datavorous/entrance-exam-dataset huggingface dataset:
A solid homogeneous sphere is moving on a rough horizontal surface, partly rolling and partly sliding. During this kind of motion of the sphere

Options:
A total kinetic energy is conserved
B angular momentum of the sphere about the point of contact with the plane is conserved
C only the rotational kinetic energy about the centre of mass is conserved
D angular momentum about centre of mass is conversed

The correct answer is:
angular momentum of the sphere about the point of contact with the plane is conserved

Solution:
Angular momentum about point of contact is conserved because torque due to friction about point of contact will be zero.

Topic: Physics, Rotational Motion, NEET
A solid is formed and it has three types of atoms P, Q and R. P forms a fcc lattice with Q atoms occupying all tetrahedral voids and R atoms occupying half of octahedral voids. The formula of solid is

Options:
A P4QR2
B P4Q2R
C PQ2R4
D P2Q4R

The correct answer is:
P2Q4R

Solution:
P is in f.c.c. lattice
Number of atoms = atom at corner ×  1 8  + atoms at faces ×  1 2
P=8Ã18+6Ã12=4
Total tetrahedral voids =8
As Q are at T.V.
So Q=8
Number of octahedral voids =4
R=12Ã4=2
So, P=4,Q=8,R=2
Hence the formula of solid is P4Q8R2 or P2Q4R

Topic: Chemistry, Solid State, NEET
A solid is formed by two elements P and Q. The element Q forms cubic close packing and atoms of P occupy one third of tetrahedral voids. The formula of the compound is

Options:
A P3Q
B P2Q3
C P3Q2
D PQ3

The correct answer is:
P2Q3

Solution:
Let octahedral voids =x= no. of atoms of Q
So, tetrahedral voids =2x
No. of atoms of P=13×2x=23x
P:Q=23x:x=2:3
Formula =P2Q3

Topic: Chemistry, Solid State, NEET
A solid metal cube of edge length 2 cm is moving in the positive y-direction at a constant speed of 6 m s-1. There is a uniform magnetic field of 0.1 T in the positive z-direction. The potential difference between the two faces of the cube perpendicular to the x-axis is

Options:
A 12 mV
B 1 mV
C 2 mV
D 6 mV

The correct answer is:
12 mV

Solution:
e=Bvl=0.12×10-2×6=12×10-3=12 mV

Topic: Physics, Electromagnetic Induction, NEET
A solid metallic cuboid of homogeneous material having geometrical dimension as shown in the figure will have a ratio of resistances for the sense of current along x, y and z directions respectively, equal to

Options:
A a:b:c
B a:c:b
C a2:b2:c2
D a2:c2:b2

The correct answer is:
a2:c2:b2

Solution:
RX =Ïabc
RY=Ïcab
RZ=Ïbac
RX:RY:RZ= abc:cab:bac
a2:c2:b2 (multiply each ratio by abc)

Topic: Physics, Current Electricity, NEET
A solid sphere and hollow sphere of the same mass and radius are given a spin about their centre of mass and then, they are placed on a rough horizontal surface. The spin angular velocity is the same for both the spheres and it is equal to Ï0. Once the pure rolling starts; let v1 and v2 be the linear speeds of their centres of mass, then

Options:
A v1=v2
B v1>v2
C v1<v2
D Data is insufficient

The correct answer is:
v1<v2

Solution:
From conservation of angular momentum about point of contact :
I Ï 0 = I Ï + mRv
or I Ï 0 = I v R + mRv
or v = I Ï 0 I R + mR
or
or v = Ï 0 1 R + mR I
Now I solid sphere < I hollow
â´ v solid < v hollow
â´ v 1 < v 2

Topic: Physics, Rotational Motion, NEET
A solid sphere has linear velocity v 0 = 4 m/s and angular velocity Ï 0 = 9 rad/s as shown. Ground on which it is moving, is smooth. It collides elastically with a rough wall of coefficient of friction Î¼ . Radius of the sphere is 1 m and mass is 2 kg.

If the sphere after colliding with the wall rolls without slipping in opposite direction, then coefficient of friction Î¼ is

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

The correct answer is:
1 4

Solution:
ω = ν R = 4 1 = 4 rad/s

Let N be perpendicular force between wall and sphere at the time of collision and Δ t the time of impact, then
N Δ t = Δ P = 8 m ( as Δ ν = 8 m/s ) ..... (1)
Further R μ N Δ t = Δ I ω = 2 5 mR 2 5 ..... (2)
As Δ ω = 5 rad/s
Dividing Eq. (2) by Eq. (1), we get
μ = 1 4

Topic: Physics, Rotational Motion, NEET
A solid sphere is in rolling motion. In rolling motion a body possesses translational kinetic energy Kt as well as rotational kinetic energy Kr simultaneously. The ratio Kt:Kt+Kr for the sphere is

Options:
A 10 : 7
B 5 : 7
C 7 : 10
D 2 : 5

The correct answer is:
5 : 7

Solution:
Transnational KE = 1 2 m v 2
Translational KE+ Rotational KE
= 1 2 m v 2 + 1 2 I ω 2
= 1 2 m v 2 + 1 2 ( 2 5 m r 2 ) ω 2
∵ for rolling motion v=rω
= 1 2 m v 2 + 1 2 × 2 5 m r 2 ( v r ) 2 = 7 10 m v 2
So, K t K t + K r = 1 2 m v 2 7 10 m v 2 = 5 7

Topic: Physics, Rotational Motion, NEET, NEET 2018
A solid sphere moves on a horizontal surface, performing pure rolling motion, with speed v0. It collides with a smooth vertical wall elastically. the velocity at which it again performs pure rolling, is

Options:
A 27v0
B 37v0
C 47v0
D 35v0

The correct answer is:
37v0

Solution:
COAM
Mv0R-25Mv0R=75MvR
v=37v0

Topic: Physics, Rotational Motion, NEET
A solid sphere of mass 2 kg is rolling on a frictionless horizontal surface with velocity 6 m s-1. It collides on the free end of an ideal spring whose other end is fixed. The maximum compression produced in the spring will be (Force constant of the spring = 36 N m-1).

Options:
A 14 m
B 2.8 m
C 1.4 m
D 0.7 m

The correct answer is:
2.8 m

Solution:
Kinetic energy of rolling solid sphere
=12mV2+12Iω2
=12mV2+12×25mr2ω2
=12mV2+15mV2
=710mV2
The potential energy of the spring on maximum compression x
=12kx2
∴ 12kx2=710mV2
x2=1410mV2k
=1410×26236
=2.8
x=2.8 m

Topic: Physics, Rotational Motion, NEET
A solid sphere of mass M and radius R is pulled by a force F as shown in figure. If the sphere does not slip over the surface, then frictional force acting on the sphere is

Options:
A 2 F7
B 3 F7
C 2 F5
D Zero

The correct answer is:
3 F7

Solution:
FÃ2R=75 MR2ÃaeR
âae=10 F7 M
â´ F+fs=mÃae
âfs=3 F7

Topic: Physics, Rotational Motion, NEET
A solid sphere of mass, m and radius, R is rotating about its diameter. A solid cylinder of the same mass and same radius is also rotating about its geometrical axis with an angular speed twice that of the sphere. The ratio of their kinetic energies of rotation EsphereEcylinder will be

Options:
A 2:3
B 1:5
C 1:4
D 3:1

The correct answer is:
1:5

Solution:
EsphereEcylinder=1225mR2ω21212mR22ω2=15.

Topic: Physics, Rotational Motion, NEET, NEET 2016 (Phase 2)
A solid sphere of radius R1and volume charge density Ï=Ï0r is enclosed by a hollow sphere of radius R2 with negative surface charge density Ï, such that the total charge in the system is zero, Ï0 is a positive constant and r is the distance from the centre of the sphere. The ratio R2R1 is

Options:
A σ ρ 0
B 2 σ ρ 0
C ρ 0 2 σ
D ρ 0 σ

The correct answer is:
ρ 0 2 σ

Solution:
For solid sphere of radius R1
q1=∫0R14πr2 drρ
=∫0R14πr2d rρ0r
q1=4πR122ρ0
q2=-4πR22σ
q1+q2=0
4πR12ρ02-4πR22σ=0
R1R22=2σρ0
R2R1=ρ02σ

Topic: Physics, Electrostatics, NEET
A solid sphere of radius R has moment of inertia I about its geometrical axis. If it is melted into a disc of radius r and thickness t. If its moment of inertia about the tangential axis (which is perpendicular to plane of the disc), is also equal to I, then the value of r is equal to

Options:
A 215 R
B 25 R
C 315 R
D 315 R

The correct answer is:
215 R

Solution:
25MR2=12Mr2+Mr2
or 25 MR2=32Mr2
∴ r=215R

Topic: Physics, Rotational Motion, NEET
A solid sphere of radius R has moment of inertia I about its geometrical axis. It is melted into a disc of radius r and thickness t. If its moment of inertia about the tangential axis (which is perpendicular to plane of the disc), is also equal to I, then the value of r is equal to

Options:
A 2 1 5 R
B 2 5 R
C 315R
D 315R

The correct answer is:
2 1 5 R

Solution:
2 5 MR 2 = 1 2 Mr 2 + Mr 2
⇒ 2 5 MR 2 = 3 2 Mr 2
⇒ r = 2 1 5 R

Topic: Physics, Rotational Motion, NEET
A solid sphere of radius R is charged uniformly. At what distance from its surface is the electrostatic potential half of the potential at the center

Options:
A R
B R2
C R3
D 2R

The correct answer is:
R3

Solution:
Voutside=12Vcentre
kQR + h=12Ã32KQR
R + h = 4R3
h = R3

Topic: Physics, Electrostatics, NEET
A solid sphere of radius $R$ is placed on a smooth horizontal surface. A horizontal ' $F$ ' is applied at height ' $h$ ' from the lowest point. For the maximum acceleration of the centre of mass, which is correct?

Options:
A $h=R$
B $h=2 R$
C $h=0$
D No relation between $h$ and $R$

The correct answer is:
No relation between $h$ and $R$

Solution:
Smooth surface is given so rolling motion is not possible. Sphere will peerform linear motion.

Topic: Physics, Rotational Motion, NEET, NEET 2002
A solid sphere of uniform density and radius R applies a gravitational force of attraction equal to F1 on a particle placed at P, distance 2 R from the centre O of the sphere. A spherical cavity of the radius R2 is now made in the sphere as shown in the figure.
The sphere with the cavity now applies a gravitational force F2 on the same particle placed at P.
The ratio F2F1 will be

Options:
A 12
B 79
C 3
D 7

The correct answer is:
79

Solution:
Gravitational force due to solid sphere, F1=GMm2R2 where M and m are mass of the solid sphere and particle respectively and R is the radius of the sphere.
The gravitational force on the particle due to sphere with cavity = force due to a solid sphere - force due to sphere creating a cavity, assumed to be present above at that position.
i.e., F2=GMm4R2-GM/8m3R/22=736GMmR2
So F2F1=7GMm36R2GMm4R2=79

Topic: Physics, Gravitation, NEET
A solid, uncharged conducting sphere of radius 3a contains a concentric hollowed spherical cavity of radius a.

A point charge +Q is placed at the center of the spheres. Taking V=0 at r→∞, the potential at position r=2a from the center is-

Options:
A KQa
B KQ2a
C KQ3a
D 2KQ3a

The correct answer is:
KQ3a

Solution:
Final Charge distribution will be as shown

Therefore potential at r=2a is given by
V=KQ2a-KQ2a+KQ3a
V=KQ3a

Topic: Physics, Electrostatics, NEET
A solid XY has NaCl structure. If radius of X+ is 100 pm. What is the radius of Y- ion:

Options:
A 120 pm
B 136.6 to 241.6 pm
C 136.6 pm
D 241.6 pm

The correct answer is:
136.6 to 241.6 pm

Solution:
For NaCl like structure.
The radius ratio should be
⇒0.414→0.732
r+r-=100rmax-=0.414 ;
100rmin-=0.732 ;
∴rmax-=1000.414=241.3 Pm ;
∴ rmin-=1000.732=136.6 Pm ;

Topic: Chemistry, Solid State, NEET
A soluble clotting factor that is made insoluble by thrombin is _________ .

Options:
A prothrombin
B fibrinogen
C fibrin
D calcium ions

The correct answer is:
fibrinogen

Solution:
Coagulation of blood is to prevent loss of excess blood by formation of clot or coagulum. This is a network of threads called fibrins in which dead and damaged formed elements are trapped. The mechanism is as follows: a) An injury or trauma stimulates the platelets to release thromboplastins which activates the process of coagulation, b) Thromboplastins help in formation of the enzyme complex thrombokinase and this begins the cascade process. c) Thrombokinase converts inactive protein prothrombin present in the plasma into thrombin. d) Thrombin converts soluble fibrinogen into insoluble fibrin. This is done with the help of Calcium ions. e) Fibrin forms a network of threads which trap the formed elements and finally forms a clot.

Topic: Biology, Body Fluids and Circulation, NEET
A solute 'A' undergoes association in a medium as
nA⇌An
The degree of association is 'α' , then van't Hoff factor may be give as

Options:
A i=1-α
B i+1+αn
C i=1-α+αn1
D i = 1

The correct answer is:
i=1-α+αn1

Solution:
In case of association, α and i can be related as follows
α=1-i1-1n
So i=1-α+αn1

Topic: Chemistry, Solutions, NEET
A solution containing 10 g per \(\mathrm{dm}^3\) of urea (molecular mass \(=60 \mathrm{~g} \mathrm{~mol}^{-1}\)) is isotonic with a \(5 \%\) solution of a nonvolatile solute. The molecular mass of this nonvolatile solution is:

Options:
A \(250 \mathrm{~g} \mathrm{~mol}^{-1}\)
B \(300 \mathrm{~g} \mathrm{~mol}^{-1}\)
C \(350 \mathrm{~g} \mathrm{~mol}^{-1}\)
D \(200 \mathrm{~g} \mathrm{~mol}^{-1}\)

The correct answer is:
\(350 \mathrm{~g} \mathrm{~mol}^{-1}\)

Solution:
The molar concentration of urea solution
$$
=\frac{10 g / d m^3}{\text { Mol. wt. of urea }}=\frac{10}{60} \mathrm{M}=\frac{1}{6} \mathrm{M}
$$
Molar concentration of $5 \%$ nonvolatile solute
$$
\begin{aligned}
& =\frac{50 \mathrm{~g} / \mathrm{dm}^3}{\text { mol. wt. of non - volatile solute }} \\
& =\frac{50}{\mathrm{~m}} \mathrm{M}
\end{aligned}
$$
Both solutions are isotonic to each other;
therefore, $\quad \frac{1}{6}=\frac{50}{m}$
$$
m=50 \times 6=300 \mathrm{~g} \mathrm{~mol}^{-1}
$$
Caution
$10 \mathrm{~g}$ per $\mathrm{dm}^3$ of urea is isotonic with $5 \%$ solution of a non-volatile solute. Hence, osmosis cannot occur between the two solutions. Thus, their molar concentrations are equal to each other.

Topic: Chemistry, Solutions, NEET, NEET 2006
A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate molar mass of the solute

Options:
A 90 g mol−1
B 72 g mol−1
C 45 g mol−1
D 23 g mol−1

The correct answer is:
23 g mol−1

Solution:
Correct Option is : (D)
23 g mol−1

Topic: Chemistry, Solutions, NEET
A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate vapour pressure of water at 298 K.

Options:
A 3.53 kPa
B 6.62 kPa
C 9.86 kPa
D 12.24 kPa

The correct answer is:
3.53 kPa

Solution:
Correct Option is : (A)
3.53 kPa

Topic: Chemistry, Solutions, NEET
A solution containing 62 g enthylene glycol in 250 g water is cooled to -10°C. If Kf for water is 1.86 K kg mol-1 , the amount of water (in g) separated as ice is

Options:
A 32
B 48
C 16
D 64

The correct answer is:
64

Solution:
ÎTf=kfÃWBMBÃ1000WA
WA(calculated=kfÎTfÃWBMBÃ1000
ÎTf=TfÂ°-Tfs=0-(-10)=10
âWA=1.8610Ã6262Ã1000
= 186 g
Amount of ice separated =250-186=64 g

Topic: Chemistry, Hydrocarbons, NEET
A solution containing 500 g of a protein per litre is isotonic with a solution containing 3.42 g sucrose per litre. The molecular mass of protein is 5×10x , hence x is

Options:
A 2
B 3
C 4
D 5

The correct answer is:
4

Solution:
In case of isotonic solution
π1=π2
C1=C2
w1m1v1protein=w2m2v2sucrose
5005×10x×1=3.42342×1
10010x=1100
10x=104
x=4

Topic: Chemistry, Solutions, NEET
A solution contains 1 g mol. each of p-toluene diazonium chloride and p-nitrophenyl diazonium chloride. To this 1 g mol. of alkaline solution of phenol is added. Predict the major product. Explain your answer.

Options:
A The major product will be p-tolyl phenyl ether.
B The major product will be p-nitrophenyl phenyl ether.
C The major product will be phenolphthalein.
D In alkaline medium, phenol generates phenoxide ion which is more electron rich than phenol and hence more reactive for electrophilic attack. The electrophile in this reaction is aryl diazonium cation. Stronger the electrophile faster is the reaction. p-Nitrophenyl diazonium cation is a stronger electrophile than p-toluene diazonium cation. Therefore, it couples preferentially with phenol.

The correct answer is:
In alkaline medium, phenol generates phenoxide ion which is more electron rich than phenol and hence more reactive for electrophilic attack. The electrophile in this reaction is aryl diazonium cation. Stronger the electrophile faster is the reaction. p-Nitrophenyl diazonium cation is a stronger electrophile than p-toluene diazonium cation. Therefore, it couples preferentially with phenol.

Solution:
Correct Option is : (D)
In alkaline medium, phenol generates phenoxide ion which is more electron rich than phenol and hence more reactive for electrophilic attack. The electrophile in this reaction is aryl diazonium cation. Stronger the electrophile faster is the reaction. p-Nitrophenyl diazonium cation is a stronger electrophile than p-toluene diazonium cation. Therefore, it couples preferentially with phenol.

Topic: Chemistry, Amines, NEET
A solution contains 410.3 g of H2SO4 per litre at 20 °C . If its density be 1.243 g ml-1 , what will be its molarity and molality respectively.

Options:
A 4.187 M, 5.03 m
B 41.87 M, 50.3 m
C 0.4187 M, 0.503
D 14.87 M, 50.3 m

The correct answer is:
4.187 M, 5.03 m

Solution:
Molarity =Wt.ingramMolecularweight×V(lit)
=41.0398=4.187M
Density = Molarity 1Molality+Molecularweight1000
1.243 = 4.187 1Molality+981000
On solving
Molality = 5.03 m.

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
A solution contains non-volatile solute of molecular mass $M_2$. Which of the following can be used to calculate the molecular mass of the solute in terms of the osmotic pressure?
Note: $m_2$ is the mass of the solute, $\mathrm{V}$ is the volume of the solution and $\pi$ is the osmotic pressure.

Options:
A $\mathrm{M}_2=\left(\frac{m_2}{\pi}\right) \mathrm{VRT}$
B $\mathrm{M}_2=\left(\frac{m_2}{\mathrm{~V}}\right) \frac{\mathrm{RT}}{\pi}$
C $\mathrm{M}_2=\left(\frac{m_2}{\mathrm{~V}}\right) \pi \mathrm{RT}$
D $\mathrm{M}_2=\left(\frac{m_2}{\mathrm{~V}}\right) \frac{\pi}{\mathrm{RT}}$

The correct answer is:
$\mathrm{M}_2=\left(\frac{m_2}{\mathrm{~V}}\right) \frac{\mathrm{RT}}{\pi}$

Solution:
For dilute solution, the van't Hoff equation is:
$\begin{aligned}
\pi & =\frac{n}{\mathrm{~V}} \mathrm{RT} \\
\pi \mathrm{V} & =\frac{m_2}{\mathrm{M}} \mathrm{RT} \\
\Rightarrow \quad \mathrm{M} & =\frac{m_2 \mathrm{RT}}{\pi \mathrm{V}}
\end{aligned}$
Van't Hoff explained that when solutes are dissolved in a solvent they dissociate into ions. Since colligative properties depend only on the number of solute particles, the dissociation of solute molecules into ions results in an increase in the number of particles and hence affects the colligative properties.

Topic: Chemistry, Solutions, NEET, NEET 2002
A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour pressures of the pure hydrocarbons at $20^{\circ} \mathrm{C}$ are 440 $\mathrm{mm} \mathrm{Hg}$ for pentane and $120 \mathrm{~mm} \mathrm{Hg}$ for hexane. the mole fraction of pentane in the vapour phase would be:

Options:
A 0.200
B 0.549
C 0.786
D 0.478

The correct answer is:
0.478

Solution:
$$
\begin{aligned}
\frac{n_{\mathrm{C}_5 \mathrm{H}_{12}}}{n_{\mathrm{C}_6 \mathrm{H}_{14}}} & =\frac{1}{4} \\
\Rightarrow \mathrm{X}_{\mathrm{C}_3 \mathrm{H}_{12}} & =\frac{1}{5} \text { and } \mathrm{X}_{\mathrm{C}_6 \mathrm{H}_{14}}=\frac{4}{5} \\
\mathrm{P}_{\mathrm{C}_5 \mathrm{H}_{12}}^{\circ} & =440 \mathrm{~mm} \mathrm{Hg} ; \\
\mathrm{P}_{\mathrm{C}_6 \mathrm{H}_{14}}^{\circ} & =120 \mathrm{~mm} \mathrm{Hg} \\
\mathrm{P}_{\mathrm{T}} & =\mathrm{P}_{\mathrm{C}_5 \mathrm{H}_{12}}^{\circ} \mathrm{X}_{\mathrm{C}_5 \mathrm{H}_{12}} \\
& +\mathrm{P}_{\mathrm{C}_6 \mathrm{H}_{14}}^{\circ} \mathrm{X}_{\mathrm{C}_6 \mathrm{H}_{14}} \\
& =440 \times \frac{1}{5}+120 \times \frac{4}{5} \\
& =184 \mathrm{~mm} \text { of } \mathrm{Hg}
\end{aligned}
$$
By Raoult's Law,
$$
\mathrm{P}_{\mathrm{C}_5 \mathrm{H}_{12}}=\mathrm{P}_{\mathrm{C}_5 \mathrm{H}_{12}}^{\circ} \mathrm{X}_{\mathrm{C}_5 \mathrm{H}_{12}}
$$
By Dalton's Law,
$$
\mathrm{P}_{\mathrm{C}_5 \mathrm{H}_{12}}=\mathrm{X}_{\mathrm{C}_5 \mathrm{H}_{12}}^{\prime} \mathrm{P}
$$
From (1) and (2),
$$
\begin{aligned}
\mathrm{P}_{\mathrm{C}_5 \mathrm{H}_{12}} & =440 \times \frac{1}{5}=88 \mathrm{~mm} \text { of } \mathrm{Hg} \\
\Rightarrow \quad 88 & =\mathrm{X}_{\mathrm{C}_5 \mathrm{H}_{12}}^{\prime} \times 184 \\
\mathrm{X}^{\prime} & =\frac{88}{184} ; \mathrm{X}^{\prime}=0.478 .
\end{aligned}
$$

Topic: Chemistry, Solutions, NEET, NEET 2005
A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate the mass percentage of solute in the resulting solution.

Options:
A 63.12%
B 52.47%
C 33.57%
D 45.85%

The correct answer is:
33.57%

Solution:
Correct Option is : (C)
33.57%

Topic: Chemistry, Solutions, NEET
A solution is prepared by adding 2 g of a substance A to 18 g of water. What is the mass per cent of the solute here?

Options:
A 20%
B 15%
C 10%
D 5%

The correct answer is:
10%

Solution:
Correct Option is : (C)
10%

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
A solution is prepared by mixing 8.5 g of CH2Cl2 and 11.95 g of CHCl3 . If vapour pressure of CH2Cl2 and CHCl3 at 298 K are 415 and 200 mm Hg respectively, the mole fraction of CHCl3 in vapour form is: Molar mass of Cl=35.5 g mol-1

Options:
A 0.162
B 0.675
C 0.325
D 0.486

The correct answer is:
0.325

Solution:
mole of CH 2 C l 2 âin liquid phase= 8.5 85 =.1
mole of CHCl 3 âin liquid phase= 11.95 119.5 =.1
mole fraction of CH 2 Cl 2 âin liquid phase= .1 .2 = 1 2
mole fraction of CHCl 3 âin liquid phase= .1 .2 = 1 2
P T = X C H 2 C l 2 Ãââ vapour pressure C H 2 C l 2 +â X CHC l 3 Ãâ vapour pressure CHC l 3
=415Ã 1 2 +200Ã 1 2 =307.5
P T ( X A ) VP = ( X A ) LP Ãâ Vapour pressure of CHCl 3
307.5Ãâ ( X CHC l 3 ) VP =200Ã 1 2
X CHC l 3 = 100 307.5 =.325

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
A solution of (-)-1-chloro-1-phenylethane in toluene racemises slowly in the presence of a small amount of SbCl5 , due to the formation of :

Options:
A Carbocation
B Free radical
C Carbanion
D Carbene

The correct answer is:
Carbocation

Solution:
(-) - 1 - chloro - 1 - phenyl ethane
with SbCl5 carbocation is formed. As the carbocation in planar, Cl â can again attack from both sides giving racemic mixture. (d, l formed)

Topic: Chemistry, General Organic Chemistry, NEET
A solution of 200 mL of 1 M KOH is added to 200 mL of 1 M HCl and the mixture is well shaken. This rise in temperature T1 is noted. The experiment is repeated by using 100 mL of each solution and increase in temperature T2 is again noted. Which of the following is correct?

Options:
A T1=T2
B T2 is twice as large as T1
C T1 is twice as large as T2
D T1 is four times as large as T2

The correct answer is:
T1=T2

Solution:
Heat produced by 200 Meq. of KOH + HCl
=13.7×2001000kcal
This is used to raise the temperature of 400 mL. Solution say T1
Thus, 13.7×2001000=400×S×T1
Now, heat produced by 100 Meq of KOH + HCl
=13.7×1001000
Thus, -13.7×1001000=200×S×T2
It is evident that T1=T2 .

Topic: Chemistry, Thermodynamics (C), NEET
A solution of acetone in ethanol:

Options:
A obeys Raoult's law
B shows a negative deviation from Raoult's law
C shows a positive deviation from Raoult's law
D behaves like a near ideal solution

The correct answer is:
shows a positive deviation from Raoult's law

Solution:
A solution of acetone in ethanol shows a positive deviation from Raoult's law because of the miscibility of these two liquids with a difference of polarity and the length of a hydrocarbon chain.
Related Theory
In an ideal solution, the solvent-solute interaction is the same as a solvent-solvent or solute - solute interaction. This implies both the solute and the solvent takes the same amount of energy to escape to the vapour phase as when they are in their pure states.
Caution
Raoult's law is apt for describing ideal solutions. Different liquids that are in the mixture do not have the same uniformity in terms of attractive forces, this type of solutions tends to deviate away from the law.

Topic: Chemistry, Solutions, NEET, NEET 2006
A solution of acetone in ethanol

Options:
A obeys Raoult's law
B shows a negative deviation from Raoult's law
C shows a positive deviation from Raoult's law
D behaves like an ideal solution

The correct answer is:
shows a positive deviation from Raoult's law

Solution:
In case of positive deviation from Raoultâs law, the intermolecular attractive forces between the solute-solvent molecules are weaker than those between the solute-solute and solvent-solvent molecules.
Mixtures of ethanol and acetone behave in this manner. In pure ethanol, molecules are hydrogen bonded. On adding acetone, its molecules get in between the host molecules and break some of the hydrogen bonds between them. Due to weakening of interactions, the solution shows positive deviation from Raoultâs law.

Topic: Chemistry, Solutions, NEET