datavorous/entrance-exam-dataset huggingface dataset:
2-Methylpent-2-ene on reductive ozonlysis will give

Options:
A Propanal only
B Propanal and ethanal
C Propanone & propanal
D Propan-2-ol and ethanal

The correct answer is:
Propanone & propanal

Solution:

Topic: Chemistry, Hydrocarbons, NEET
2-propanol will be the product of which one of the following reactions? Multiple can be true.

I. CH2=CH-CH3+H2O â¶H+

II. CH3-CHO âiiH3O+iCH3Mgl

III. CH2O âiiH3O+iC2H5Mgl

IV.CH2=CH-CH3âNeutral KMnO4

Options:
A I and II
B II and III
C I and III
D II and IV

The correct answer is:
I and II

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
$3 \mathrm{~A} \rightarrow 2 \mathrm{~B}$, then the rate of reaction $\frac{+d[\mathrm{~B}]}{d t}$ is equal to:

Options:
A $-\frac{3}{2} \frac{d[\mathrm{~A}]}{d t}$
B $-\frac{2}{3} \frac{d[\mathrm{~A}]}{d t}$
C $-\frac{1}{2} \frac{d[\mathrm{~A}]}{d t}$
D $2 \frac{d[A]}{d t}$

The correct answer is:
$-\frac{2}{3} \frac{d[\mathrm{~A}]}{d t}$

Solution:
For the reaction,
$\begin{gathered}
3 \mathrm{~A} \rightarrow 2 \mathrm{~B} \\
\text { Rate }=-\frac{1}{3} \frac{d[\mathrm{~A}]}{d t}=+\frac{1}{2} \frac{d[\mathrm{~B}]}{d t} \\
\therefore+\frac{d[\mathrm{~B}]}{d t}=-\frac{2}{3} \frac{d[\mathrm{~A}]}{d t}
\end{gathered}$

Topic: Chemistry, Chemical Kinetics, NEET, NEET 2002
3d4 configuration may have the exchange of

Options:
A four electrons
B three electrons
C sixteen electrons
D six electrons

The correct answer is:
six electrons

Solution:
Possible exchanges of electrons are
1-2,1-3,1-4
2-3,2-4
3-4
Hence total 6 exchanges are possible.

Topic: Chemistry, Structure of Atom, NEET
3 Faradays of electricity was passed through an aqueous solution of iron (II) bromide. The weight of iron metal (at. wt. = 56) deposited at the cathode (in g) is

Options:
A 65
B 84
C 112
D 168

The correct answer is:
84

Solution:
Fe2++2e-âFe
EFe=562=28
wFe=EFeÃ number of Faraday
=28Ã3=84 g

Topic: Chemistry, Electrochemistry, NEET
3 Fe₃O₄ (s) + 8 Al (s) → 9 Fe (s) + 4 Al₂O₃ (s) Identify the species undergoing oxidation and reduction:

Options:
A Oxidation: Fe₃O₄, Reduction: Al
B Oxidation: Al, Reduction: Fe₃O₄
C Oxidation: Al, Reduction: Fe
D Oxidation: Fe₃O₄, Reduction: Fe

The correct answer is:
Oxidation: Al, Reduction: Fe₃O₄

Solution:
Correct Option is : (B)
Oxidation: Al, Reduction: Fe₃O₄

Topic: Chemistry, Redox Reactions, NEET
3 g of acetic acid is added to 250 mL of 0.1 M HCl and the solution made up to 500 mL. To 20 mL of this solution 12 mL of 5 M NaOH is added. The pH of the solution is?
[Given: pKa of acetic acid =4.75 , molar mass of acetic acid =60 g/mol,log3=0.4771 ] Neglect any changes in volume.

Options:
A 2.61
B 4.52
C 5.22
D 8.78

The correct answer is:
5.22

Solution:
mili mole of acetic acid in 20 ml =2
mili mole of HCl in 20 ml =1
mili mole of NaOH 20 ml =2.5
NaOH(left)1.50+CH3COOH20.5→CH3COONa01.5+H2O0
pH=pKa+logsaltacid=4.74+log1.50.5
=4.74+log3=4.74+0.4771=5.22

Topic: Chemistry, Ionic Equilibrium, NEET
3 g of H2 react with 29 g of O2 to form H2O . Find the amount of the reactant left unreacted?

Options:
A 6 g
B 8 g
C 5.7 g
D 5.2 g

The correct answer is:
5.2 g

Solution:
2H22×2.016=4.032+O232 g→2H2O2×2.016+16=36.032
3 g of H2 require O2=324.032×3=23.8 g
O2 left unreacted =29-23.8=5.2 g

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
3O2( g)⇌2O3( g)

for the above reaction at 298 K, Kc is found to be 3.0×10-59. If the concentration of O2 at equilibrium is 0.040 M then concentration of O3 in M is

Options:
A 1.9×10-63
B 2.4×1031
C 1.2×1021
D 4.38×10-32

The correct answer is:
4.38×10-32

Solution:
Kc=3.0×10-59
O2g=4×10-2.
The given reaction 3O2g⇌2O3g
Kc=product of molar concentration of productsproduct of molar concentration of reactants=O3g2O2g3

Put all the given values in the expression of Kc

3.0×10-59=O3g24×10-23
⇒[O3(g)]2=3.0×10-59×(4×10-2)3
⇒O3g2=192×10-65
Hence, the concentration of O3 = 4.38×10-32 M

Topic: Chemistry, Chemical Equilibrium, NEET, NEET 2022 (Phase 1)
3-PGA is first stable product in

Options:
A Carbon-reduction cycle
B Photorespiration
C Light reaction
D All of these

The correct answer is:
Carbon-reduction cycle

Solution:
3-phosphoglyceric acid (3-PGA) is the first stable product of the Calvin-Benson cycle or C3 cycle or carbon-reduction cycle.

Topic: Biology, Photosynthesis in Higher Plants, NEET
3-PGA is the first stable product in ......(i).... and is formed when initial CO2 acceptor ......(ii)... take CO2 in presence of enzyme RuBisCO.

Options:
A (i) carbon-reduction cycle, (ii) PEP
B (i) Calvin cycle, (ii) RuBP
C (i) Calvin Cycle, (ii) Malic acid
D (i) Hatch and slack cycle , (ii) PEP

The correct answer is:
(i) Calvin cycle, (ii) RuBP

Solution:
In Calvin cycle 5C RuBP is the initial CO2 acceptor and and 3-PGA is the first stable compound formed, the reaction is catalysed by the enzyme RuBisCO.

Topic: Biology, Photosynthesis in Higher Plants, NEET
4Ã1010 electrons are removed from a neutral metal sphere of diameter 20âcm placed in air. The magnitude of the electric field in (N C â1) at a distance of 20âcm from its centre is

Options:
A 640
B 5760
C Zero
D 1440

The correct answer is:
1440

Solution:
Given,
Number of electrons n=4Ã1010 m -3
Diameter of sphere D=20 cm
Radius of sphere r=202=10 cm=0.01 m
According to equation,
Eoutside=q4ÏÎµ0r2
=ne4ÏÎµ0r2
Here, r=20Ã10-2m
=4Ã1010Ã1.6Ã10-19Ã9Ã10920Ã10-22
=4Ã1.6Ã920Ã20Ã10-4
=4Ã1.6Ã9Ã10420Ã20
=1.6Ã9Ã102
=1.44Ã103=1440 NC-1

Topic: Physics, Electrostatics, NEET
4AgNO3+2Cl2(dry)→4X+2Y+Z

In the above sequence of reaction y and z are respectively

Options:
A NO2 and N2O4
B N2O5,O2
C AgCl and O2
D AgCl, N2

The correct answer is:
N2O5,O2

Solution:
4AgNO3+2Cl2→4AgCl(X)+2N2O5(Y)+O2(Z)
Hence Y and Z are N2O5 and O2 respectively.

Topic: Chemistry, d and f Block Elements, NEET
4d, 5p, 5f and 6p orbitals are arranged in the order of decreasing energy. The correct option is:

Options:
A 5f>6p>5p>4d
B 6p>5f>5p>4d
C 6p>5f>4d>5p
D 5f>6p>4d>5p

The correct answer is:
5f>6p>5p>4d

Solution:
n+l values:

4d=4+2=6

5p=5+1=6

5f=5+3=8

6p=6+1=7

Sub-shell which has higher n+l value, has more energy.

If n+l is same for two sub-shells, then the one having more n value has more energy.

5f>6p>5p>4d

Topic: Chemistry, Structure of Atom, NEET, NEET 2019 (All India)
4 g of NaOH present in 0.1 dm3 solution has specific gravity 1.038 gmL. The normality of NaOH solution is?

Options:
A 1 N
B 2 N
C 3 N
D 4 N

The correct answer is:
1 N

Solution:
Volume of solution =0.1 Litre
Normality of NaOH solution=Equivalent of NaOHVolume in litre
N= 4 40 0.1 =1 N

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
4 ml of HCl solution of pH=2 is mixed with 6ml of NaOH solution of pH=12 . What would be the final pH of solution?
log2=0.3

Options:
A 10.3
B 11.3
C 11
D 4.3

The correct answer is:
11.3

Solution:
[HCl]=0.01[NaOH]=0.01
Mmixt=6Ã0.01-4Ã0.0110=0.0210
[OÎH]=0.002
pOH=−log(0.002)=2.7

Topic: Chemistry, Ionic Equilibrium, NEET
4 ml of HCl solution of pH = 2 is mixed with 6 ml of NaOH solution of pH = 12. What would be the final pH of solution? (log 2 = 0.3)

Options:
A 10.3
B 11.3
C 11
D 4.3

The correct answer is:
11.3

Solution:
pH=2, HCl=10-2M & pOH=2
or NaOH=10-2 M
4 ml of 10-2 M HClâ¡4Ã10-5 moles HCl.
6 ml of 10-2 MNaOHâ¡6Ã10-5 moles NaOH
After mixing total moles of OH â = moles of NaOHâ moles of HCl
=6Ã 10 â5 â4Ã 10 â5
After mixing excess moles of OH-=2Ã10-5
Total volume =6+4=10âmlâ=â10âÃ 10 â3 âlitre
OH-=2Ã10-510Ã103=2Ã10-3
or pOH=3-logâ¡2=3-0.3=2.7
or pH=11.3

Topic: Chemistry, Ionic Equilibrium, NEET
4Al + 3O2→

Options:
A 2Al2O3
B 3Al2O3
C Al2O3
D 4Al2O3

The correct answer is:
2Al2O3

Solution:
Correct Option is : (A)
2Al2O3

Topic: Chemistry, p Block Elements (Group 13 & 14), NEET
4s13d5is the electronic configuration of which of the following elements?

Options:
A Cr
B Mn
C Sc
D Cu

The correct answer is:
Cr

Solution:
Correct Option is : 1
Cr

Topic: Chemistry, d and f Block Elements, NEET
0.5 g of fuming H2SO4 (Oleum) is diluted with water. This solution is completely neutralized by 26.7 mL of 0.4 N NaOH. The percentage of free SO3 in the sample is

Options:
A 30.6%
B 40.6%
C 20.6%
D 50.6%

The correct answer is:
20.6%

Solution:
2NaOH+ H 2 S O 4 âN a 2 S O 4 +2 H 2 O
2NaOH+S O 3 âN a 2 S O 4 +2 H 2 O
Meq of H2SO4+Meq of SO3=Meq of NaOH
(0.5-a)49Ã1000+a80/2Ã1000=26.7Ã0.4
So a=0.103
So % of SO3=0.1030.5Ã100=20.6% .

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
5 g of zinc is treated separately with an excess of
(1) dilute hydrochloric acid and
(2) aqueous sodium hydroxide.
The ratio of the volumes of H2 evolved in these two reactions is

Options:
A 1:4
B 1:1
C 1:2
D 2:1

The correct answer is:
1:1

Solution:
Zn+2HClâZnCl2+H2
Zn+2NaOHâNa2ZnO2+H2
Mole of Zn=Mole of dil.HCl2=Mole of NaOH2
As volume of a gas â moles of gas
Volume of   H 2  from HCl Volume of H 2 from NaoH = 1 1

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
$5 \mathrm{~L}$ of $1 \mathrm{M} \mathrm{NaOH}$ solution is mixed with $3 \mathrm{~L}$ of $0.5 \mathrm{M} \mathrm{NaOH}$ solution. Find out the molarity of the resultant solution:

Options:
A $0.80 \mathrm{M}$
B $1.0 \mathrm{M}$
C $0.73 \mathrm{M}$
D $0.50 \mathrm{M}$

The correct answer is:
$0.73 \mathrm{M}$

Solution:
2.5 L $1 \mathrm{M}$ Naoh with 3 L $0.5 \mathrm{M}$ $\mathrm{NaOH}$.
Intially, Number of moles in $2.5 \mathrm{~L}$,
$\begin{aligned}
1 \mathrm{M} \mathrm{NaOH} & =2.5 \times 1 \\
& =2.5 \text { moles }
\end{aligned}$
Number of moles in $3 \mathrm{~L}, 0.5 \mathrm{M} \mathrm{NaOH}$ $=3 \times 0.5=1.5$ moles.
$\Rightarrow$ Total number of moles in final mixture $=2.5+1.5=4.0$ moles .
The total volume of solution $=2.5+3=5.5 \mathrm{~L}$
$\begin{aligned}
& \Rightarrow \text { Molarity of final solution }=\frac{4}{5.5} \\
& =0.728 \mathrm{M}
\end{aligned}$

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET, NEET 2002
0.005 mol of Ba OH 2 is dissolved in 100 mL of water. Assuming complete ionisation of Ba OH 2 the pOH of the solution will be

Options:
A 1
B 5
C 2
D 14

The correct answer is:
1

Solution:
100 mL of solution contain 0.005 moles of Ba OH 2
â´â1âL 1000 mL of solution contain
= 10âÃâ0.005 mole of Ba OH 2
Concentration of Ba OH 2 (i.e. moles on litres)
= 0.05 M
Each Ba OH 2 give 2OH â ions
Thus, moles of OH â per L =2Ã0.05=0.1
iii ââµpOHâ=ââlog OH â
pOHâ=ââlog 0.1 â=ââ log10 â1
pOHâ=âlogâ10â=â1

Topic: Chemistry, Ionic Equilibrium, NEET
0.5 molal aqueous solution of a weak acid $(\mathrm{HX})$ is $20 \%$ ionised. If $\mathrm{K}_f$ for water is $1.86 \mathrm{Kg} \mathrm{mol}^{-1}$, the lowering in freezing point of the solution is:

Options:
A $0.56 \mathrm{~K}$
B $1.12 \mathrm{~K}$
C $-0.56 \mathrm{~K}$
D $-1.12 \mathrm{~K}$

The correct answer is:
$1.12 \mathrm{~K}$

Solution:
$\operatorname{As} \Delta \mathrm{T}_f=i k_f \mathrm{~m}$ For
$\begin{aligned}
& \mathrm{HX} \rightleftharpoons \mathrm{H}^{+}+\mathrm{X}^{-} \\
& t=0 \quad 1 \quad 0 \quad 0 \\
& \begin{array}{llll}
t_{\text {eq }} & (1-0.20) & 0.20 & 0.20
\end{array} \\
&
\end{aligned}$
Total no. of moles $=1-0.20+0.20$
$\begin{aligned}
+0.20 & =1+0.20=1.2 \\
\therefore \Delta \mathrm{T}_f & =1.2 \times 1.86 \times 0.5 \\
& =1.1160
\end{aligned}$
Related Theory
The boiling point and freezing point are colligative properties. Due to the addition of solutes, the boiling point tends to increase, and freezing point tends to decrease.

Topic: Chemistry, Solutions, NEET, NEET 2007
0.5 molal solution of ethylene glycol in water is used as a coolant in car. If the freezing point depression constant of water is 1.86oCkg per mole, the mixture will freeze at

Options:
A 0.93oC
B -0.93oC
C 1.86oC
D -1.86oC

The correct answer is:
-0.93oC

Solution:
ΔTf=kf×m
=1.86×0.5=0.93oC
∴ΔTf= Tfo-Tf
∴Tf=Tfo-ΔTf
=0-0.93= -0.93oC

Topic: Chemistry, Solutions, NEET
5 moles of an ideal gas at 27Â°C expands isothermally and reversibly from a volume of 5 L to 50 L. The work done by the system is

Options:
A -14.7 KJ
B -28.72 KJ
C 27.72 KJ
D -56.72 KJ

The correct answer is:
-28.72 KJ

Solution:
W=-2.303nRTlogV2V1
=-2.303×5×8.314×300×log505 = –28.72 KJ

Topic: Chemistry, Thermodynamics (C), NEET
5N H2SO4 was diluted from 1 litre to 10 litres . Normality of the solution obtained is

Options:
A 10 N
B 5Â N
C 1Â N
D 0.5 N

The correct answer is:
0.5 N

Solution:
N1V1=N2V2 5N Ã1 lit = x Ã 10 lit.
â´ x=0.5 N

Topic: Chemistry, Redox Reactions, NEET
6Ã 10 â3 mole K 2 Cr 2 O 7 reacts completely with 9Ã 10 â3 mole of X n+ to give XO 3 - and Cr 3+ . Value of n will be

Options:
A 1
B 3
C 2
D 4

The correct answer is:
1

Solution:
X n + → X + 5 O 3 - + 5 - n e -
Cr 2 O 7 2 - + 1 4 H + + 6 e - → 2 Cr 3 + + 7 H 2 O
Making no. of moles of electrons same for oxidation and reduction reaction
6 × 1 0 - 3 × 6 = 5 - n × 9 × 1 0 - 3
Implies n=1

Topic: Chemistry, Redox Reactions, NEET
0.6 mL of acetic acid (CH3COOH), having density 1.06 g mL–1, is dissolved in 1 litre of water. The depression in freezing point observed for this strength of acid was 0.0205°C. What are the van’t Hoff factor and the dissociation constant of acid?

Options:
A Vant Hoff factor is 0.098 and Dissociation constant of acid is 3.54× 10–5
B Vant Hoff factor is 0.098 and Dissociation constant of acid is 1.86 × 10–5
C Vant Hoff factor is 1.041 and Dissociation constant of acid is 3.54 × 10–5
D Vant Hoff factor is 1.041 and Dissociation constant of acid is 1.86 × 10–5

The correct answer is:
Vant Hoff factor is 1.041 and Dissociation constant of acid is 1.86 × 10–5

Solution:
Correct Option is : (D)
Vant Hoff factor is 1.041 and Dissociation constant of acid is 1.86 × 10–5

Topic: Chemistry, Solutions, NEET
6 mol of a mixture of Mohr's salt and Fe2(SO4)3 requires 500 ml of 1 M of K2Cr2O7 for complete oxidation in acidic medium. The mole % of the Mohr's salt in the mixture is

Options:
A 75
B 50
C 60
D 25

The correct answer is:
50

Solution:
Cr2O72-+6Fe2+(n=1) (Mohr's salt) +14H+→2Cr3++6Fe3++7H2O
Equivalent of Fe2+= moles of Mohr's salt
= equivalent of K2Cr2O7
=500×10-3×6×1=3
Hence mole percent of Mohr's salt
=36×100=50

Topic: Chemistry, Redox Reactions, NEET
0.6 mole of PCl5,0.3 mole of PCl3 and 0.4 mole of Cl2 are taken in a 1 L flask to obtain the following equilibrium
PCl5(g)⇌PCl3(g)+Cl2(g)
If the equilibrium constant Kc for the reaction is 0.15. Predict the direction of the reaction.

Options:
A Forward direction
B Backward direction
C Direction of the reaction can not be predicted
D Reaction does not move in any direction

The correct answer is:
Backward direction

Solution:
PCl5(g)⇌PCl3(g)+Cl2(g)
Qc=0.4×0.30.6=0.2
Kc=0.15 since Qc>Kc reaction will proceed in backward direction.

Topic: Chemistry, Chemical Equilibrium, NEET
8 : 8 coordination of CsCl is found to change into 6 : 6 coordination on

Options:
A Applying pressure
B Increasing temperature
C Both A and B
D None of these

The correct answer is:
Increasing temperature

Solution:
CsCl8:8âHigh temperatureHigh pressureNaCl6:6
High temperature changes 8 : 8 coordination to 6 : 6, whereas high pressure changes 6 : 6 coordination to 8 : 8.

Topic: Chemistry, Solid State, NEET
8th, 9th and 10th pair of ribs

Options:
A articulate with the sternum anteriorly
B articulate with 7th rib with the help of a hyaline cartilage
C articulate with 7th rib with the help of a fibrocartilage
D do not articulate arteriorly

The correct answer is:
articulate with 7th rib with the help of a hyaline cartilage

Solution:
Ribs are the part of the thoracic skeletal system that protects the lungs and heart. Ribs are attached to the vertebral column and sternum. There are 12 pairs of ribs attached to the vertebral column posteriorly to the T1-T12 bone.
The first seven sets of ribs attached to the sternum are referred to as "true ribs". They are attached by costal cartilages. The last five sets of ribs are referred to as "false ribs" of which three pairs share a common cartilaginous connection to the sternum by articulating with 7th rib with hyaline cartilage and the last two (eleventh and twelfth ribs) are termed floating ribs. They are attached to the vertebrae only and not to the sternum.

Topic: Biology, Locomotion and Movement, NEET
1.0 g of $\mathrm{H}_2$ has same number of molecules as in:

Options:
A 14 g of $\mathrm{N}_2$
B 18 g of $\mathrm{H}_2 \mathrm{O}$
C 16 g of CO
D 28 g of $\mathrm{N}_2$

The correct answer is:
14 g of $\mathrm{N}_2$

Solution:
Number of moles of $\mathrm{H}_2=\frac{1}{2}=0.5$
Number of molecules of $\mathrm{H}_2=0.5 \mathrm{~N}_{\mathrm{A}}$.
(1) Number of moles of $\mathrm{N}_2=\frac{14}{28}=0.5$
Number of molecules of $\mathrm{N}_2=0.5 \mathrm{~N}_{\mathrm{A}}$
(2) Number of moles of $\mathrm{H}_2 \mathrm{O}=\frac{18}{18}=1$
Number of molecules of $\mathrm{H}_2 \mathrm{O}=1 \times \mathrm{N}_{\mathrm{A}}=\mathrm{N}_{\mathrm{A}}$
(3) Number of moles of $\mathrm{CO}=\frac{16}{28}=\frac{4}{7}$
Number of molecules of $\mathrm{CO}=\frac{4}{7} \mathrm{~N}_{\mathrm{A}}$
(4) Number of moles of $\mathrm{N}_2=\frac{28}{28}=1$
Number of molecules of $\mathrm{N}_2=1 \times \mathrm{N}_{\mathrm{A}}=\mathrm{N}_{\mathrm{A}}$

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET, NEET 2024 (Re-NEET)
$10 \mathrm{~g}$ of hydrogen and 64 of oxygen were filled in a steel vessel and exploded. Amount of water produced in this reaction will be

Options:
A $2 \mathrm{~mol}$
B $3 \mathrm{~mol}$
C $4 \mathrm{~mol}$
D $1 \mathrm{~mol}$

The correct answer is:
$4 \mathrm{~mol}$

Solution:
Key Idea
(i) $\mathrm{H}_2+\frac{1}{2} \mathrm{O}_2 \longrightarrow \mathrm{H}_2 \mathrm{O}$
(ii) Amount of water produced is decided by limited reactant (ie, the reactant which is used in small amount)
$$
\begin{aligned}
& \mathrm{H}_2+\frac{1}{2} \mathrm{O}_2 \longrightarrow \mathrm{H}_2 \mathrm{O} \\
& 1 \mathrm{~mol} \quad \frac{1}{2} \mathrm{~mol} \quad 1 \mathrm{~mol} \\
& \frac{10}{2} \mathrm{~mol} \quad \frac{64}{32} \mathrm{~mol} \\
& =5 \mathrm{~mol}=2 \mathrm{~mol} \\
& \because \frac{1}{2} \mathrm{molO}_2 \text { gives }=1 \mathrm{~mol} \mathrm{H}_2 \mathrm{O} \\
& \therefore 2 \mathrm{~mol} \mathrm{O}_2 \text { will give }=1 \times 2 \times 2=4 \mathrm{~mol} .
\end{aligned}
$$

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET, NEET 2009 (Screening)
1.0 g of magnesium is burnt with 0.56 g O2 in a closed vessel. Which reactant is left in excess and how much?
(At. wt. Mg = 24; O = 16)

Options:
A O2 , 0.28 g
B O2 , 0.16 g
C Mg, 0.16 g
D Mg, 0.44 g

The correct answer is:
Mg, 0.16 g

Solution:
nMg=weightatomicweight=124mole, nO2=0.5632mole

∴ mass of Mg=0.0066×24g=0.16g
O2 is Limiting reagent
Mg is excess of reagent

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
1.0 g of magnesium is burnt with 0.56 g O2 in a closed vessel. Which reaction is left in excess and how much? (At, wt. Mg=24; O=16)

Options:
A Mg, 0.16 g
B O2,Â 0.16Â g
C Mg,Â 0.44Â g
D O2,Â O.28Â g

The correct answer is:
Mg, 0.16 g

Solution:
Mg+12O2→MgO
1.024 0.5632
0.512 0.074
Oxygen is limiting reagent so 0.074-x2=0
x=0.072
excess Mg=0.512-0.072 mole
mass of Mg is =1-0.7×12=0.16 gram.

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET, NEET 2014
10 mL citric acid (H3C6H5O7) is neutralised completely by 35.6 mL of 0.312 M NaOH solution. The molarity of the solution of citric acid is

Options:
A 1.11 M
B 0.45 M
C 0.11 M
D 0.37 M

The correct answer is:
0.37 M

Solution:
H3C6H5O7+3NaOHâNa3C6H5O7+3H2O
M1V11=M2V23 or M1=M2V23V1
= 0.312×35.6 3×10 =0.37 M

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
1.0 mol of a monoatomic ideal gas is expanded from state (1) to state (2) as shown in Fig. 6.4. Calculate the work done for the expansion of gas from state (1) to state (2) at 298 K.

Options:
A –1320.46 J
B –2214.46 J
C –1717.46 J
D –1967.46 J

The correct answer is:
–1717.46 J

Solution:
Correct Option is : (C)
–1717.46 J

Topic: Chemistry, Thermodynamics (C), NEET
10 resistors, each of resistance R are connected in series to a battery of emf E and negligible internal resistance. Then those are connected in parallel to the same battery, the current is increased n times. The value of n is:

Options:
A 1000
B 10
C 100
D 1

The correct answer is:
100

Solution:
When the resistors are connected in series with the battery, the equivalent resistance Rs can be written as

Rs=R+R+....10 times=10R ...1

Hence, the current is is given by

is=ERs=E10R ...2

When the resistors are connected in parallel, the equivalent resistance Rp is given by

Rp=11R+1R+.... 10 times= R10 ...3

Hence, the current ip is given by

ip=ERp=ER10=10ER ...4

Divide equation (4) by equation (2) to obtain the required value of the multiplier n.

ipis=10ERE10R= 100

Topic: Physics, Current Electricity, NEET, NEET 2023 (All India)
1123Na and Mg1224 are

Options:
A Isotopes
B Isobars
C Isodiaphers
D Isotones

The correct answer is:
Isotones

Solution:
∵ No. of neutrons are same

Topic: Chemistry, Structure of Atom, NEET
1.2 L of hydrogen and 1.12 L of chlorine are reacted. The composition by volume of mixture is

Options:
A 0.08 L of H2 and 2.24 L of HCl
B 2.24 L of H2 and 2.24 L of HCl
C 0.08 L of Cl2 and 20.8 L of HCl
D 22.4 L of Cl2 and 20.8 L of HCl

The correct answer is:
0.08 L of H2 and 2.24 L of HCl

Solution:
H21vol1.2L+Cl21vol1.12L→2HCl2vol2.24L
The unreacted H2=1.2L-1.12L=0.08L
Volume of HCl formed = 2.24 L.

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
1.4 moles of phosphorus trichloride are present in a sample. How many atoms are there in the sample?

Options:
A 5.6×1023
B 34×1024
C 2.4×1023
D 3.372×1024

The correct answer is:
3.372×1024

Solution:
Number of atoms in 1 mole of PCl3=4
Number of atoms in 1.4 moles of PCl3
=4×1.4×6.022×1023
=3.372×1024

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
0.15 mol of pyridinium chloride has been added into 500 cm3 of 0.2 M pyridine solution. The pH and OH- ion concentration in the resulting solution assuming no change in volume is (kb=1.5Ã10-9M)

Options:
A 5,10-8mol/litre
B 5,10-9mol/litre
C 6,10-9mol/litre
D 7,10-9mol/litre

The correct answer is:
5,10-9mol/litre

Solution:
pOH =-logkb+log[Salt][Base]
=-log1.5×10-9+log2×0.150.2
=-log1.5-log10-9+log1.5=9
pH=14-pOH=14-9=5
[OH-]=10-pOH=10-9moll-1

Topic: Chemistry, Ionic Equilibrium, NEET
15 moles of H 2 and 5.2 moles of I 2 are mixed and allowed to attain equilibrium at 500 â o C . At equilibrium, the number of moles of HI is found to be 10 mole. The equilibrium constant for the formation of HI is

Options:
A 50
B 15
C 100
D 25

The correct answer is:
50

Solution:
H2 + I2 â 2HI
15 5.2 0
15â5 5.2â5 10
Equilibrium constant
Kc=HI2H2[I2]=10Ã1010Ã0.2=50

Topic: Chemistry, Chemical Equilibrium, NEET
1.6 g of an organic compound gave 2.6 g of magnesium pyrophosphate. The percentage of phosphorus in the compound is

Options:
A 45.38%
B 54.38%
C 37.76%
D 19.02%

The correct answer is:
45.38%

Solution:
MassâofâPhosphorus =31u
Sinceâ1âmoleâof âMg2P2O7 âhas â2 gâatomsâofâP
MassâofâPhosphorusâin âMg2P2O7=62u
Massâof âMg2P2O7=222.5u
Weightâof âMg2P2O7=2.6âgâ
Weightâofâorganicâcompound =1.6âg
Percentageâofâphsphorus=Mass of phsphorus Mass of Mg2P2O7ÃWeight of Mg2P2O7 Weight of organic compoundÃ100
Percentageâofâphsphorus=62222.5Ã2.61.6Ã100=45.38%

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
18 g of glucose, C6H12O6, is dissolved in 1 kg of water in a saucepan. At what temperature will water boil at 1.013 bar? (Kbfor water is 0.52 K kg mol-1)

Options:
A 124.568 K
B 453.685 K
C 373.202 K
D 521.471 K

The correct answer is:
373.202 K

Solution:
Correct Option is : (C)
373.202 K

Topic: Chemistry, Solutions, NEET
19 g of molten SnCl2 is electrolysed for sometime using inert electrodes. 0.119 g of Sn is deposited at the cathode. No substance is lost during the electrolysis. The ratio of the weights of SnCl2 : SnCl4 after electrolysis [Atomic weight of Sn = 119]

Options:
A 71.34 : 1
B 31.34 : 1
C 7.134 : 1
D None of these

The correct answer is:
71.34 : 1

Solution:
2SnCl 2 → SnCl 4  + Sn 2×190   ≡   261    ≡   119 0.38  ≡   0.261    ≡   0.119
When 0.119 g Sn deposited then
(i) SnCl 2 Consumed =0.38 g
SnCl 2 Left =19−0.38 =18.62 g
(ii) SnCl 2 Formed =0.261
W SnCl 2 W SnCl 2 = 18.62 0.261 =71.34:1.

Topic: Chemistry, Electrochemistry, NEET