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NEET • CHEMISTRY • STATES OF MATTER

States of matter practice · Mixed difficulty

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Showing 361-390 of 394 questions

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361 Medium
361 of 394
Analyzing +4 | -1
Statement - I: At 273 K, hydrogen and helium gases have positive slope at all pressure in PV
vs P graph.

Statement - II: Complete combustion of any substance requires more moles of oxygen than
the moles of substance burnt.
View answer and explanation
Correct option: A
If both Statement-I and Statement-II are true, and Statement-II is the correct
explanation of Statement-I
Option B (If both Statement-I and Statement-II are true but Statement-II is not the correct explanation of Statement-I) is incorrect. Statement I is true because hydrogen and helium have positive deviations from ideal behavior at all pressures at 273 K. Statement II is true, but it doesn't explain Statement I.
362 Medium
362 of 394
Remembering +4 | -1
Which of the following represents total kinetic energy of one mole of gas?
View answer and explanation
Correct option: B
3/2RT3/2RT
Option B (3/2RT3/2RT) is correct because the total kinetic energy of one mole of a gas is given by the formula 3/2 * R * T.
363 Medium
363 of 394
Remembering +4 | -1
Find the number of allotropes which contain discrete molecules held together by London forces:

Diamond, Graphite, Fullerenes, White phosphorus, Rhombic Sulphur, Black Phosphorus.
View answer and explanation
Correct option: A
3
Option A (3) is correct because Fullerenes, White phosphorus, and Rhombic Sulphur contain discrete molecules held together by London forces.
364 Medium
364 of 394
Understanding +4 | -1
Following properties will decrease with increase in temperature except
View answer and explanation
Correct option: D
Vapour pressure
Option D (Vapour pressure) is correct because vapor pressure increases with increasing temperature as more molecules have sufficient energy to escape into the gas phase.
365 Easy
365 of 394
Remembering +4 | -1
The transport of matter in the absence of bulk flow is known as :
View answer and explanation
Correct option: A
Diffusion
Option A (Diffusion) is correct because diffusion is the movement of matter from a region of high concentration to a region of low concentration without bulk flow.
366 Medium
366 of 394
Application +4 | -1
If pressure of a gas contained in a closed vessel is increased by 0.4%0.4\% when heated by 1∘C1^\circ C its initial temperature must be:
View answer and explanation
Correct option: A
250  K250\,\,\mathrm{K}
Option A (250 K250\,\mathrm{K}) is correct because it is calculated using the relationship between pressure and temperature for a gas in a closed vessel.
367 Easy
367 of 394
Application +4 | -1
The partial pressure of hydrogen in a flask containing 2 g H2   and   32 g SO22\,g\,{H_2}\,\,\,{\text{and}}\,\,\,32\,g\,S{O_2} is
View answer and explanation
Correct option: C
2/32/3 of total pressure
Option C (2/32/3 of total pressure) is correct because the partial pressure of a gas is proportional to its mole fraction, and the mole fraction of H2 is 2/3.
368 Easy
368 of 394
Remembering +4 | -1
Normal temperature and pressure (NTP)\mathrm{(NTP}) of gases refers to:
View answer and explanation
Correct option: A
273 K273\,\mathrm{K} and 760\mathrm{760} mm Hg
Option A (273 K and 760 mm Hg) is correct because Normal Temperature and Pressure (NTP) is defined as 273 K (0°C) and 760 mm Hg (1 atm).
369 Easy
369 of 394
Understanding +4 | -1
Which of the following property of water can be used to explain the spherical shape of rain droplets
View answer and explanation
Correct option: B
Surface tension
Option B (Surface tension) is correct because surface tension minimizes the surface area of a liquid, causing rain droplets to form spherical shapes.
370 Medium
370 of 394
Understanding +4 | -1
If volume containing gas is compressed to half, how many moles of gas remained in the vessel?
View answer and explanation
Correct option: C
Same
Option C (Same) is correct because the number of moles of gas remains constant when the volume is compressed, assuming no gas is added or removed.
371 Easy
371 of 394
Understanding +4 | -1
Which is lighter than dry air?
View answer and explanation
Correct option: A
Moist air
Option A (Moist air) is correct because water vapor has a lower molar mass than the average molar mass of dry air.
372 Easy
372 of 394
Understanding +4 | -1
According to kinetic theory of gases for a diatomic molecule
View answer and explanation
Correct option: D
The average translational kinetic energy of the molecule is proportional to the absolute temperature
Option D (The average translational kinetic energy of the molecule is proportional to the absolute temperature) is correct because the average translational kinetic energy of a gas molecule is directly proportional to the absolute temperature, as described by the kinetic theory of gases.
373 Medium
373 of 394
Analyzing +4 | -1
Statement I: When temperature of one mole of an ideal gas is increased from 27∘C27^{\circ} \mathrm{C} to 127∘C127^{\circ} \mathrm{C} under isobaric condition, volume increases by 100 L100 \mathrm{~L}.

Statement II: Under isobaric condition, volume varies directly with temperature (K)(\mathrm{K}).
View answer and explanation
Correct option: D
Statement II is correct but Statement I is incorrect
Option D is correct. Statement II is correct because volume is directly proportional to temperature in Kelvin under isobaric conditions, while Statement I is incorrect because the volume increase is not simply 100L.
374 Medium
374 of 394
Analyzing +4 | -1
For one mole of a van der Waals gas, the compressibility factor Z(=PVRT)Z\left(=\frac{P V}{R T}\right) at a fixed volume will certainly decrease if [Given: "a", "b" are standard parameters for van der Waals gas]
View answer and explanation
Correct option: B
b" decreases and "a" increases at constant temperature
Option B (b\" decreases and \"a\" increases at constant temperature) is correct because a larger 'a' (attraction) and smaller 'b' (repulsion) will decrease the compressibility factor Z.
375 Medium
375 of 394
Remembering +4 | -1
The relationship between coefficient of viscosity of a liquid and temperature can be expressed as
View answer and explanation
Correct option: B
η=AeE/RT\eta = A{e^{E/RT}}
Option B is correct because the relationship between viscosity and temperature is typically expressed as an exponential decay, represented by the Arrhenius-type equation η=AeE/RT\eta = A{e^{E/RT}}.
376 Medium
376 of 394
Application +4 | -1
A perfect gas of a given mass is heated first in a small vessel and then in a large vessel, such that their volume remains unchanged. The P−TP - T curves are:
View answer and explanation
Correct option: D
Linear with different slopes
Option D (Linear with different slopes) is correct because the pressure is directly proportional to temperature (P=kT) at constant volume, and the slope depends on the number of moles and volume.
377 Easy
377 of 394
Understanding +4 | -1
The rate of diffusion of a gas is proportional to
View answer and explanation
Correct option: A
P/dP/\sqrt d
Option A (P/dP/\sqrt d) is correct because, according to Graham's law modified for pressure, the rate of diffusion is directly proportional to the pressure and inversely proportional to the square root of the density.
378 Medium
378 of 394
Application +4 | -1
The rmsrms speed of gas molecules at a temperature 27K27\mathrm{K} and pressure 1.5 bar is 1×104cm/sec⁡1 \times {10^4}\mathrm{cm}/\sec. If both temperature and pressure are raised three times, the rms\mathrm{rms} speed of the gas will be:
View answer and explanation
Correct option: D
≈1×104cm/sec⁡\approx 1 \times {10^4}\mathrm{cm}/\sec
Option D (≈1×104cm/sec⁡\approx 1 \times {10^4}\mathrm{cm}/\sec) is correct because the RMS speed depends on the square root of temperature and is independent of pressure.
379 Easy
379 of 394
Remembering +4 | -1
The gas with the highest critical temperature is
View answer and explanation
Correct option: D
CO2{CO}_2
Option D is correct. CO2 has the highest critical temperature among the given options due to its stronger intermolecular forces.
380 Medium
380 of 394
Application +4 | -1
The numerical value of RTPV\frac{{RT}}{{PV}} for a gas at critical condition is … times of RTPV\frac{{RT}}{{PV}} at normal condition
View answer and explanation
Correct option: C
8/38/3
Option C (8/38/3) is correct because the compressibility factor at the critical point is 3/8, and at normal conditions, it's approximately 1.
381 Hard
381 of 394
Application +4 | -1
A vessel is filled with a mixture of O2\mathrm{O}_2 and N2\mathrm{N}_2. At what ratio of partial pressures will be the mass of gases be identical
View answer and explanation
Correct option: C
P(O2)=0.875P(N2)\mathrm{P}_{\left(\mathrm{O}_2\right)}=0.875 \mathrm{P}_{\left(\mathrm{N}_2\right)}
The correct answer is P(O2)=0.875P(N2)\mathrm{P}_{\left(\mathrm{O}_2\right)}=0.875 \mathrm{P}_{\left(\mathrm{N}_2\right)}. This is derived by setting the masses of oxygen and nitrogen equal and using the ideal gas law to relate pressure and moles.
382 Easy
382 of 394
Understanding +4 | -1
At normal conditions, Dalton's law of partial pressure is not applicable to which gaseous mixture
View answer and explanation
Correct option: A
H2\mathrm{H}_2 and Cl2\mathrm{Cl}_2 mixture
Option A (H2\mathrm{H}_2 and Cl2\mathrm{Cl}_2 mixture) is correct because hydrogen and chlorine react with each other, violating the assumption of non-reactivity in Dalton's law.
383 Easy
383 of 394
Remembering +4 | -1
The temperature at which real gases obey the Boyle’s law over a wide range of pressure is called
View answer and explanation
Correct option: B
Boyle’s temperature
Option B is correct. Boyle's temperature is the temperature at which a real gas obeys Boyle's law over a wide range of pressures.
384 Medium
384 of 394
Remembering +4 | -1
The gases in the liquid form are held together by a weak attraction among the molecules, called as:
View answer and explanation
Correct option: C
Van der Waals’ attraction
Option C (Van der Waals’ attraction) is correct because gases in liquid form are held together by weak Van der Waals' forces.
385 Medium
385 of 394
Understanding +4 | -1
If the rms speed of a gaseous molecule is x  ms−1x\,\,\mathrm{m}{\mathrm{s}^{ - 1}} at a pressure pp atm, then what will be the rms speed at a pressure 2p2p atm and constant temperature
View answer and explanation
Correct option: A
xx
Option A (xx) is correct because the root-mean-square (rms) speed of a gas depends only on temperature and molar mass, and since the temperature is constant, the rms speed remains the same.
386 Hard
386 of 394
Application +4 | -1
Gaseous benzene reacts with hydrogen gas in presence of a nickel catalyst to form gaseous cyclohexane according to the reaction:

C6H6(9)+3H2(9)⟶C6H12(9)\mathrm{C}_6 \mathrm{H}_{6(9)}+3 \mathrm{H}_{2(9)} \longrightarrow \mathrm{C}_6 \mathrm{H}_{12(9)}

A mixture of C6H6\mathrm{C}_6 \mathrm{H}_6 and excess H2\mathrm{H}_2 has a pressure of 60 mm60 \mathrm{~mm} of Hg\mathrm{Hg} in an unknown volume. After the gas has been passed over a nickel catalyst and all the benzene converted to cyclohexane, the pressure of the gas was 30 mm30 \mathrm{~mm} of Hg\mathrm{Hg} in the same volume and temperature. The fraction of C6H5\mathrm{C}_6 \mathrm{H}_5 (by volume) present in the original volume is:
View answer and explanation
Correct option: D
1/61 / 6
Option D (1/6) is correct because it represents the fraction of benzene in the original mixture, calculated from the pressure change after the reaction.
387 Easy
387 of 394
Remembering +4 | -1
The ratio among most probable velocity, mean velocity and RMS velocity is given by
View answer and explanation
Correct option: D
2:8π:3\sqrt 2 :\sqrt {\frac{8}{\pi }} :\sqrt 3
Option D (2:8π:3\sqrt 2 : \sqrt {\frac{8}{\pi }} : \sqrt 3) is correct because it represents the correct ratio of most probable velocity, mean velocity, and RMS velocity.
388 Easy
388 of 394
Application +4 | -1
density of Neon is maximum at
View answer and explanation
Correct option: B
00C, 2 atm{0^0}C,\,2\,atm
Option B is correct because density is directly proportional to pressure and inversely proportional to temperature. Higher pressure and lower temperature result in maximum density.
389 Easy
389 of 394
Remembering +4 | -1
The ratio among most probable velocity, average velocity and root mean square velocity is given by
View answer and explanation
Correct option: D
2   : 8/π :  3\sqrt 2 \,\,\,:\,\sqrt {8/\pi } \,:\,\,\sqrt 3
Option D (2:8/π:3\sqrt 2 : \sqrt {8/\pi} : \sqrt 3) is correct because it represents the correct ratio of most probable velocity, average velocity, and root mean square velocity.
390 Medium
390 of 394
Application +4 | -1
The rates of diffusion of SO2,CO2,PCl3\mathrm{S}{\mathrm{O}_2},\mathrm{C}{\mathrm{O}_2},\mathrm{PC}{\mathrm{l}_3} and SO3\mathrm{S}{\mathrm{O}_3} are in the following order
View answer and explanation
Correct option: D
CO2>SO2>SO3>PCl3\mathrm{C}{\mathrm{O}_2} > S{\mathrm{O}_2} > S{\mathrm{O}_3} > PC{\mathrm{l}_3}
Option D (CO2>SO2>SO3>PCl3\mathrm{C}{\mathrm{O}_2} > S{\mathrm{O}_2} > S{\mathrm{O}_3} > PC{\mathrm{l}_3}) is correct because the rate of diffusion is inversely proportional to the square root of the molar mass.
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