datavorous/entrance-exam-dataset huggingface dataset:
22 g of CO2 at 27 Â°C is mixed with 16 g of O2 at 37 Â°C. If both gases are considered as ideal then the temperature of the mixture is (ignore vibrational degree of freedom)

Options:
A 30°C
B 30.5°C
C 31.5°C
D 32°C

The correct answer is:
32°C

Solution:
Increase in energy of CO2=Loss in energy of O2
n1Cv1T-T1=n2Cv2T2-T
125T-27=12537-T
T=32 °C

Topic: Physics, Thermodynamics, NEET
22 g of carbon dioxide at 27 Â°C is mixed in a closed container with 16 g of oxygen at 37 Â°C. If both gases are considered as ideal gases, then the temperature of the mixture is ( neglect vibrational degree of freedom )

Options:
A 24.2 °C
B 28.5 °C
C 31.5 Â°C
D 33.5 °C

The correct answer is:
31.5 Â°C

Solution:
For carbon dioxide, number of moles n1=2244=12
Molar specific heat of CO2 at constant volume CV1=5R2
For oxygen, number of moles n2=1632=12
Molar specific heat of O2 at constant volume CV2=5R2
Let T K be the temperature of mixture.
Heat lost by O2 = Heat gained by CO2
n2CV2âT2=n1CV1âT1
12 52 R310-T=12Ã5R2T-300
2T=610
T=6102=305 K = 32 Â°C

Topic: Physics, Thermal Properties of Matter, NEET
23% amount of CO2 is transported in the blood as

Options:
A bicarbonate ions
B carbaminohaemoglobin
C carboxyhaemoglobin
D carbonates

The correct answer is:
carbaminohaemoglobin

Solution:
Carbon dioxide is transported in the blood from the tissue to the lungs in three ways: 1 (i) 7% dissolved in solution; (ii) 70 % as bicarbonates; (iii) 23% bound to proteins, particularly haemoglobin. Approximately 20 to 25% of carbon dioxide is transport in the red blood cell and 75% in the plasma.

Topic: Biology, Breathing and Exchange of Gases, NEET
2,3-Dimethyl-2-butene can be prepared by heating which of the following compounds with a strong acid?

Options:
A
B CH33C-CH=CH2
C CH32C=CH-CH2-CH3
D CH32CH-CH2-CH=CH2

The correct answer is:
CH33C-CH=CH2

Solution:

Topic: Chemistry, Hydrocarbons, NEET, NEET 2015 (Phase 2)
2.4 g of pure Mg (at. mass = 24) is dropped in 100 mL of 1 M HCl. Which of the following statement is wrong?

Options:
A 1.12 L of hydrogen is produced at S.T.P.
B 0.05 mol of magnesium is left behind
C HCl is the limiting reagent
D 0.005 mol of magnesium is left behind

The correct answer is:
0.005 mol of magnesium is left behind

Solution:
Mg2.4g0.1mol-0.050.05mol+2HCl(1M,100ml)0.1mol→MgCl2+H2
HCl is a limiting reagent
0.1 mol HCl give 0.12 mol H2 or 1.12 lit H2 at STP
0.05 mol Mg left
So option D is wrong.

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
2,4-dichlorophenoxyacetic acid is used to eliminate _____ weeds from a ____ crop field.

Options:
A monocotyledonous; dicot
B dicotyledonous; monocot
C biennial; annual
D biennial; perennial

The correct answer is:
dicotyledonous; monocot

Solution:
Auxin, 2,4-Dichlorophenoxyacetic acid is used to kill dicotyledonous or broadleaf weeds. It does not affect mature monocotyledonous plants. It is used to prepare weed-free lawns by gardeners.

Topic: Biology, Plant - Growth and Development, NEET
2,4-DNP is obtained by reacting hydrazine hydrate with which of the following?

Options:
A
B
C
D

The correct answer is:

Solution:
The major component of Borsch reagent is 2, 4- dinitrophyenyl hydrazine which can be obtained by reaction of 2, 4-dinitrochloro benzene and hydrazine

Topic: Chemistry, Amines, NEET
2.4-DNP test is used to detect the presence of which functional group?

Options:
A Aldehydes/Ketones
B Amine
C Amide
D Carboxylic acid

The correct answer is:
Aldehydes/Ketones

Solution:
Correct Option is : 1
Aldehydes/Ketones

Topic: Chemistry, Aldehydes and Ketones, NEET
2.5 Litre of 1 M NaOH solution is mixed with another 3 litre solution of 0.5 M NaOH. The molarity of resultant solution is

Options:
A 1 M
B 0.84 M
C 0.73 M
D 0.56 M

The correct answer is:
0.73 M

Solution:
M f = M 1 V 1 + M 2 V 2 V 1 + V 2
= 25+1.5 5.5
∴   M f =0.73M

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
28 g N2 and 6 g H2 were mixed. At equilibrium 17 g NH3 was formed. The weight of N2 and H2 of equilibrium are respectively.

Options:
A 11 g, 20 g
B 28 g, 6 g
C 14 g, 3 g
D None of these

The correct answer is:
14 g, 3 g

Solution:
Moles of N2=2828=1 ,
Moles of H2=62=3
Moles of NH3=1717=1
Moles of N2=12=14 g
Moles of H 2 = 3 2 ⇒ 3×2 2 =3 g .

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
2.9 g of a gas at 95°C occupied the same volume as 0.184 g of dihydrogen at 17 °C, at the same pressure. What is the molar mass of the gas?

Options:
A 35 g mol-1
B 40 g mol-1
C 45 g mol-1
D 50 g mol-1

The correct answer is:
40 g mol-1

Solution:
Correct Option is : (B)
40 g mol-1

Topic: Chemistry, States of Matter, NEET
30 mL of 0.1 M BaCl2 is mixed with 40 mL of 0.2 MAl2SO43 . What is the weight of BaSO4 formed ? BaCl2 +Al2SO43â BaSO4 + AlCl3 [Molecular Mass of BaSO4 = 233]

Options:
A 0.999 g
B 0.699 g
C 0.799 g
D 0.99 g

The correct answer is:
0.699 g

Solution:
First balance it 3 BaCl 2 + Al 2 SO 4 3 â 3 BaSO 4 + 2 AlCl 3 mM before reaction 3 0 Ã 0 . 1 4 0 Ã 0 . 2 = 3 = 8 0 0 mM after reaction 0 8 - 1 3 2
âµ 3 mM of BaCl2 reacts with one mole of Al2SO43 .
â´ mM of BaSO4 formed = 3 =wt.mol wt.Ã1000;
â´ wt. of BaSO4 formed =3 Ã2331000= 0.699 g

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
30 mL of 0.2 N BaCl2 is mixed with 40 mL of 0.3 N Al2SO43 . How many g of BaSO4 are formed?

Options:
A 0.10 g
B 0.60 g
C 0.90 g
D 0.70 g

The correct answer is:
0.70 g

Solution:
BaCl 2 + Al 2 SO 4 3 → BaSO 4 + AlCl 3 Meq. Before Reaction 3 0 × 0 . 2 4 0 × 0 . 3 0 0 = 6 = 1 2 = 0 = 0 Meq. After Reaction 0 6 6 6
6Â Meq.Â ofÂ BaCl2 reacts with 6Â Meq.Â ofÂ Al2SO43 to produce 6Â Meq.Â ofÂ BaSO4 and 6Â Meq.Â ofAlCl3.Â
â´Â Â Â Meq.Â ofÂ BaSO4Â formedÂ =Â 6Â =Wt.EÂ ÃÂ 1000; E=2332
wt.Â ofÂ BaSO4Â formedÂ =6Â ÃE1000=6Â Ã2332Â ÃÂ 1000=Â Â 0.70Â gÂ Â

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
4.0 g of a gas occupies 22.4 liters at NTP. The specific heat capacity of the gas at constant volume is 5.0 J K-1 mol-1 . If the speed of sound in this gas at NTP is 952 m s-1 , then the heat capacity at constant pressure is
(Take gas constant R=8.3 J K-1 mol-1 )

Options:
A 7.5 J K-1 mol-1
B 7.0 J K-1 mol-1
C 8.5 J K-1 mol-1
D 8.0 J K-1 mol-1

The correct answer is:
8.0 J K-1 mol-1

Solution:
No. of mole of gas = 1 so molar mass = 4g/mole
v=γRTm ⇒952×952=γ×8.3×2734×10-3⇒γ=1.6=1610=85

CP=8.0 J K-1 mol-1

Topic: Physics, Kinetic Theory of Gases, NEET, NEET 2015 (Phase 2)
40 ml N10HCl solution is mixed with 60 ml of N20KOH solution. The resulting mixture will be –

Options:
A Acidic
B Basic
C Neutral
D cannot be predicted

The correct answer is:
Acidic

Solution:
Number of meq of HCl=40×110=4
Number of meq of KOH=60×120=3
∴ meq of HCl > meq of KOH
∴ Solution will be acidic.

Topic: Chemistry, Ionic Equilibrium, NEET
40 mL sample of 0.1 M solution of nitric acid is added to 20 mL of 0.3 M aqueous ammonia. What is the pH of the resulting solution? (pKb=4.7447)

Options:
A 8.95
B 89.5
C 0.895
D 4.48

The correct answer is:
8.95

Solution:
HNO34 mmol+NH36 mmolâNH4NO34 mmol
[Since WB(NH3) is left and salt (NH4NO3) is formed so it is a basic buffer]
Base left = 6 â 4 = 2 mmol
[Acid] = 40Ã0.1 = 4 mmol
[Base] =20Ã0.3=6
pOH =pKb+log10[Salt][Base]
So, pOH =4.7447+log4/2=5.0457
and pH = 14 â 5.0457 = 8.95.

Topic: Chemistry, Ionic Equilibrium, NEET
40% (w/v) of NaCl solution (specific gravity = 1.12) is equivalent to

Options:
A 3.57×105 ppm
B 3.57×106 ppm
C 1×106 ppm
D 4×105 ppm

The correct answer is:
3.57×105 ppm

Solution:
Volume of solution = 100 mL
Mass of solution = Volume × density
Mass of solution =100×1.12=112 g
Concentration in ppm =WsoluteWsolution×106
ppm in NaCl solution =40112×106
=3.57×105 ppm.

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
0.44 g of a colourless oxide of nitrogen occupies 224 ml at STP. The molecular formula is

Options:
A NO
B NO2
C N2O
D N2O5

The correct answer is:
N2O

Solution:
Molecular weight =  Weight Volume  × 22400 ml
=0.44Ã22400224=44
Hence molecular formula =N2O .

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
$4.5 \mathrm{~g}$ of aluminium (at mass $27 \mathrm{amu}$ ) is deposited at cathode from $\mathrm{Al}^{3+}$ solution by a certain quality of electric charge. The volume of hydrogen produced at STP from $\mathrm{H}^{+}$ions in solution by the same quantity of electric charge will be:

Options:
A $44.8 \mathrm{~L}$
B $22.4 \mathrm{~L}$
C $11.2 \mathrm{~L}$
D $5.6 \mathrm{~L}$

The correct answer is:
$5.6 \mathrm{~L}$

Solution:
We know that,
I Faraday charge liberates 1 eq. of substance.
eq. wt. of $\mathrm{AI}=\frac{\text { wt.of } \mathrm{Al}}{\text { eq. } w t}=\frac{4.5}{9}=0.5$
no. of Faraday required $=0.5$
no. of eq. of $\mathrm{H}_2$ produce $=0.5$ eq.
Volume occupied by 1 eq. of $\mathrm{H}_2=$ $\frac{22.4}{2}=11.2$
Volume occupied by $0.5 \mathrm{eq}=11.2 \times$ 0.5
$=5.6$ litre at STP

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET, NEET 2005
45 g of ethylene glycol (C2H6O2) is mixed with 600 g of water. Calculate (i) the freezing point depression and (ii) the freezing point of the solution.

Options:
A (i) 2.2 K and (ii) 270.95 K
B (i) 4.8 K and (ii) 270.95 K
C (i) 2.2 K and (ii) 369.21 K
D (i) 4.8 K and (ii) 369.21 K

The correct answer is:
(i) 2.2 K and (ii) 270.95 K

Solution:
Correct Option is : (A)
(i) 2.2 K and (ii) 270.95 K

Topic: Chemistry, Solutions, NEET
50 g ice at 0 Â°C in kept in an insulating vessel and 50 g water at 100 Â°C is mixed in it. Then the final temperature of the mixture is (neglect the heat loss)

Options:
A 10 Â°C
B 0 Â°C<Tm<20 Â°C
C 20 Â°C
D above 20 Â°C

The correct answer is:
10 Â°C

Solution:
Heat taken = Heat given
miLf+miswt-0=mwswtn-t
50Ã80+50Ã1Ãt=50Ã1Ã100-t
4000=50100-2t
80=100-2t
2t=100-80
2t=20
t=10 Â°C

Topic: Physics, Thermal Properties of Matter, NEET
50 g of ice at 0 Â°C is mixed with 50 g of water at 80 Â°C. The final temperature of the mixture is (latent heat of fusion of ice =80 cal/g, sw=1cal/g Â°C)

Options:
A 0 °C
B 40 Â°C
C 60 Â°C
D less than 0 Â°C

The correct answer is:
0 °C

Solution:
The heat required by the ice Q=m×L=50×80=4000 cal
the heat released by water Q=msΔθ=50×1×80=4000 cal
both are equal. Final temperature =0 °C

Topic: Physics, Thermal Properties of Matter, NEET
0.50 g sample of impure CaCO3 is dissolved in 50 ml of 0.0985 (N) HCl. After the reaction is complete, the excess HCl required 6 ml of 0.105 N NaOH for neutralisation. The percentage purity of CaCO3 in the sample is

Options:
A 42.95
B 429.5
C 4.295
D 21.86

The correct answer is:
42.95

Solution:
Miliequivalents = Normality × V(inml)
Miliequivalents of HCl = Miliequivalents of NaOH + Miliequivalents of CaCO3
0.0985×50=0.105×6+WCaCO3100×2×1000
4.925-0.630=WCaCO3×20
WCaCO3=0.21475g
%CaCO3=0.214750.50×100=42.95%

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
$50 \mathrm{~mL}$ of each gas $A$ and of gas $B$ takes 150 and $200 \mathrm{~s}$ respectively for effusing through a pin hole under the similar conditions. If molecular mass of gas $B$ is 36 , the molecular mass of gas $A$ will be

Options:
A 96
B 128
C 32
D 20.25

The correct answer is:
20.25

Solution:
Given,
$\begin{aligned}
V_A & =V_B=50 \mathrm{~mL} \\
T_A & =150 \mathrm{~s} \\
T_B & =200 \mathrm{~s} \\
M_B & =36 \\
M_A & =?
\end{aligned}$
From Graham's law of effusion
$\begin{aligned}
\frac{r_B}{r_A} & =\sqrt{\frac{M_A}{M_B}}=\frac{V_B T_A}{T_B \cdot V_A} \\
\Rightarrow \sqrt{\frac{M_A}{36}} & =\frac{V_A \times 150}{200 \times V_A} \\
\text {or } \sqrt{\frac{M_A}{36}} & =\frac{15}{20}=\frac{3}{4} \\
\frac{M_A}{36} & =\frac{9}{16} \\
M_A & =\frac{9 \times 36}{16}=\frac{9 \times 9}{4}=\frac{81}{4}=20.25
\end{aligned}$

Topic: Chemistry, States of Matter, NEET, NEET 2012 (Screening)
56 g of nitrogen and 96 g of oxygen are mixed isothermally and at a total pressure of 10 atm. The partial pressures of oxygen and nitrogen (in atm) are respectively.

Options:
A 5, 5
B 2, 8
C 6, 4
D 8, 2

The correct answer is:
6, 4

Solution:
According to Dalton’s law,
Partial pressure of a component = Mole fraction × Total pressure
As mass of N2=56g
Mass of O2=96g
Total pressure = 10 atm
nN2=5628=2,nO2=9632=3
xN2=22+3=0.4
xO2=32+3=0.6
pN2=0.4×10=4atm
pO2=0.6×10=6atm

Topic: Chemistry, States of Matter, NEET
64 g non-volatile solute is added to 702 g benzene. The vapour pressure of benzene has decreased from 200 mm of Hg to 180 mm of Hg . Molecular weight of the solute is

Options:
A 128
B 64
C 96
D 256

The correct answer is:
64

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

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
70 calories of heat is required to raise the temperature of 2 moles of an ideal diatomic gas at constant pressure from 30°C to 35°C. The amount of heat required (in calories) to raise the temperature of the same gas through the same range ( 30°C to 35°C ) at constant volume is

Options:
A 30
B 50
C 70
D 90

The correct answer is:
50

Solution:
Heat is required to raise the temperature at constant pressure,
Q1=nCPΔT,
Heat is required to raise the temperature at constant volume,
Q2=nCVΔT
Q2Q1=CVCP=1γ
Q2=Q1γ=701.4=50 cal

Topic: Physics, Thermodynamics, NEET
0.73 g of organic compound on oxidation gave 1.32 g of carbon dioxide. The percentage of carbon in the given compound will be

Options:
A 49.32
B 59.32
C 29.32
D 98.64

The correct answer is:
49.32

Solution:
%ofcarbon=1244×wt.ofCO2wt.ofsample×100=1244×1.320.73×100=49.32

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
75% of the first order reaction completed in 32 min. 50% of the reaction was completed in

Options:
A 24Â min
B 8Â min
C 16 min
D 4Â min

The correct answer is:
16 min

Solution:
for first order reactions
t 75 =2Ã t 50
t 1/2 = 32 2 =16 min

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

Options:
A C2H2
B C4H4
C C6H6
D C8H8

The correct answer is:
C6H6

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

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
80 g of water at 30 Â°C is poured on a large block of ice at 0 Â°C. The mass of ice that melts is

Options:
A 30 g
B 80 g
C 1600 g
D 150 g

The correct answer is:
30 g

Solution:
mice L=mwSΔQ
mice80=80130-0
mice=30 g

Topic: Physics, Thermal Properties of Matter, NEET
80 railway wagons all of same mass 5Ã103 kg are pulled by an engine with a force of 4Ã105 N. The tension in the coupling between 30th and 31st wagon from the engine is

Options:
A 25Ã104 N

B 40 ×104 N
C 20 Ã104 N
D 32 Ã104 N

The correct answer is:
25Ã104 N

Solution:
Railway waygons to be pulled = 80 - 30 = 50
∴m=50×5×103 kg
a=F80 M=4×10580×5×103
⇒ F 30 = 50 × 5 × 10 3 × 4 × 10 5 80 × 5 × 10 3 =25×104 N

Topic: Physics, Laws of Motion, NEET
0.85% aqueous solution of NaNO3 is apparently 90% dissociated. The osmotic pressure of solution at 300 K is

Options:
A 4.674 atm
B 46.74 atm
C 2.46 atm
D 4.674 mm Hg

The correct answer is:
4.674 atm

Solution:
M=W×1000mol.wt.×V
Molarity =1000cm3×0.85g100cm3×85gmol-1=0.1molL-1
NaNO3→Na++NO3-
So i=1-α+α-α1
=1+α=1+0.90=1.90
π=iCRT=1.90×0.1×0.082×300
=4.674 atm.

Topic: Chemistry, Solutions, NEET
89Ac231 after emission of some α and β- particles gives 82Pb207 . The number of such α and β- particles are respectively

Options:
A 5,7
B 6,5
C 7,5
D 5,6

The correct answer is:
6,5

Solution:
Number of α- particles =231-2074=6
Number of β−=12− 89−2 =5 .

Topic: Chemistry, Chemical Kinetics, NEET
90 g non-volatile, non-dissociative solute is added to 1746 g water to form a dilute, ideal solution. The vapour pressure of water has decreased from 300 mm of Hg to 291 mm of Hg . The molecular weight of solute is.

Options:
A 90
B 60
C 30
D 15

The correct answer is:
30

Solution:
P-PsP=XsoluteÃi i=1â´solve is non-ioinsable
300-291300=90/M90M+174618
9300=90M90M+97
3Ã90MÃ100+97Ã3100=90M
1810M+97Ã3100=90M
90M-1810M-97Ã3100
900-1810M=97Ã310
M=882Ã1097Ã3=30.3â30

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
${ }_{92} \mathrm{U}^{235}$ nucleus absorbs a neutron and disintegrates in ${ }_{54} \mathrm{Xe}^{139},{ }_{38} \mathrm{Sr}^{94}$ and $x$. What will be the product $x$ ?

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

The correct answer is:
2 neutrons

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

Topic: Physics, Nuclear Physics, NEET, NEET 2002
92U235 undergoes successive disintegrations with the end product of 82Pb203. The number of Î± and Î²- particles emitted are

Options:
A Î±=6Â ,Â Î²=4
B Î±=6Â ,Â Î²=0
C α=8 , β=6
D Î±=3,Â Î²=3

The correct answer is:
α=8 , β=6

Solution:
Let number of α particles decayed be x and number of β particles decayed be y.
Then equation for the decay is given by
92U235→ xα24+ yβ-10+Pb82203
Equating the mass number on both sides
235=4x+203 …(i)
Equating atomic number on both sides
92=2x-y+82 …(ii)
Solving Equ.(i) and(ii), we get
x=8,y=6
∴8α particles and 6β particles are emitted in disintegration.

Topic: Physics, Nuclear Physics, NEET
99% of a first order reaction was completed in 32 min. When will 99.9% of the reaction complete?

Options:
A 24 min
B 8 min
C 4 min
D 48 min

The correct answer is:
48 min

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

Topic: Chemistry, Chemical Kinetics, NEET
1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. The freezing point depression constant of benzene is 5.12 K kg mol–1. Find the molar mass of the solute.

Options:
A 422 g mol-1
B 342 g mol-1
C 256 g mol-1
D 582 g mol-1

The correct answer is:
256 g mol-1

Solution:
Correct Option is : (C)
256 g mol-1

Topic: Chemistry, Solutions, NEET
1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. The freezing point depression constant of benzene is 5.12 K kg mol-1. The molar mass of the solute is

Options:
A 256 kg/mol
B 256 g mol
C 256 g/mol
D 256 mg/mol

The correct answer is:
256 g/mol

Solution:
We have M 2 = K fâ¡ Ã wâ¡ 2 Ã 1 0 0 0 Î T fâ¡ Ã wâ¡ 1
when wâ¡ 1 = weight of solvent = 50 g
wâ¡ 2 = weight of solute = 1.00 g
K fâ¡ = molal depression constant
= 5.12 K kg mol - 1
Substituting the values of various terms involved in the above equation, we get
M 2 = 5.12 K kg mol - 1 Ã 1.00 g Ã 1 0 0 0 g kg - 1 0.40 Ã 5 0 g
=256 g mol −1
Thus, molar mass of the solute = 256 g/âmol

Topic: Chemistry, Solutions, NEET
100 g of liquid A (molar mass 140 g mol–1) was dissolved in 1000 g of liquid B (molar mass 180 g mol–1). The vapour pressure of pure liquid B was found to be 500 torr. Calculate (i) the vapour pressure of pure liquid A and (ii) its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.

Options:
A 280.7 torr, 49 torr
B 364.5 torr, 13 torr
C 280.7 torr, 32 torr
D 364.5 torr, 49 torr

The correct answer is:
280.7 torr, 32 torr

Solution:
Correct Option is : (C)
280.7 torr, 32 torr

Topic: Chemistry, Solutions, NEET
$1.00 \mathrm{~g}$ of non-eletrolyte solute (molar mass $250 \mathrm{~g} \mathrm{~mol}^{-1}$ ) was dissolved in $51.2 \mathrm{~g}$ of benzene. If the freezing point depression constant, $\mathrm{K}_f$ of benzene is 5.12 $\mathrm{K} \mathrm{kg} \mathrm{mol}^{-1}$, the freezing point of benzene will be lowered by:

Options:
A $0.2 \mathrm{~K}$
B $0.4 \mathrm{~K}$
C $0.3 \mathrm{~K}$
D $0.5 \mathrm{~K}$

The correct answer is:
$0.4 \mathrm{~K}$

Solution:
Molality of a non-electrolyte solute
$$
\begin{aligned}
& \frac{\frac{\text { weight of solute in gram }}{\text { molecular weight of solute }}}{\text { weight of solvent in kg }} \\
& =\frac{\frac{1}{250}}{0.0512}=\frac{1}{250 \times 0.0512} \\
& =0.0781 \mathrm{~m} \\
\Delta \mathrm{T}_f & =k_f \times \text { molality of solution } \\
& =5.12 \times 0.0781=0.4 \mathrm{~K}
\end{aligned}
$$

Topic: Chemistry, Solutions, NEET, NEET 2006
100 ml of 5 mâ H 2 SO 4 of density 1 gm/ml is mixed with 100 ml of 8 mâ H 2 SO 4 of density 1.25 g/mL. If there is no change in volume of resulting solution due to mixing, the molarity of the resulting mixture is-

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

The correct answer is:
4.5 M

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

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
100 mL of a solution with pH = 6 is diluted to 1000 mL by adding water. pH will

Options:
A decrease by 0.7 unit
B increase by 1 unit
C increase by 9 units
D increase by 0.7 unit

The correct answer is:
increase by 0.7 unit

Solution:
pH = 6
[H+]=10-6M
100×10-6=M×1000 or M=10-7
[H+] already present in water =10-7
Total [H+]=10-7+10-7=2×10-7
pH=-log2×10-7=7-log2
pH=7-0.3010=6.7

Topic: Chemistry, Ionic Equilibrium, NEET
100 mL of liuqid 'A' was mixed with 25 ml of a liquid 'B' to give a non-ideal solution of A-B mixture showing (-) ve deviation from Raoult's law. The volume of this mixture would be

Options:
A 75 mL
B 125 mL
C close to 125 mL, but not exceeding 125 mL
D just more than 125 mL

The correct answer is:
close to 125 mL, but not exceeding 125 mL

Solution:
When two liquids form a non-ideal solution which exhibits a (-) ve deviation from Raoult's law, the volume change on mixing the two components is always negative, i.e., ΔVmix<0 . Hence, the volume of solution cannot be equal to or greater than 125 mL, but is slightly less than 125 mL.

Topic: Chemistry, Solutions, NEET
108 g of silver (molar mass 108 g mol-1 ) is deposited at cathode from AgNO3 (aq) solution by a certain quantity of electricity. The volume (in L) of oxygen gas produced at 273 Kand 1 bar pressure from water by the same quantity of electricity is

Options:
A 2.82 L
B 5.675 L
C 1.41 L
D 0.746 L

The correct answer is:
5.675 L

Solution:
[nAg]deposit=108108=1 mole
Ag++1e-âAg
1F charge is required for 1 mole of Ag
H2Oâ12O2+2H++2e-
2F charge deposit =12 mole
1F charge deposit =14 mole
PV=nRT
V=14Ã0.0821Ã2731
V=5.675 L

Topic: Chemistry, Electrochemistry, NEET
116 mg of a compound on vapourisation in a Victor Meyer's apparatus displaces 44.8 mL of air measured at STP. The molecular mass of the compound is

Options:
A 58 g/mol
B 0.48 g/mol
C 116 g/mol
D 44.8 g/mol

The correct answer is:
58 g/mol

Solution:
44.8 mL of air displaced by
=116 mg=1161000g
So, 22400 mL of air displaced by
=1161000Ã2240044.8=58 g

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
14.2 g Na2SO4 is present in 102 kg water. Its concentration in ppm is

Options:
A 10 ppm
B 100 ppm
C 142 ppm
D 14.2 ppm

The correct answer is:
142 ppm

Solution:
concentration in ppm =Weight of soluteWeight of solutionÃ106
ppm. of Na2SO4 in solution =14.2Ã106102Ã103=142 .

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET