datavorous/entrance-exam-dataset huggingface dataset:
Calculate the energy of the reaction
Li7+pâ¶2 2He4
If the binding energy per nucleon in Li7 and He4 nuclei are 5.60 MeV and 7.06 MeV, respectively.

Options:
A 19.6 MeV

B 2.4 MeV

C 8.4 MeV

D 17.28 MeV

The correct answer is:
17.28 MeV

Solution:
The binding energy of Li=39.20 MeV
The binding energy of He4=28.24 MeV
The binding energy of 2He4=56.48 MeV
â´ Energy =56.48-39.20=17.28 MeV

Topic: Physics, Atomic Physics, NEET
Calculate the energy required for the process: He+(g) ➔He2+(g) + e–The ionization energy for the H atom in the ground state is 2.18 × 10–18J atom–1

Options:
A 8.72 × 10–18J per atom
B 5.46 × 10–18J per atom
C 6.58 × 10–18J per atom
D 9.63 × 10–18J per atom

The correct answer is:
8.72 × 10–18J per atom

Solution:
Correct Option is : (A)
8.72 × 10–18J per atom

Topic: Chemistry, Structure of Atom, NEET
Calculate the enthalpy change for the process: CCl4(g) → C(g) + 4Cl(g) and calculate bond enthalpy of C–Cl in CCl4(g). ∆vapH⊖(CCl4) = 30.5 kJ mol–1; ∆fH⊖(CCl4) = –135.5 kJ mol–1. ∆aH⊖(C) = 715.0 kJ mol–1, where ∆aH is enthalpy of atomisation ∆aH⊖(Cl2) = 242 kJ mol–1

Options:
A 412 kJ mol–1
B 326 kJ mol–1
C 369 kJ mol–1
D 442 kJ mol–1

The correct answer is:
326 kJ mol–1

Solution:
Correct Option is : (B)
326 kJ mol–1

Topic: Chemistry, Thermodynamics (C), NEET
Calculate the enthalpy change of the following reaction
H2C=CH2(g)+H2(g)→H3C-CH3(g)
The bond energy of C-H,C-C,C=C,H-H are 414, 347, 615 and 453 KJ mol-1 respectively.

Options:
A +125 kJ
B -12.5 kJ
C -125 kJ
D +12.5 kJ

The correct answer is:
-125 kJ

Solution:
ΔH=Σ bond energy of reactant -Σ bond energy of product
to write bond energy use proper symbols
E C = C  or  Δ C = C H =[1(C=C)+4(C-H)+1(H-H)]-[1(C-C)+6(C-H)]
=-1(C-C)-2(C-H)+1(C=C)+1(H-H)
=-347-2(414)+1(615)+1(435)
=-125 kJ.

Topic: Chemistry, Thermodynamics (C), NEET
Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at –10.0°C. (∆fusH = 6.03 kJ mol–1at 0°C , Cp[H2O(l)] = 75.3 J mol–1K–1, Cp[H2O(s)] = 36.8 J mol–1K–1)

Options:
A –8.543 kJ mol-1
B –7.151 kJ mol–1
C –6.582 kJ mol-1
D –5.236 kJ mol-1

The correct answer is:
–7.151 kJ mol–1

Solution:
Correct Option is : (B)
–7.151 kJ mol–1

Topic: Chemistry, Thermodynamics (C), NEET
calculate the equilibrium constant for

NH3 + H2O âNH4++OH-

if, for the equilibrium,

NH4++H2Oâ NH4OH + H+

the equilibrium constant is 5.5 Ã 10â10

Options:
A 1.8Â ÃÂ 10â4
B 1.8 Ã 10â5
C 1.8Â ÃÂ 10â6
D 1.8Â ÃÂ 10â7

The correct answer is:
1.8 Ã 10â5

Solution:
NH4++H2Oâ NH4OH + H+
K eq = NH 4 OH OH â H + H 2 O NH 4 + OH â
Keq=KwKb
â5.5Ã10-10=10-14Kb
Kb=10-145.5Ã10-10=1.8Ã10-5

Topic: Chemistry, Ionic Equilibrium, NEET
Calculate the equivalent conductance of the electrolytic cell.
Given: l/a = 0.5  cm −1 , R = 50 ohm, N = 1.0

Options:
A 10â ohm â1 â cm 2 âgâ eq â1
B 20  ohm −1   cm 2  g  eq −1
C 300  ohm −1   cm 2  g  eq −1
D 100  ohm −1   cm 2  g  eq −1

The correct answer is:
10  ohm −1   cm 2  g  eq −1

Solution:
l/aâ=â0.5â cm â1 ,âRâ=â50âohm
Ï= Ra l = 50 0.5 =100
Î=kÃ 1000 N = 1 Ï Ã 1000 N = 1 100 Ã 1000 1
10  ohm −1   cm 2  g  eq −1

Topic: Chemistry, Electrochemistry, NEET
Calculate the equivalent weight of metal, if the oxide of a metal contains 32% oxygen.

Options:
A 34
B 32
C 17
D 8

The correct answer is:
17

Solution:
Let weight of metal oxide =100âg
Weight of oxygen =32âg
â´ Weight of metal =100−32=68 g
Equivalent weight of metal in oxide = wt. of metal wt. of oxygen ×8= 68 32 ×8=17.

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
Calculate the final temperature of the gas, if one mole of an ideal gas is allowed to expand reversibly and adibatically from a temperature of 27°C and the work done during the process is 3 kJ .
Given C V  = 20 J/K

Options:
A 100 K
B 150 K
C 195 K
D 255 K

The correct answer is:
150 K

Solution:
Given number of moles =1
Initial temperature = 27 o C=300âK
Work done by the system =3âkJ=3000âK
It will be (-) because work is done by the system.
Heat capacity at constant volume Cv â=â20âJ/k
We know that work done
W=-nCV(T2-T1) ;
3000=-1Ã20(T2-300)
3000=-20T2+6000
20 T 2 =3000 T 2 = 3000 20 =150 K

Topic: Chemistry, Thermodynamics (C), NEET
Calculate the gas constant for 1 g of gas from the following data: Cp = 0.245 cal g-1 Â°C-1; Cv = 0.165 cal g-1 Â°C-1 and J = 4.2 Ã 107 erg cal-1.

Options:
A

3.36×106 erg g-1 °C-1

B

2.13×106 erg g-1 °C-1

C

4.26×106 erg g-1 °C-1

D

2.57×106 erg g-1 °C-1

The correct answer is:

3.36×106 erg g-1 °C-1

Solution:
Since we know

Cp-Cv=rJorr=JCp-Cv

∴ r=4.2×107(0.245-0.165)

or r=3.36×106 erg g-1 °C-1

Topic: Physics, Thermodynamics, NEET
Calculate the half-life of a first order reaction from their rate constants given below: 2 min–1

Options:
A 0.89 minutes
B 0.35 minutes
C 1.22 minutes
D 1.43 minutes

The correct answer is:
0.35 minutes

Solution:
Correct Option is : (B)
0.35 minutes

Topic: Chemistry, Chemical Kinetics, NEET
Calculate the half-life of a first order reaction from their rate constants given below: 4 years–1

Options:
A 0.173 years
B 0.645 years
C 1.248 years
D 1.405 years

The correct answer is:
0.173 years

Solution:
Correct Option is : (A)
0.173 years

Topic: Chemistry, Chemical Kinetics, NEET
Calculate the half-life of a first order reaction from their rate constants given below: 200 s–1

Options:
A 8.23 x 10–3seconds
B 5.68 x 10–3seconds
C 3.47 x 10–3seconds
D 9.96 x 10–3seconds

The correct answer is:
3.47 x 10–3seconds

Solution:
Correct Option is : (C)
3.47 x 10–3seconds

Topic: Chemistry, Chemical Kinetics, NEET
Calculate the heat absorbed by a system in going through the cyclic process shown in diagram.

Options:
A 31.7 J

B 31.4 J

C 41.7 J

D 44.3 J

The correct answer is:
31.4 J

Solution:
The cycle i clockwise; thus net work done by the gas. As in a cyclic process no change in internal energy takes place, heat supplied is equal to the work done by the gas in one complete cycle. So in this case heat supplied to the gas is given as
Q = work done by the gas, W
= area of ellipse
=Ï ab
where a and b are semi-major and semi-minor axes of the ellipse, respectively, which are given from the diagram as
a = 1.0 x 105 N/m and b = 100 x 10-6 m3
Thus area of ellipse is
Q = W = Ï Ã 1.0 Ã 1 0 5 Ã 1 0 0 Ã 1 0 - 6
= 3.14 Ã 1 0 = 31.4 J

Topic: Physics, Thermodynamics, NEET
Calculate the heat of formation of CO using given equations
C+O2→CO2; ΔH=X
CO+12O2→CO2; ΔH=Y

Options:
A X-Y
B X+Y
C Y-2X
D 2X-Y

The correct answer is:
X-Y

Solution:
C+12O2âCO , on substracting equation (ii) from equation (i) ,we get X-Y .

Topic: Chemistry, Thermodynamics (C), NEET
Calculate the heat of reaction for the following reaction SO2+12O2→SO3 . Given the values of heat of formation of SO2 and SO3 are -298.2 kJ and -98.2kJ .

Options:
A -200 kJ
B -356.2 kJ
C +200 kJ
D -396.2 kJ

The correct answer is:
+200 kJ

Solution:
S+O2→SO2ΔH=-298.2kJ ...........(i)
SO2+12O2→SO3ΔH=-98.2kJ .........(ii)
eq. (i)-(ii) and find the required result.

Topic: Chemistry, Thermodynamics (C), NEET
Calculate the heat required to make 6.4 kg of CaC2 from CaO(s) and C(s) from the reaction CaO(s)+3C(s)→CaC2(s)CO(g) give that, ΔHfo(CaO)=-151.6 kcal, ΔHfo(CaC2)=-14.2 kcal, ΔHfo(CO)=-26.4 kcal

Options:
A 5624 kcal
B 86.24×103 kcal
C 1.11×104 kcal
D 1100 kcal

The correct answer is:
1.11×104 kcal

Solution:
CaO(s)+3C(s)→CaC2(s)+CO(g)
ΔH=ΣH(Product)-ΣH(reactant)
=[-14.2-(-26.4)]-[-151.6)+3×0]
=111 kcal
Molecular weight of CaC2=64
64gCaC2=111 kcal
So 6.4 kg CaC2=111×6.4×10364=1.11×104 kcal.

Topic: Chemistry, Thermodynamics (C), NEET
Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below (a) Human muscle-fluid, pH = 6.83 (b) Human stomach fluid, pH = 1.2 (c) Human blood, pH = 7.38 (d) Human saliva, pH = 6.4.

Options:
A (a) [H+] = 1.60 × 10-7, (b) [H+] = 1.00 × 10-1, (c) [H+] = 4.00 × 10-8, (d) [H+] = 4.00 × 10-7
B (a) [H+] = 2.50 × 10-7, (b) [H+] = 1.00 × 10-2, (c) [H+] = 5.00 × 10-8, (d) [H+] = 2.50 × 10-7
C (a) [H+] = 2.50 × 10-8, (b) [H+] = 1.00 × 10-3, (c) [H+] = 5.00 × 10-7, (d) [H+] = 2.50 × 10-6
D (a) [H+] = 1.60 × 10-8, (b) [H+] = 1.00 × 10-2, (c) [H+] = 4.00 × 10-7, (d) [H+] = 4.00 × 10-6

The correct answer is:
(a) [H+] = 1.60 × 10-7, (b) [H+] = 1.00 × 10-1, (c) [H+] = 4.00 × 10-8, (d) [H+] = 4.00 × 10-7

Solution:
Correct Option is : (A)
(a) [H+] = 1.60 × 10-7, (b) [H+] = 1.00 × 10-1, (c) [H+] = 4.00 × 10-8, (d) [H+] = 4.00 × 10-7

Topic: Chemistry, Ionic Equilibrium, NEET
Calculate the interionic distance for cesium chloride crystal structure

Options:
A 3a2
B a
C a2
D 2a3

The correct answer is:
3a2

Solution:
As CsCl is having body-centred arrangenent of particles, Its body diagonal is 3a ,
Hence the interionic distance is d=3a/2 .

Topic: Chemistry, Solid State, NEET
Calculate the ionisation energy of sodium in kJ mol â1 if electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom.

Options:
A 4.946Ã102
B 4.946×103
C 4.946×104
D 4.946×105

The correct answer is:
4.946×102

Solution:
Ionisation energy is the amount of energy required to remove most loosely bound electron from the isolated gaseous atom.

For the sodium atom, it is equal to the electromagnetic radiation of wavelength 242 nm .

As per the Planck's quantum theory.

The energy required for removing the one-mole electron from the one-mole sodium atoms.

E= hc Î» Ã N A J mol â1

E= 6.626Ã 10 â34 Ã3Ã 10 8 242Ã 10 â9 Ã6.022Ã 10 23 J mol â1

E=0.4946Ã 10 6 J mol â1 âE= 4.946Ã 10 2 kJ mol â1

Topic: Chemistry, Structure of Atom, NEET
Calculate the ionisation enthalpy of atomic hydrogen in terms of J mol⁻¹ given that the energy of an electron in the ground state of the hydrogen atom is (-2.18×10⁻¹⁸) J.

Options:
A 2.18×106J mol⁻¹
B -4.36×106J mol⁻¹
C 1.31×106J mol⁻¹
D 3.46×106J mol⁻¹

The correct answer is:
1.31×106J mol⁻¹

Solution:
Correct Option is : (C)
1.31×106J mol⁻¹

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
Calculate the (i)wavelength, (ii)frequency and (iii)wavenumber of a light wave whose period is 2.0 × 10–10s.

Options:
A (i) 5.0 × 10–2m, (ii) 6.0 × 109s–1and (iii) 16.66 m–1
B (i) 6.0 × 10–2m, (ii) 5.0 × 109s–1and (iii) 16.66 m–1
C (i) 4.0 × 10–2m, (ii) 5.0 × 109s–1and (iii) 25.66 m–1
D (i) 6.0 × 10–2m, (ii) 8.0 × 109s–1and (iii) 12.66 m–1

The correct answer is:
(i) 6.0 × 10–2m, (ii) 5.0 × 109s–1and (iii) 16.66 m–1

Solution:
Correct Option is : (B)
(i) 6.0 × 10–2m, (ii) 5.0 × 109s–1and (iii) 16.66 m–1

Topic: Chemistry, Structure of Atom, NEET
Calculate the mass of 2 g atom of Mg.

Options:
A 48 g
B 84 g
C 78 g
D 87 g .

The correct answer is:
48 g

Solution:
âµ1 g atom of Mg has mass=24 g
âµ2 g atom of Mg has mass=2Ã24 g=48 g

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
Calculate the mass of a non-volatile solute (molar mass 40 g mol–1) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.

Options:
A 36 g
B 57 g
C 10 g
D 44 g

The correct answer is:
10 g

Solution:
Correct Option is : (C)
10 g

Topic: Chemistry, Solutions, NEET
Calculate the mass of a photon with wavelength 3.6 Å?

Options:
A 7.546 × 10–29kg
B 3.189 × 10–29kg
C 4.356 × 10–29kg
D 6.135 × 10–29kg

The correct answer is:
6.135 × 10–29kg

Solution:
Correct Option is : (D)
6.135 × 10–29kg

Topic: Chemistry, Structure of Atom, NEET
Calculate the mass of anhydrous Na2CO3 required to prepare 250 ml of 0.25 M solution.

Options:
A 6.225 g
B 66.25 g
C 6.0 g
D 6.625 g

The correct answer is:
6.625 g

Solution:
Molarity of solution is given by M=wm×V(l) ;
Hence, 0.25=w106×0.25 ; w=6.625 g

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
Calculate the mass of ascorbic acid (Vitamin C, C6H8O6) to be dissolved in 75 g of acetic acid to lower its melting point by 1.5°C. (Kf= 3.9 K kg mol-1)

Options:
A 5.077 g
B 8.452 g
C 6.968 g
D 2.142 g

The correct answer is:
5.077 g

Solution:
Correct Option is : (A)
5.077 g

Topic: Chemistry, Solutions, NEET
Calculate the mass of BaCO3 produced when excess CO2 is bubbled through a solution containing 0.205 moles of BaOH2 .

Options:
A 40.5g
B 48.5g
C 4.5g
D 60.5g

The correct answer is:
40.5g

Solution:
BaOH2+CO2→ BaCO3+H2O
Mole of BaOH2:Mole of CO2 :: Mole of BaCO3:Mole of H2O
Mole of BaCO3 formed=0.205
Weight of BaCO3 formed=0.205×197.4=40.5g

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
Calculate the mass of Mg  atomic mass = 24 g/mol that combines with 1.5 mol of O2 to form magnesium oxide MgO .

Options:
A 24 g
B 36 g
C 48 g
D 72 g

The correct answer is:
72 g

Solution:
Mg 1 mole  +  1 2 O 2 0.5 mole →MgO
0.5 mole of oxygen react with 1 mole of Mg
1.5 mole of oxygen react with 1.50.5=3 mole
24Ã3=72 gm .

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
Calculate the mass of sodium carbonate required to prepare 500 ml of a semi­- normal solution

Options:
A 13.25 g
B 26.5 g
C 53 g
D 6.125 g

The correct answer is:
13.25 g

Solution:
N =  w × 1000 eq.wt. × volume in ml.
eq.wt.=1062=53
w =  0.5 × 53 × 500 1000 =13.25 g .

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
Calculate the mass of urea (NH2CONH2) required in making 2.5 kg of 0.25 molal aqueous solution.

Options:
A 78.652 g
B 58.465 g
C 21.652 g
D 36.946 g

The correct answer is:
36.946 g

Solution:
Correct Option is : (D)
36.946 g

Topic: Chemistry, Solutions, NEET
Calculate the mass per cent of different elements present in sodium sulphate (Na2SO4)

Options:
A Mass% of Na is 32.4%, Mass% of S is 22.5%, Mass% of O is 45%
B Mass% of Na is 32.4%, Mass% of S is 45%, Mass% of O is 22.5%
C Mass% of Na is 45%, Mass% of S is 22.5%, Mass% of O is 32.5%
D Mass% of Na is 22.5%, Mass% of S is 32.5%, Mass% of O is 45%

The correct answer is:
Mass% of Na is 32.4%, Mass% of S is 22.5%, Mass% of O is 45%

Solution:
Correct Option is : (A)
Mass% of Na is 32.4%, Mass% of S is 22.5%, Mass% of O is 45%

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
Calculate the mass percentage of aspirin (C9H8O4) in acetonitrile (CH3CN) when 6.5 g of C9H8O4is dissolved in 450 g of CH3CN.

Options:
A 7.582 %
B 3.689 %
C 1.424 %
D 9.639 %

The correct answer is:
1.424 %

Solution:
Correct Option is : (C)
1.424 %

Topic: Chemistry, Solutions, NEET
Calculate the mass percentage of benzene (C6H6) and carbon tetrachloride (CCl4) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.

Options:
A C6H6= 36.47% , CCl4= 84.72%
B C6H6= 15.28% , CCl4= 36.48%
C C6H6= 15.28% , CCl4= 84.72%
D C6H6= 36.47% , CCl4= 36.48%

The correct answer is:
C6H6= 15.28% , CCl4= 84.72%

Solution:
Correct Option is : (C)
C6H6= 15.28% , CCl4= 84.72%

Topic: Chemistry, Solutions, NEET
Calculate the molar solubility of \(\mathrm{Ni}(\mathrm{OH}) 2\) in \(0.10 \mathrm{M} \mathrm{NaOH}\). The ionic product of \(\mathrm{Ni}(\mathrm{OH}) 2\) is \(2.0 \times 10^{-15}\).

Options:
A \(4 \times 10^{-9}\)
B \(1 \times 10^{-11}\)
C \(2 \times 10^{-13}\)
D \(8 \times 10^{-16}\)

The correct answer is:
\(2 \times 10^{-13}\)

Solution:
Correct Option is : (C)
\(2 \times 10^{-13}\)

Topic: Chemistry, Ionic Equilibrium, NEET
Calculate the molarity of a solution containing 5 g of NaOH in 450 mL solution.

Options:
A 0.278 mol L–1
B 0.683 mol L-1
C 0.523 mol L-1
D 0.352 mol L-1

The correct answer is:
0.278 mol L–1

Solution:
Correct Option is : (A)
0.278 mol L–1

Topic: Chemistry, Solutions, NEET
Calculate the molarity of a solution of ethanol in water, in which the mole fraction of ethanol is 0.040 (assume the density of water to be one).

Options:
A 1.84 M
B 2.85 M
C 2.31 M
D 3.45 M

The correct answer is:
2.31 M

Solution:
Correct Option is : (C)
2.31 M

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
Calculate the molarity of the following solution:30 g of Co(NO3)2.6H2O in 4.3 L of solution

Options:
A 0.052M
B 0.095M
C 0.024M
D 0.065M

The correct answer is:
0.024M

Solution:
Correct Option is : (C)
0.024M

Topic: Chemistry, Solutions, NEET
Calculate the molarity of the following solution:30 mL of 0.5 M H2SO4diluted to 500 mL.

Options:
A 0.08M
B 0.03M
C 0.06M
D 0.01M

The correct answer is:
0.03M

Solution:
Correct Option is : (B)
0.03M

Topic: Chemistry, Solutions, NEET
Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.

Options:
A 0.46
B 0.68
C 0.79
D 0.92

The correct answer is:
0.46

Solution:
Correct Option is : (A)
0.46

Topic: Chemistry, Solutions, NEET
Calculate the mole fraction of ethylene glycol (C2H6O2) in a solution containing 20% of C2H6O2by mass.

Options:
A 0.032
B 0.068
C 0.098
D 0.010

The correct answer is:
0.068

Solution:
Correct Option is : (B)
0.068

Topic: Chemistry, Solutions, NEET
Calculate the mole fraction of water in a mixture containing 18g water and 414g ethanol (assume ideal behaviour of the mixture).

Options:
A 0.1
B 0.4
C 0.7
D 0.9

The correct answer is:
0.1

Solution:
Molecular weight of C2H5OH=24+5+16+1=46
Molecular mass of H2O=18
414g of C2H5OH has 41446=9 mole
18g of H2O has =1818=1 mole
Mole fraction of water =n1n1+n2=11+9

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
Calculate the momentum transferred to a surface when a radiation of energy E falls normally on it. Assume that the reflectivity of the surface is unity.

Options:
A Ec
B 2Ec
C Ec
D Ec2

The correct answer is:
2Ec

Solution:
Pinitial=Ec

Pfinal=-Ec

The change in momentum is the impulse, which is the momentum transferred to the surface

∆P=Pfinal-Pinitial=Ec--Ec=2Ec

Topic: Physics, Dual Nature of Matter, NEET
Calculate the momentum transferred to a surface when a radiation of energy E falls normally on it. Assume that the reflectivity of the surface is unity.

Options:
A E / c
B 2 E / c
C Ec
D E / c 2

The correct answer is:
2 E / c

Solution:
Initial momentum = E / c
Final momentum = - E / c
∴ Change of momentum = E c - - E c = 2 E c
∴ Momentum transferred to surface = 2 E c .

Topic: Physics, Dual Nature of Matter, NEET
Calculate the net force acting on the charge present at the origin.

Options:
A q q 1 4π ε 0 a 2 × 2
B q q 1 4π ε 0 2 a 2 + q q 1 8π ε 0 a 2
C q q 1 4Ï Îµ 0 a 2 ( 2 + 1 2 )
D q q 1 4π ε 0 a 2 ( 1 2 − 2 )

The correct answer is:
q q 1 4π ε 0 a 2 ( 2 + 1 2 )

Solution:
F 1 =k q q 1 a 2 ;     F 2 =k q q 1 ( 2 a ) 2
Resultant force F= 2 F 1 + F 2
= 2 kq q 1 a 2 + kq q 1 2 a 2
= kq q 1 a 2 ( 2 + 1 2 )
= q q 1 4π ε 0 a 2 ( 2 + 1 2 )

Topic: Physics, Electrostatics, NEET
Calculate the number of atoms in each of the following: (i) 52 moles of Ar (ii) 52 u of He and (iii) 52 g of He.

Options:
A (i) 3.131×1022atoms (ii) 15 atoms and (iii) 7.83×1020atoms
B (i) 3.131×1025atoms (ii) 13 atoms and (iii) 7.83×1024atoms
C (i) 3.131×1025atoms (ii) 10 atoms and (iii) 7.83×1024atoms
D (i) 3.541×1025atoms (ii) 13 atoms and (iii) 8.546×1024atoms

The correct answer is:
(i) 3.131×1025atoms (ii) 13 atoms and (iii) 7.83×1024atoms

Solution:
Correct Option is : (B)
(i) 3.131×1025atoms (ii) 13 atoms and (iii) 7.83×1024atoms

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
Calculate the number of atoms in each of the following.
(i) 52 moles of Ar (ii) 52 u of He (iii) 52 g of He
{NA = 6.022Ã1023}

Options:
A 3.130 x 1023, 12, 6.8284 x 1020
B 3.138 x 1022, 12, 6.7854 x 1028
C 3.131 x 1025, 13, 7.8286 x 1024
D 3.135 x 1028, 15, 6.7288 x 1020

The correct answer is:
3.131 x 1025, 13, 7.8286 x 1024

Solution:
(i) 1 mol of Ar = 6.022 x 1023 atoms.
52 moles of Ar = 52 x 6.022 x 1023 atoms
= 313.144 x 1023 atoms
= 3.131 x 1025 atoms
(ii) 4 u of He = 1 He atom
∴ 52 u of He = 5 2 4 He atoms = 13 He atoms.
(iii) 1 mol atom of He = 4 g = 6.022 x 1023 atoms.
52 g of He = 5 2 × 6 . 0 2 2 × 1 0 2 3 4 atoms
= 78.286 x 1023 atoms
= 7.8286 x 1024 atoms.

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
Calculate the number of equivalents in 10 litre of 0.5 MBa(OH)2 Solution (Ba=137) .

Options:
A 0.1
B 10
C 100
D 1

The correct answer is:
10

Solution:
Normality of Ba(OH)2=m×n-factor=0.5×2=1N .
Equivalent of the solution =N×Vlitre=1×10=10

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
Calculate the number of geometrical isomers possible for the following compound CH3-CH=CH-CH=CH-C2H5 is :

Options:
A 4
B 3
C 2
D 5

The correct answer is:
4

Solution:
Polyene containing more than one double bond and terminal are different so
no. of geometrical isomer =2n⇒22=4

Topic: Chemistry, General Organic Chemistry, NEET
Calculate the number of oxygen atoms required to combine with 7g of N2 to form N2O3 when 80% of N2 is converted to N2O3 .

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

The correct answer is:
3.6×1023

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

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
Calculate the number of protons, neutrons and electrons in80Br35

Options:
A Number of protons = number of electrons= 35 and Number of neutrons = 55
B Number of protons = number of electrons= 40 and Number of neutrons = 60
C Number of protons = number of electrons= 35 and Number of neutrons = 45
D Number of protons = number of electrons= 48 and Number of neutrons = 45

The correct answer is:
Number of protons = number of electrons= 35 and Number of neutrons = 45

Solution:
Correct Option is : (C)
Number of protons = number of electrons= 35 and Number of neutrons = 45

Topic: Chemistry, Structure of Atom, NEET
Calculate the number of protons, neutrons and electrons in the element 89231Y

Options:
A 89, 89, 242
B 89, 71, 89
C 89, 142, 89
D 89, 231, 89

The correct answer is:
89, 142, 89

Solution:
In 89231Y number of protons and electrons is 89 and No. of neutrons =A-Z=231-89=142 .

Topic: Chemistry, Chemical Kinetics, NEET
Calculate the number of sigma bonds and pi bonds in CH3CH2COCH2CH3

Options:
A 16 sigma bonds and 1 pi bond
B 15 sigma bonds and 1 pi bond
C 16 sigma bonds and 2 pi bond
D 14 sigma bonds and 1 pi bond

The correct answer is:
15 sigma bonds and 1 pi bond

Solution:
Correct Option is : (B)
15 sigma bonds and 1 pi bond

Topic: Chemistry, General Organic Chemistry, NEET
Calculate the number of sigma bonds and pi bonds in CH3CH=CH(CH2)3CH3

Options:
A 21 sigma bonds and 1 pi bond
B 19 sigma bonds and 1 pi bond
C 20 sigma bonds and 2 pi bond
D 20 sigma bonds and 1 pi bond

The correct answer is:
20 sigma bonds and 1 pi bond

Solution:
Correct Option is : (D)
20 sigma bonds and 1 pi bond

Topic: Chemistry, General Organic Chemistry, NEET
Calculate the number of unpaired electrons in [Mn(H2O)6]2+ , Considering H2O as a weak field ligand (At.No.of Mn=25)

Options:
A Two
B Three
C Four
D Five

The correct answer is:
Five

Solution:
Mn25→3d54s2
Mn2+→3d5
In presence of weak ligand field, there will be no pairing of electrons. So, it will form a high spin complex. i.e. the number of unpaired electrons = 5.

Topic: Chemistry, Coordination Compounds, NEET
Calculate the number of weak hydrogen bonds in a DNA strand 204 â« long with 15% thymine bases.

Options:
A 154
B 162
C 138
D 120

The correct answer is:
162

Solution:
DNA length of one turn is 34 â«. It means one turn contains 10 base pairs. So 34 â«=10 base pairs.
Hence, 204 â«=204Ã10/34= 60 base pairs. As given 15% is thymine =15/100Ã 60=9 thymine.
Thymine pairs with adenine and hence adenine is also 9. So AT is total of 18. Total GC=60-18=42.
Total hydrogen bonds = 2x AT+3x GC (AT shows double bond and GC shows triple bond) = 2 x 18 + 3 x 42=36+126=162.

Topic: Biology, Molecular Basis of Inheritance, NEET
Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at 37°C.

Options:
A 64.23 Pa
B 48.24 Pa
C 16.25 Pa
D 30.96 Pa

The correct answer is:
30.96 Pa

Solution:
Correct Option is : (D)
30.96 Pa

Topic: Chemistry, Solutions, NEET
Calculate the osmotic pressure of 10% solution of cane sugar in atmosphere at 69oC is

Options:
A 724 atm
B 824 atm
C 8.21 atm
D 7.21 atm

The correct answer is:
8.21 atm

Solution:
π=CST
π=wmRTV=10342×0.0821×(273+69)0.1
∴ π=8.21 atm

Topic: Chemistry, Solutions, NEET