battery at t=0 and t=∞. ( t=0 is the time of closing of the switch)

Options:
A 1
B 2
C 12
D 0

The correct answer is:
1

Solution:
At t=0 capacitor is short circuit and inductor is open circuit.
i1=205+5=2A
At t=∞ capacitor is open circuit and inductor is a short circuit.
i2=205+5=2 A
∴ i1i2=22=1

Topic: Physics, Current Electricity, NEET
Calculate the ratio of the capacitance of two capacitors filled with dielectrics of the same dimensions but of different values K and K4 arranged in two ways as shown in figure (i) and (ii)

Options:
A 5:2

B 25:16
C 5:4
D 2:5

The correct answer is:
25:16

Solution:
Here, C 1 = Îµ 0 K 1 A 2 d + Îµ 0 K 2 A 2 d
= Îµ 0 KA 2 d + Îµ 0 KA 8 d Given : K 1 = K and K 2 = K 4
= Îµ 0 KA d 1 2 + 1 8 = Îµ 0 KA d 4 + 1 8
= 5 Îµ 0 KA 8 d ...... (i)
and C 2 = Îµ 0 A d 2 K 1 + d 2 K 2 = Îµ 0 A d 2 K + d 2 K 4
= 2 K Îµ 0 A 5 d .... (ii)
Dividing eqs. (i) by (ii)
C 1 C 2 = 5 Îµ 0 KA 8 d 2 K Îµ 0 A 5 d = 5 Ã 5 2 Ã 8 = 2 5 1 6

Topic: Physics, Electrostatics, NEET
Calculate the ratio of the speed of sound in neon to that in water vapour at any temperature. Given that the molecular weight of neon gas =2.02Ã10-2 kg mole-1 and molecular weight of water vapour =1.8Ã10-2 kg mole-1. Î³neon=53 and Î³ for water vapour =43.

Options:
A 1.1
B 1.7
C 1.9
D 1.3

The correct answer is:
1.1

Solution:
Let v1 and v2 be the speeds of sound in neon and water vapour respectively. Suppose that M1 and M2 are the molecular weights of neon and water vapour respectively. If γ is the ratio of the two specific heats, then
υ 1 υ 2 = RT γ neon M 1 × M 2 RT γ water vapour = M 2 M 1 × γ neon γ water vapour
Here, M1=2.02×10-2 kg mole-1 ;
M2=1.8×10-2 kg mole-1
∴ υ 1 υ 2 = 1 · 8 × 1 0 - 2 × 5 / 3 2 · 0 2 × 1 0 - 2 × 4 / 3 = 1 . 1

Topic: Physics, Waves and Sound, NEET
Calculate the real angle of dip, if a magnet is suspended at an angle of 30° to the magnetic meridian, the dip needle makes an angle of 45° with horizontal.

Options:
A tan-1⁡32
B tan-1⁡3
C tan-1⁡32
D tan-1⁡23

The correct answer is:
tan-1⁡32

Solution:
Let θ=dip angle
BH= Horizontal component of magnetic field of earth
BV= Vertical component of magnetic field of earth
Also, tan⁡θ=BvBH
But tan⁡45o=BVBHcos⁡30o
∴ tan⁡θtan⁡45o=cos⁡30o
θ=tan-1⁡32

Topic: Physics, Magnetic Properties of Matter, NEET
Calculate the resonance energy of N2O from the following data; Δ f H of N2O is 82 KJ mol-1N≡N, 946 KJ mol-1 , N=N , 418 KJ mol-1 , 0 = 0, 498 KJ mol-1 , N=0,607 KJ mol-1 .

Options:
A +8.85 kJ  mol −1
B â88âkJâ mol â1
C −8.8 kJ  mol −1
D +88 kJ  mol −1

The correct answer is:
−88 kJ  mol −1

Solution:
Reaction N 2 â+â 1 2 O 2 â N 2 O;âÎHâ=â82âkJâ mol â1
The structure of N2O is N=N=O
Î f H (calculated) = Bond energy of reactant â Bond energy of product
=(Nâ¡N)+12(O=O)-[(N=N)+(N=O)]
=â 946â+â 1 2 498 â 418â+â607 â=â170âkJ
Resonance energy = experimental Δ f H N 2 O − calculated Δ f H N 2 O
=â82â170â=ââ88âkJâ mol â1 .

Topic: Chemistry, Thermodynamics (C), NEET
Calculate the solubility of A2X3in pure water, assuming that neither kind of ion reacts with water. The solubility product of A2X3, Ksp = 1.1 × 10-23.

Options:
A 1 × 10-5M
B 1 × 10-6M
C 1 × 10-4M
D 1 × 10-7M

The correct answer is:
1 × 10-5M

Solution:
Correct Option is : (A)
1 × 10-5M

Topic: Chemistry, Ionic Equilibrium, NEET
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, NEET
Calculate the ‘spin only’ magnetic moment of M2+(aq) ion (Z = 27)

Options:
A 5.92BM
B 4.87BM
C 3.87BM
D 2.54BM

The correct answer is:
3.87BM

Solution:
Correct Option is : (C)
3.87BM

Topic: Chemistry, d and f Block Elements, NEET
Calculate the strength of 5 volume H2O2 solution.

Options:
A 1.5%
B 2.5%
C 5%
D 10%

The correct answer is:
1.5%

Solution:
Correct Option is : (A)
1.5%

Topic: Chemistry, Hydrogen, NEET
Calculate the temperature at which a perfect black body radiates at the rate of 1 W cm-2, Value of Stefan's constant, Ï=5.67Ã10-5 erg s-1 cm-1 K-4.

Options:
A 576 K
B 648 K
C 695 K
D 766 K

The correct answer is:
648 K

Solution:
Here, σ=5.67×10-5 erg s-1 cm-1 K-4 ;
E=1 W cm-2=107 erg s-1 cm-2
Let T be the temperature of the perfect black body.
âNow,E=σT4orT=(Eσ)1/4
∴T=(1075.67×10-5)1/4=648 K

Topic: Physics, Thermal Properties of Matter, NEET
Calculate the temperature at whivh the RMS velocity of SO2 becomes equal to that of O2 at 303 K

Options:
A 273 K
B 606 K
C 303 K
D 403 K

The correct answer is:
606 K

Solution:
U SO 2 U O 2 = M O 2 T SO 2 M SO 2 T O 2 = 32× 64× T SO 2 303 =1
1= 32× 64 T SO 2 303 ⇒ T SO 2 =606 K

Topic: Chemistry, States of Matter, NEET
Calculate the temperature of 4.0 mol of a gas occupying 5 dm3at 3.32 bar. (R = 0.083 bar dm3K–1mol–1).

Options:
A 50 K
B 65 K
C 35 K
D 48 K

The correct answer is:
50 K

Solution:
Correct Option is : (A)
50 K

Topic: Chemistry, States of Matter, NEET
Calculate the the change in its potential energy of an electron in an atom which jumps in such a way that its kinetic energy changes from x to x9.

Options:
A x9 eV
B 16x9eV
C 9x eV
D 916 eV

The correct answer is:
16x9eV

Solution:
Change in kinetic energy
x-x9=8x9
From Bohr's model of the atom:
The kinetic energy in a state is equal to half of the potential energy in magnitude.
â´Potential energy P.E Change
=2 Ã Kinetic energy K.E Change
= 2Ã 8x 9 â= 16x 9 eV

Topic: Chemistry, Structure of Atom, NEET
Calculate the total number of electrons present in one mole of methane

Options:
A 6.022 × 1022electrons
B 6.022 × 1024electrons
C 6.022 × 1025electrons
D 6.022 × 1023electrons

The correct answer is:
6.022 × 1024electrons

Solution:
Correct Option is : (B)
6.022 × 1024electrons

Topic: Chemistry, Structure of Atom, NEET
Calculate the total number of possible isomers (including stereoisomers) for the complex compound CuII(NH3)4PtIICl4 (including given compound) are

Options:
A 3
B 4
C 5
D 6

The correct answer is:
6

Solution:
1.[Cu(NH3)4].[PtCl4]
2.[Cu(NH3)3Cl].[PtCl3(NH3)]
3.[Cu(NH3)2Cl2].[PtCl2(NH3)2]cis
4.[Cu(NH3)2Cl2].[PtCl2(NH3)2]Trans
5.[Cu(NH3)Cl3].[Pt(Cl)(NH3)3]
6.[Pt(NH3)3Cl].[CuCl3NH3]

Topic: Chemistry, Coordination Compounds, NEET
Calculate the total number of unpaired electrons in the following gaseous ions: Mn3+, Cr3+, V3+and Ti3+combined

Options:
A 15
B 11
C 10
D 12

The correct answer is:
10

Solution:
Correct Option is : (C)
10

Topic: Chemistry, d and f Block Elements, NEET
Calculate the total pressure in a 10.0 L cylinder which contains 0.4 g helium, 1.6 g oxygen and 1.4 g nitrogen at 27â.

{R = 0.082 L amp K-1 mol-1}

Options:
A 0.492 atm
B 49.2 atm
C 4.92 atm
D 0.0492 atm

The correct answer is:
0.492 atm

Solution:
Number of moles of He=0.44=0.1
Number of moles of oxygen=1.632=0.05
Number of moles of nitrogen=1.428=0.05
Total moles in the 10.0 L cylinder at 27â
=0.1+0.05+0.05
=0.2 mol
pT=nRTV=0.2Ã0.082Ã30010=0.492 atm

Topic: Chemistry, States of Matter, NEET
Calculate the unpaired electrons for a metal ion of first transition series having the magnetic moment of 2.83 BM.

Options:
A 6
B 4
C 3
D 2

The correct answer is:
2

Solution:
μ=n(n+2) ( magnetic moment)
( where n = no. of unpaired electron).
2.83=n(n+2)
n(n+2)=8
n2+2n-8=0
n=2.

Topic: Chemistry, Coordination Compounds, NEET
Calculate the value of equilibrium constant when 2 moles of PCl5 is heated in a closed vessel of 2 litre capacity. At equilibrium, 40% of PCl5 is dissociated into PCl3 and Cl2 .

Options:
A 0.266
B 0.53
C 2.66
D 5.3

The correct answer is:
0.266

Solution:
Volume of container = 2 litre .
Kc=2Ã40100Ã2Ã2Ã40100Ã22Ã60100Ã2=0.266

Topic: Chemistry, Chemical Equilibrium, NEET
Calculate the value of magnetic moment for 26Fe3+.

Options:
A 1.73 BM
B 3.87 BM
C 4.90 BM
D 5.92 BM

The correct answer is:
5.92 BM

Solution:
Fe : 3 d 6 4 s 2
Fe3+: 3d54so
âââââ

n=5

The formula of magnetic moment Âµ :

n n+2 BM

Whereas:- n = Number of unpaired electron
BM= Bohr Magneton
Âµ= 35 BM
=5.92 BM

Topic: Chemistry, d and f Block Elements, NEET
Calculate the volume of O2 liberated at STP by passing 5A current for 193 sec through acidified water:

Options:
A 56 mL
B 112 mL
C 158 mL
D 965 mL

The correct answer is:
56 mL

Solution:
As per the first law of electrolysis, The mass deposited or volume liberated at the electrode is directly proportional to the charge passing between the electrode.

m=ZQ

Where Z=EF (electrovalent equivalent)

Q=i×t.

mass of O2 = E×i×tF (Where F=96500 C)

=8×5×19396500 g

=0.08 g
∵ At STP, volume of 32 g O2 = 22400 mL
∴ The volume of 0.08 g O2 =22400×0.0832

=56 ml

Topic: Chemistry, Electrochemistry, NEET
Calculate the volume of water required to dissolve 0.1 g lead (II) chloride to get a saturated solution. (Kspof PbCl2= 3.2 × 10–8, atomic mass of Pb = 207 u)

Options:
A 0.5l
B 0.4l
C 0.3l
D 0.2l

The correct answer is:
0.2l

Solution:
Correct Option is : (D)
0.2l

Topic: Chemistry, Ionic Equilibrium, NEET
Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. Also name the series to which this transition belongs and the region of the spectrum.

Options:
A 538 nm and this transition belongs to Balmer series and comes in the visible region of the spectrum.
B 628 nm and this transition belongs to Balmer series and comes in the visible region of the spectrum.
C 434 nm and this transition belongs to Paschen series and comes in the visible region of the spectrum.
D 434 nm and this transition belongs to Balmer series and comes in the visible region of the spectrum.

The correct answer is:
434 nm and this transition belongs to Balmer series and comes in the visible region of the spectrum.

Solution:
Correct Option is : (D)
434 nm and this transition belongs to Balmer series and comes in the visible region of the spectrum.

Topic: Chemistry, Structure of Atom, NEET
Calculate the wavelength in nanometres of a quantum of light with frequency of 8×1015s-1 . The given value of Planck’s constant is 6.63×10-34Js. The velocity of light is 3.0×108ms-1 .

Options:
A 3.2×107
B 2.5×10-25
C 5×10-18
D 40

The correct answer is:
40

Solution:
λ=cυ=3×1088×1015=3.75×10-8
=3.75×10-8×109nm=4×101nm .

Topic: Chemistry, Structure of Atom, NEET
Calculate the wavelength of an electron moving with a velocity of 2.05 × 107m s–1.

Options:
A 2.31 × 10–11m
B 4.62 × 10–11m
C 3.55 × 10–11m
D 5.78 × 10–11m

The correct answer is:
3.55 × 10–11m

Solution:
Correct Option is : (C)
3.55 × 10–11m

Topic: Chemistry, Structure of Atom, NEET
Calculate the wavelength of light required to break the bond between two chlorine atoms in a chlorine molecule. The Cl – Cl bond energy is 243 kJ mol-1(h=6.6×10-34Js;c=3×108m/s ; Avogadro’s number =6.02×1023mol-1 )

Options:
A 8.18Ã10-31 m
B 6.26Ã10-21 m
C 4.905Ã10-7 m
D 4.1Ã10-6m

The correct answer is:
4.905Ã10-7 m

Solution:
Energy required to break one Cl â Cl bond =bond energy per moleAvogadroâs number=243Ã1036.02Ã1023 J
Let the wavelength of the photon to cause rupture of one Cl â Cl bond be Î»,
Î»=hcE=6.6Ã10-34Ã3Ã108Ã6.02Ã1023243Ã103=119.196243Ã10-34Ã1031Ã10-3=4.905Ã10-7 m.

Topic: Chemistry, Structure of Atom, NEET
Calculate the wavelength of the radiation emitted when in a hydrogen atom electron falls from infinity to stationary state 1 (Rydberg constant =1.097×107m-1 )

Options:
A 91 nm
B 192 nm
C 406 nm
D 9.1×10-8nm

The correct answer is:
91 nm

Solution:
1Î»=R1n12-1n221Î»=1.097Ã107m-1112-1â2
â´Î»=91Ã10-9m
We know 10-9=1nm So Î»=91 nm

Topic: Chemistry, Structure of Atom, NEET
Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen.

Options:
A 1.523 × 106m–1
B 2.421 × 106m–1
C 3.745 × 106m–1
D 0.251 × 106m–1

The correct answer is:
1.523 × 106m–1

Solution:
Correct Option is : (A)
1.523 × 106m–1

Topic: Chemistry, Structure of Atom, NEET
Calculate the weight of Al deposited when a current of 5 Faradays is passed through a solution of Al+++ ions, (The atomic weight of Al is 27)

Options:
A 27 gm
B 36 gm
C 45 gm
D 39 gm

The correct answer is:
45 gm

Solution:
Al3++3e-âAl
EAl=273=9
WAl=EAlÃNo.offaradays
=9Ã5=45 gm.

Topic: Chemistry, Electrochemistry, NEET
Calculate the work done during combustion of 0.138 kg of ethanol, C2H5OH(l) at 300 K. (Given: R=8.314JK-1mol-1 , molar mass of ethanol =46gmol-1 .)

Options:
A – 7482 J
B 7482 J
C – 2494 J
D 2494 J

The correct answer is:
7482 J

Solution:
C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l) …(i)
0.138 kg of ethanol ≡138g46gmol-1=3 moles
By multiplying equation (i) by 3, we get
3C2H5OH(l)+9O2(g)→6CO2(g)+9H2O(l)
Δng=6-9=3
w=-PΔV
w=-ΔngRT=-(-3)×8.314×300 (∵PV=nRT)
=7482.6 J

Topic: Chemistry, Thermodynamics (C), NEET
Calculate the work done during isothermal expansion of one mole of an ideal gas from 10 atm to 1 atm at 300 K

Options:
A -4138.8J
B -4938.8J
C -5744.1J
D -6257.2J

The correct answer is:
-5744.1J

Solution:
Given that
P1=10atm ,
P2=1atm ,
T=300K ,
n=1
R=8.314J/K/mol
Now, by using
W=2.303nRTlog10P2P1
=2.303×1×8.314×300log10110
W= -5744.1Joule

Topic: Chemistry, Thermodynamics (C), NEET
Californium (atomic number = 98) is a member of

Options:
A s-block
B p-block
C d-block
D f-block

The correct answer is:
f-block

Solution:
Its configuration is [Rn] 5f107s2 . Since the last electron enters the 5f-orbital, therefore, it is a member of f-block.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
Callus can form plantlets by altering the concentration of

Options:
A Phytohormones
B Amino sugars
C Vitamins
D Sugars

The correct answer is:
Phytohormones

Solution:
Plant callus is a growing mass of unorganized plant parenchyma cells. In biological research and biotechnology, callus formation is induced from plant tissue samples (explants) after surface sterilization and plating onto tissue culture medium in vitro (in a closed culture vessel such as a petri dish). The culture medium is supplemented with plant growth hormones, such as auxin, cytokinin, and gibberellin, to initiate callus formation or somatic embryogenesis or morphogenesis.

Topic: Biology, Strategies for Enhancement in Food Production, NEET
Calotropis contains which of the following types of phyllotaxy?

Options:
A Alternative
B Opposite
C Whorled
D Superimposed

The correct answer is:
Opposite

Solution:
In opposite type, a pair of leaves arise at each node and lie opposite to each other as in Calotropis and guava plants.

Topic: Biology, Morphology of Flowering Plants, NEET
Calvin cycle or dark reaction of photosynthesis is a light-independent reaction in which products of the light reaction are consumed. In which step of the Calvin cycle these products are not consumed?

Options:
A Carboxylation
B Reduction
C Regeneration
D Both B and C

The correct answer is:
Carboxylation

Solution:
Calvin cycle or dark reaction of photosynthesis in which fixation of carbondioxide occurs to form glucose. It occurs in three steps carboxylation, reduction and regeneration. In carboxylation product of light reaction ATP and NADPH2 are not consumed, it is enzyme dependent process in which CO2 is trapped into organic form.

Topic: Biology, Photosynthesis in Higher Plants, NEET
CAM pathway operates in

Options:
A Drought escaping mesophytes
B Drought resisting xerophytes
C Drought enduring hydrophytes
D None of these

The correct answer is:
Drought resisting xerophytes

Solution:
Drought tolerant plants/Drought resisting xerophytes make use of either C4 carbon fixation or crassulacean acid metabolism (CAM) to fix carbon during photosynthesis. CAM is particularly good for arid conditions because carbon dioxide can be taken up at night, allowing the stomata to stay closed during the heat of day and thus reducing water loss. The CO2 is stored as the four-carbon acid malate, and then used during photosynthesis during the day. The pre-collected CO2 is concentrated around the enzyme RuBisCO, increasing photosynthetic efficiency.

Topic: Biology, Photosynthesis in Higher Plants, NEET