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IIT JEE exam which consists of JEE Main and JEE Advanced is one of the most important entrance exams for engineering aspirants. The exam is held for candidates who are aspiring to pursue a career in the field of engineering and technical studies. Chemistry is important because everything you do is chemistry! Even your body is made of chemicals. Chemical reactions occur when you breathe, eat, or just sit there reading. All matter is made of chemicals, so the importance of chemistry is that it's the study of everything.

Q1. Statement 1: The numerical value of a for H2O is higher than C6H6.
Statement 2: H2O has H-bonding.
•  Statement 1 is True, Statement 2 is True; Statement 2 is correct explanation for Statement 1
•  Statement 1 is True, Statement 2 is True; Statement 2 is not correct explanation for Statement 1
•  Statement 1 is True, Statement 2 is False
•  Statement 1 is False, Statement 2 is True
Q2. Statement 1: Crystalline solids are anisotropic
Statement 2: Crystalline solids are not as closely packed as amorphous solids
•  Statement 1 is True, Statement 2 is True; Statement 2 is correct explanation for Statement 1
•  Statement 1 is True, Statement 2 is True; Statement 2 is not correct explanation for Statement 1
•  Statement 1 is True, Statement 2 is False
•  Statement 1 is False, Statement 2 is True
Q3.  Statement 1: The solid NaCl is a bad conductor of electricity
Statement 2: In solid NaCl there is no velocity of ions
•  Statement 1 is True, Statement 2 is True; Statement 2 is correct explanation for Statement 1
•  Statement 1 is True, Statement 2 is True; Statement 2 is not correct explanation for Statement 1
•  Statement 1 is True, Statement 2 is False
•  Statement 1 is False, Statement 2 is True
Q4.  Sketch shows the plot of Z vs p for a hypothetical gas for one mole at three distinct temperature   Boyle’s temperature is the temperature at which a gas shows ideal behaviour over a pressure range in the low pressure region. Boyle’s temperature (Tb)=a/Rb. If a plot is obtained at temperature below Boyle’s temperature then the curve will show negative deviation in low pressure region and positive deviation in the high pressure region. Near critical temperature, the curve is more likely as CO2 and the temperature above critical temperature curve is more like H2 at 0℃

For 500 K plot value of Z changes from 2 to 2.2 if pressure is varied from 1000 atm to 1200 atm (high pressure) then the value of b/RT will be
•  10-4 atm-1
•  10-3 atm-1
•  10-5 atm-1
•  0.10 atm-1
Q5. The pressure volume relationship for gases helps to explain the mechanics of breathing. When we breathe in, the diaphragm is lowered and the chest wall is expanded, increasing the volume of the chest cavity. Boyle’s law tells us that the pressure inside the cavity must decrease outside air enters the lungs because it is at a higher pressure than the air in the chest cavity. When we breathe out the diaphragm rises and the chest will contract decreasing the volume of chest cavity.

A 15.0 L cylinder of Ar gas is connected to an evacuated 235.0 L tank. If the final pressure is 750 mm Hg. What have been the original gas pressure in the cylinder?
•  76 atm
•  12.56 atm
•  16.45 atm
•  23 atm
Q6. Van der Waals’ equation for calculating the pressure of a non ideal gas is (p+an2/V2)(V-nb)=nRT Van der Waals’ suggested that the pressure exerted by an ideal gas, pideal is related to the experimentally measured pressure, preal by the equation, pideal = preal + an2/V2   Constant 'a' is measure of intermolecular interaction between gaseous molecules that gives rise to non-ideal behaviour depends on how frequently any two molecules approach each other closely. Another correction concerns the volume occupied by the gas molecules. In the ideal gas equation, V represents the volume of the container. However each molecules does occupy a finite although small, intrinsic volume, so the effective volume of the gas becomes (V-nb), where n is the number of moles of the gas and b is a constant.

Which of the following represents a plot of compressibility factor (Z)vs p at room temperature for He?
Q7. In simple cubic lattice, the spheres are packed in the form of a square array by laying down a base of spheres and then piling upon the base other layers in such a way that each sphere is immediately above the other sphere. In this structure, each sphere is in contact with six nearest neighbours. The percentage of occupied volume in this structure can be calculated as follows V=4/3 πr3 or V=4/3 π(a/2)3 The fraction of the total volume occupied by the sphere is ϕ=4/3 π(a/2)3/a3 =π/6=0.5236 or 52.36%

In a simple cubic cell, an atom at the corner contributes to the unit cell
•  1/4 part
•  1/2 part
•  1 part
•  1/8 part
Q8. While dealing with X-ray diffraction, it is more convenient to express higher order reflections in terms of the first order reflection from planes of higher (hkl). For example, a second order reflection from (111) planes may be considered equivalent to the first order reflection from (222) planes. Similarly a third order reflection from (111) planes may be considered as the first order reflection from (333) planes. This fact can be introduced into the Bragg equation nλ=2d sin⁡θ by rewriting it as
λ=2(d/n) sin⁡θ =2dhkl sin⁡θ Where, dhkl is the perpendicular distance between adjacent planes having the indices (hkl)
(hkl) represents
•  Crystal faces
•  Lattice parameter
•  Crystal systems
•  Miller indices
Q9.The figure given below shows three glass chambers that are connected by valves of negligible volume. At the outset of an experiment, the valves are closed and the chambers contain the gases as detailed in the diagram. All the chambers are at the temperature of 300 K and external pressure of 1.0 atm Pext = 1.0 atm
What will be the work done by N2 gas when valve 2 is opened and value 1 remains closed?
•  8.2 L atm
•  -8.2 atm
•  0
•  -3.28 L atm
Q10.  The distribution of the molecular velocities of gas molecules at any temperature T is shown below. (The plot below is known as Maxwell’s distribution of molecular speeds.) Where
v is molecular velocity
n is number of molecules having velocity v
Let us define ∆Nv, which is equal to the number of molecules between the velocity range v and v+∆v, given by
∆Nv = 4πNa3e-bv2 v2 ∆v
Where
N is total number of molecules
a=√M0/2πRT and b=√M0/2RT
R is universal gas constant
T is temperature of the gas
M0 is molecular weight of the gas

SI units of a are
•  M3
•  m-1 s
•  m2 s-2
•  m s2 #### Written by: AUTHORNAME

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BEST NEET COACHING CENTER | BEST IIT JEE COACHING INSTITUTE | BEST NEET, IIT JEE COACHING INSTITUTE: STATES OF MATTER Quiz-16
STATES OF MATTER Quiz-16
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