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NEET • CHEMISTRY • STATES OF MATTER

States of matter practice · Mixed difficulty

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Showing 301-330 of 394 questions

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301 Easy
301 of 394
Application +4 | -1
The ratio between kinetic energies of 16 g of O2{O_2} and 28 g of N2{N_2} respectively at 300 K will be
View answer and explanation
Correct option: B
0.043055556
Option B (0.043055556) is correct because the kinetic energy is directly proportional to the number of moles, and the ratio of moles of O2 to N2 multiplied by their respective molar masses gives the correct ratio.
302 Easy
302 of 394
Understanding +4 | -1
Dalton’s law of partial pressure is not applicable to
View answer and explanation
Correct option: C
NH3+HCl\mathrm{N}{\mathrm{H}_3} + \mathrm{HCl}
Option C (NH3+HCl\mathrm{N}{\mathrm{H}_3} + \mathrm{HCl}) is correct because NH3 and HCl react to form a solid (NH4Cl), thus violating the assumption of non-reacting gases in Dalton's law.
303 Hard
303 of 394
Remembering +4 | -1
The mean free path (λ)\left( \lambda \right) of a gas sample is given by:
View answer and explanation
Correct option: B
λ=12πσ2N\lambda = \frac{1}{{\sqrt 2 \pi {\sigma ^2}N}}
Option B (λ=12πσ2N\lambda = \frac{{1}}{{\sqrt 2 \pi {{\sigma ^2}}N}}) is correct because it represents the correct formula for the mean free path of a gas molecule.
304 Easy
304 of 394
Application +4 | -1
A mixture of N2{\mathrm{N}_\mathrm{2}} and Ar\mathrm{Ar} gases in a cylinder contains 7g\mathrm{7 g}of N2{\mathrm{N}_\mathrm{2}} and 8g\mathrm{8 g} of Ar\mathrm{Ar}. If the total pressure of the mixture of
the gases in the cylinder is 27\mathrm{27} bar, the partial pressure of N2{\mathrm{N}_\mathrm{2}} is :

[Use atomic masses N=14, Ar =40\mathrm{N = 14,}\,\mathrm{Ar}\,\mathrm{ = 40}]
View answer and explanation
Correct option: C
15\mathrm{15} bar
Option C (15\mathrm{15} bar) is correct because the partial pressure of a gas is proportional to its mole fraction in the mixture, and the mole fraction of N2 is 0.555.
305 Easy
305 of 394
Application +4 | -1
Pressure remaining constant, the volume of a given sample of gas at 127 ∘C127\,^\circ C will be doubled at:
View answer and explanation
Correct option: B
527 ∘C527\,^\circ C
Option B is correct because, according to Charles's Law, if the volume doubles at constant pressure, the absolute temperature must also double. Converting to Celsius gives 527°C.
306 Hard
306 of 394
Application +4 | -1
Choose the correct option for the total pressure (in atm.) in a mixture of 4 g O2{O_2} and 2 g H2{H_2} confined in a total volume of one litre at 00C{0^0}C is
[Given R=0.082L  atm  mol−1 K−1, T=273KR = 0.082\mathrm{L}\,\,\mathrm{atm}\,\,\mathrm{mo}{\mathrm{l}^{ - 1}}\,{K^{ - 1}},\,T = 273\mathrm{K}]
View answer and explanation
Correct option: A
25.1825.18
Option A (25.1825.18) is correct because it is calculated by determining the number of moles of each gas, using the ideal gas law to find the partial pressures, and summing them to find the total pressure.
307 Easy
307 of 394
Understanding +4 | -1
Stability of α\alpha-Helix structure of proteins depends upon
View answer and explanation
Correct option: B
H-bonding interaction
Option B (H-bonding interaction) is correct because the alpha-helix structure of proteins is stabilized by hydrogen bonds between the carbonyl oxygen of one amino acid and the amide hydrogen of another amino acid four residues down the chain.
308 Medium
308 of 394
Remembering +4 | -1
The inter molecular interaction that is dependent on the inverse cube of distance between the molecules is
View answer and explanation
Correct option: D
Dipole-dipole interaction
Option D (Dipole-dipole interaction) is correct because dipole-dipole interactions have an interaction energy that is dependent on the inverse cube of the distance between the molecules.
309 Hard
309 of 394
Application +4 | -1
If the ratio of critical temperature to critical pressure for a real gas is mb/R\mathrm{mb} / \mathrm{R}, then what is the integer value of mm ?
View answer and explanation
Correct option: A
8
Option A (8) is correct because the critical temperature and pressure are related to the van der Waals constants a and b.
310 Easy
310 of 394
Remembering +4 | -1
When universal gas constant (R) is divided by Avogadro number (N0)\left( {{N_0}} \right), then the value ofRN0\frac{R}{{{N_0}}} is equivalent to
View answer and explanation
Correct option: B
Boltzmann’s constant
Option B (Boltzmann’s constant) is correct because Boltzmann's constant is defined as the universal gas constant divided by Avogadro's number.
311 Hard
311 of 394
Application +4 | -1
2 g2 \mathrm{~g} of gas A{A} introduced in a evacuated flask at 25∘C25^{\circ} \mathrm{C}. The pressure of the gas is 1 atm1 \mathrm{~atm}. Now 3 g3 \mathrm{~g} of another gas BB is introduced in the same flask so total pressure becomes 1.5 atm1.5 \mathrm{~atm}. The ratio of molecular mass of AA and BB is
View answer and explanation
Correct option: B
13\frac{1}{3}
Option B (13\frac{1}{3}) is correct because the ratio of molecular masses can be found by comparing the partial pressures and masses of the two gases using the ideal gas law.
312 Easy
312 of 394
Remembering +4 | -1
The compressibility factor ‘z’ for the gas is given by
View answer and explanation
Correct option: B
z=PVobsnRTz = \frac{{P{V_{obs}}}}{{nRT}}
Option B is correct because the compressibility factor (z) is defined as the ratio of the actual volume of a real gas to the ideal volume, which is expressed as z = (PVobs)/(nRT).
313 Hard
313 of 394
Understanding +4 | -1
Longest mean free path stands for :
View answer and explanation
Correct option: A
H2\mathrm{H}_2
Option A is correct because Hydrogen (H2H_2) has the smallest molecular size among the given options, leading to the longest mean free path.
314 Medium
314 of 394
Understanding +4 | -1
An ideal gas expands according to PV=PV = constant. On expansion, the temperature of gas:
View answer and explanation
Correct option: C
Will remain constant
Option C (Will remain constant) is correct because the given condition PV=PV = constant implies an isothermal process, where the temperature remains constant.
315 Hard
315 of 394
Analyzing +4 | -1
Consider 1 cm3c{m^3} sample of air at absolute temperature T0{T_0} at sea-level and another 1  cm31\,\,c{m^3} sample of air at a height where the pressure is one third atmosphere. The absolute temperature TT of the sample at the height is :
View answer and explanation
Correct option: D
Cannot be determined in terms of T0{T_0} from the above data
Option D is correct because the temperature at the height depends on the process (adiabatic, isothermal, etc.) the air undergoes as it rises, which is not specified.
316 Medium
316 of 394
Understanding +4 | -1
Which of the following gases can be liquefied most easily?
View answer and explanation
Correct option: D
NH3N{H_3}
Option D is correct because NH3NH_3 has the strongest intermolecular forces (hydrogen bonding) among the given options, making it easiest to liquefy.
317 Medium
317 of 394
Understanding +4 | -1
Different gases at the same temperature have same
View answer and explanation
Correct option: D
Average kinetic energy
Option D (Average kinetic energy) is correct because the average kinetic energy of gas molecules depends only on the temperature, and if the gases are at the same temperature, their average kinetic energies are the same.
318 Easy
318 of 394
Remembering +4 | -1
The speed possessed by majority of gaseous molecules is:
View answer and explanation
Correct option: B
Most probable speed
Option B (Most probable speed) is correct because the most probable speed is the speed possessed by the largest number of molecules in a gas.
319 Easy
319 of 394
Understanding +4 | -1
Dipole – Induced dipole interactions are present in which of the following pairs
View answer and explanation
Correct option: B
HClHCl and He atoms
Option B (HClHCl and He atoms) is correct because HCl is a polar molecule (dipole) and He is a nonpolar atom, leading to a dipole-induced dipole interaction.
320 Hard
320 of 394
Application +4 | -1
The ratio of the rate of diffusion of Helium and methane under identical condition of pressure and temperature will be
View answer and explanation
Correct option: B
2
Option B (2) is correct because the ratio of diffusion rates is the square root of the inverse ratio of molar masses.
321 Easy
321 of 394
Application +4 | -1
The rate of diffusion of hydrogen is about
View answer and explanation
Correct option: B
1.4 times that of helium
Option B (1.4 times that of helium) is correct because the rate of diffusion is inversely proportional to the square root of the molar mass. The molar mass of hydrogen is approximately 2 g/mol, and the molar mass of helium is approximately 4 g/mol. Therefore, the rate of diffusion of hydrogen is approximately 4/2=2≈1.4\sqrt{4/2} = \sqrt{2} \approx 1.4 times that of helium.
322 Easy
322 of 394
Understanding +4 | -1
If PP is pressure and ρ\rho is density of a gas, then PP and ρ\rho are related as
View answer and explanation
Correct option: A
P∝ρP \propto \rho
Option A (P∝ρP \propto \rho) is correct because, according to the ideal gas law, pressure is directly proportional to density when temperature and molar mass are constant.
323 Hard
323 of 394
Analyzing +4 | -1
At 700 K700 \mathrm{~K}, equilibrium is established between molecular deuterium and atomic deuterium as follows.
D2⇌2D\mathrm{D}_2 \rightleftharpoons 2 \mathrm{D}
It was found that, at a pressure of 100 torr and 700 K700 \mathrm{~K} temperature, the equilibrium mixture effuses through a pin hole 1.1 times as fast as helium (under similar temperature and pressure conditions).
The percentage of D2D_2 molecules dissociated into ions is
View answer and explanation
Correct option: A
21
The correct answer is 21. The percentage of D2 molecules dissociated into ions is calculated using Graham's law of effusion and the average molar mass of the mixture.
324 Easy
324 of 394
Application +4 | -1
Hydrogen diffuses six times faster than gas A. The molar mass of gas A is
View answer and explanation
Correct option: A
72 g/mol
Option A (72 g/mol) is correct because Graham's law of diffusion relates diffusion rate to the square root of molar mass.
325 Hard
325 of 394
Understanding +4 | -1
The gases showing heating and cooling effect during Joule-Thomson’s experiment have Joule-Thomson coefficient:
View answer and explanation
Correct option: B
−ve\mathrm{ - ve} and +ve+ \mathrm{ve} respectively
The correct answer is B (−ve\mathrm{ - ve} and +ve+ \mathrm{ve} respectively) because heating occurs when the Joule-Thomson coefficient is negative, and cooling occurs when it is positive.
326 Easy
326 of 394
Remembering +4 | -1
The critical temperature, Boyle’s temperature and inversion temperature respectively are given as
View answer and explanation
Correct option: B
8a27Rb,aRb, 2aRb\frac{{8a}}{{27Rb}},\frac{a}{{Rb}},\,\frac{{2a}}{{Rb}}
Option B is correct. The critical temperature, Boyle's temperature, and inversion temperature are 8a27Rb\frac{{8a}}{{27Rb}}, aRb\frac{a}{{Rb}}, and 2aRb\frac{{2a}}{{Rb}} respectively.
327 Hard
327 of 394
Application +4 | -1
Oxygen gas generated by the decomposition of potassium chlorate is collected. The volume of oxygen collected at 24 ∘C24\,^\circ C and atmospheric pressure of 760 mm Hg is 128 mL. Calculated the mass of oxygen gas obtained. The pressure of the water vapour at 24 ∘C24\,^\circ C is 22.4 mm Hg
View answer and explanation
Correct option: B
0.163 g0.163\mathrm{ }\,g
Option B is correct because, using the ideal gas law and accounting for the water vapor pressure, the mass of oxygen gas is calculated to be approximately 0.163 g.
328 Hard
328 of 394
Understanding +4 | -1
The ratio of Boyle’s temperature and critical temperature for a gas is:
View answer and explanation
Correct option: B
27/827/8
Option B (27/827/8) is correct because the ratio of Boyle's temperature to critical temperature is TB/Tc=(a/Rb)/(8a/27Rb)=27/8T_B/T_c = (a/Rb) / (8a/27Rb) = 27/8.
329 Hard
329 of 394
Analyzing +4 | -1
A mixture of 0.50 mole of H2{\mathrm{H}_\mathrm{2}} and 0.50 mole of SO2\mathrm{S}{\mathrm{O}_\mathrm{2}} is introduced into a 10.00 L container at 250C\mathrm{2}{\mathrm{5}^\mathrm{0}}\mathrm{C}. The container has a pinhole leak. After a period of time, the partial pressure of H2{\mathrm{H}_2} in the remaining mixture
View answer and explanation
Correct option: C
Is less than that of SO2\mathrm{S}{\mathrm{O}_2}
Option C (Is less than that of SO2\mathrm{S}{\mathrm{O}_2}) is correct because H2 diffuses faster than SO2 through the pinhole, leading to a relatively lower partial pressure of H2 in the remaining mixture.
330 Hard
330 of 394
Application +4 | -1
The average molecular speed is greatest in which of the following gas samples?
View answer and explanation
Correct option: D
2.0 mole of HeHe at 140 K
Option D (2.0 mole of HeHe at 140 K) is correct because Helium has the lowest molar mass and, combined with a relatively low temperature, results in the highest average molecular speed.
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