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NEET • CHEMISTRY • STATES OF MATTER

States of matter practice · Mixed difficulty

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Showing 331-360 of 394 questions

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331 Easy
331 of 394
Understanding +4 | -1
Critical temperature of a gas is the temperature
View answer and explanation
Correct option: B
Above which it cannot be liquefied
Option B is correct because the critical temperature is the temperature above which a gas cannot be liquefied, regardless of pressure.
332 Hard
332 of 394
Application +4 | -1
A closed container containing O2\mathrm{O}_2 and some liquid water was found to exert 740 mm740 \mathrm{~mm} pressure at 27∘C.27^{\circ} \mathrm{C}.

(a) Then calculate the pressure exerted by O2\mathrm{O}_2 if aqueous tension at 27∘C27^{\circ} \mathrm{C} is 20 mm20 \mathrm{~mm}

(b) What will be the final pressure if volume is reduced to half.

(consider volume of liquid water negligible)
View answer and explanation
Correct option: A
(a) 720 mm;720 \mathrm{~mm}; (b) 1460 mm1460 \mathrm{~mm}
Option A ((a) 720 mm;720 \mathrm{~mm}; (b) 1460 mm1460 \mathrm{~mm}) is correct because it accurately calculates the partial pressure of O2 and the final pressure after volume reduction, considering the vapor pressure of water.
333 Medium
333 of 394
Understanding +4 | -1
In which molecule the van der Waals’ force is likely to be the most important in determining the m.p. and b.p.?
View answer and explanation
Correct option: A
Br2\mathrm{B}{\mathrm{r}_2}
Option A (Br2\mathrm{B}{\mathrm{r}_2}) is correct because bromine is a large, nonpolar molecule with a significant number of electrons, leading to strong London dispersion forces (a type of Van der Waals force) that significantly influence its melting and boiling points.
334 Easy
334 of 394
Understanding +4 | -1
At absolute zero:
View answer and explanation
Correct option: D
All of the above
Option D (All of the above) is correct because at absolute zero (0 K or -273°C), all molecular motion ceases, and theoretically, the gaseous phase does not exist as substances would be in their solid or liquid states.
335 Easy
335 of 394
Understanding +4 | -1
Which of the following gases will have highest rate of diffusion
View answer and explanation
Correct option: B
NH3N{H_3}
Option B (NH3N{H_3}) is correct because it has the lowest molar mass among the given gases, and according to Graham's law, the rate of diffusion is inversely proportional to the square root of the molar mass.
336 Hard
336 of 394
Application +4 | -1
The volume of gas A is twice than that of gas B. The compressibility factor of gas A is thrice then that of gas B at same temperature. The pressures of the gases for equal number of moles are
View answer and explanation
Correct option: B
2pA=3pB2{p_A} = 3{p_B}
Option D (3pA=2pB3{p_A} = 2{p_B}) is correct because it is derived from the compressibility factor equation and the given conditions.
337 Medium
337 of 394
Understanding +4 | -1
Which one of the following is correct about surface tension (ST) and viscosity (η)(\eta )?
View answer and explanation
Correct option: A
Both decrease with temperature
Option A (Both decrease with temperature) is correct because increasing temperature weakens intermolecular forces, leading to a decrease in both surface tension and viscosity.
338 Medium
338 of 394
Remembering +4 | -1
The forces present in the crystals of naphthalene are:
View answer and explanation
Correct option: A
Van der Waals’ forces
Option A (Van der Waals’ forces) is correct because naphthalene is a nonpolar molecule, and the primary intermolecular forces holding the crystal together are van der Waals' forces.
339 Hard
339 of 394
Remembering +4 | -1
The temperature at which the second virial coefficient of a real gas is zero is called
View answer and explanation
Correct option: D
Boyle’s temperature
Option D (Boyle’s temperature) is correct because Boyle's temperature is defined as the temperature at which the second virial coefficient is zero.
340 Medium
340 of 394
Understanding +4 | -1
The partial pressure of a dry gas is:
View answer and explanation
Correct option: A
Less than that of wet gas
Option A (Less than that of wet gas) is correct because the presence of water vapor in wet gas reduces the partial pressure of the dry gas components.
341 Medium
341 of 394
Application +4 | -1
A bottle of dry NH3\mathrm{N}{\mathrm{H}_3} and bottle of dry HCl\mathrm{HCl} connected through a long tube are opened simultaneously at both ends, the white (NH4Cl)\left( {\mathrm{N}{\mathrm{H}_4}\mathrm{Cl}} \right) ring first formed will be:
View answer and explanation
Correct option: B
Near the HCl\mathrm{HCl} bottle
Option B (Near the HCl\mathrm{HCl} bottle) is correct because, according to Graham's Law of Diffusion, HCl diffuses slower than NH3 due to its higher molecular weight, so they will meet closer to the HCl source.
342 Easy
342 of 394
Understanding +4 | -1
How does the surface tension of a liquid vary with increase in temperature
View answer and explanation
Correct option: B
Decreases
Option B (Decreases) is correct because as temperature increases, the kinetic energy of molecules increases, reducing the effect of intermolecular forces and thus decreasing surface tension.
343 Hard
343 of 394
Application +4 | -1
Iron crystallises in a bcc system with a lattice parameter of 2.861 Å. Calculate the density of iron in the bcc system (atomic weight of Fe=56,NA=6.02×1023 mol−1{Fe}=56, N_A=6.02 \times 10^{23} \mathrm{~mol}^{-1} )
View answer and explanation
Correct option: A
7.94 g mL−17.94 \mathrm{~g} \mathrm{~mL}^{-1}
Option A (7.94 g mL−17.94 \mathrm{~g} \mathrm{~mL}^{-1}) is correct because it is the density calculated using the bcc structure formula with the given lattice parameter and atomic weight.
344 Medium
344 of 394
Remembering +4 | -1
S.I units of gas constant R is
View answer and explanation
Correct option: C
8.314  J. K−1.mol−18.314\,\,J.\,{K^{ - 1}}.mo{l^{ - 1}}
Option C is correct because the S.I. unit of the gas constant R is Joules per Kelvin per mole.
345 Hard
345 of 394
Application +4 | -1
If for two gases of molecular weights MAM_A and MBM_B at temperature TAT_A and TB,TAMB=TBMAT_B, T_A M_B=T_B M_A, then which property has the same magnitude for both the gases.
View answer and explanation
Correct option: D
Vms\mathrm{V}_{\mathrm{ms}}
Option D (Vrms) is correct because the root mean square velocity depends only on temperature and molar mass, and the given condition ensures they are equal.
346 Hard
346 of 394
Analyzing +4 | -1
When a sample of gas is compressed at constant temperature from 15 atm to 60 atm, its volume changes from 76  cm3  to  20.5  cm376\,\,\mathrm{c}{\mathrm{m}^3}\,\,\mathrm{to}\,\,20.5\,\,\mathrm{c}{\mathrm{m}^3}. Which of the following statements are possible explanations of this behaviour? 1. The gas behaves non-ideally. 2. The gas dimerises. 3. The gas is absorbed into the vessel walls.
View answer and explanation
Correct option: D
1 only1\mathrm{ }\,only
Option D (1 only1\mathrm{ }\,only) is correct because the observed volume change deviates from ideal gas behavior, indicating non-ideality.
347 Medium
347 of 394
Application +4 | -1
Two identical cylinders contain helium at 2.5atm2.5atm and argon at 1atm1atm respectively. if both the gases are filled in one of the cylinders, the pressure would be:
View answer and explanation
Correct option: A
3.5  atm3.5\,\,\mathrm{atm}
Option A is correct because, according to Dalton's Law of Partial Pressures, the total pressure is the sum of the partial pressures of each gas.
348 Medium
348 of 394
Application +4 | -1
Ratio of root mean square speed of O2{O_2} at 200 K and N2{N_2} at 400 K will be
View answer and explanation
Correct option: B
7:4\sqrt 7 :4
Option B (7:4\sqrt 7 :4) is correct because the RMS velocity is proportional to the square root of (T/M). Taking the ratio and simplifying gives the correct answer.
349 Medium
349 of 394
Application +4 | -1
32 g of O2{\mathrm{O}_2} 2 g of H2{H_2} and 28 g of N2{N_2} at STP occupy separately a volume of
View answer and explanation
Correct option: C
22.4L22.4\mathrm{ }L
Option C (22.4L22.4\mathrm{ }L) is correct because one mole of any gas at STP occupies 22.4 L.
350 Easy
350 of 394
Understanding +4 | -1
The gas which has lowest critical temperature is
View answer and explanation
Correct option: A
O2{O_2}
Option C is correct because Helium (He) has the weakest intermolecular forces and thus the lowest critical temperature.
351 Hard
351 of 394
Application +4 | -1
The most probable velocity (in cm/s) of hydrogen molecule at 27 ∘C27\,^\circ C will be
View answer and explanation
Correct option: B
17.8×10417.8 \times {10^4}
Option B (17.8×10417.8 \times {10^4}) is correct because the most probable velocity is calculated using the formula vp=2RTMv_p = \sqrt{\frac{{2RT}}{M}}, and substituting the given values yields this result.
352 Medium
352 of 394
Application +4 | -1
If 1  litre1\,\,\mathrm{litre} of a gas AA at 600 mm and 0.5 litre of gas BB at 800 mm are taken in a 2  litre2\,\,\mathrm{litre} bulb. The resulting pressure is:
View answer and explanation
Correct option: D
500mm500\mathrm{ }mm
Option D is correct because, according to Dalton's Law of Partial Pressures, the total pressure is the sum of the partial pressures. P_total = (P1V1 + P2V2)/V_total = (600*1 + 800*0.5)/2 = 500 mm
353 Medium
353 of 394
Remembering +4 | -1
The term that corrects for the attractive forces present in a real gas in the van der Waals’ equation is
View answer and explanation
Correct option: B
n2a/V2{n^2}a/{V^2}
Option B is correct because the term n2a/V2{n^2}a/{V^2} in the van der Waals equation corrects for the intermolecular attractive forces present in a real gas.
354 Hard
354 of 394
Understanding +4 | -1
For the type of interactions; (I) Covalent bond, (II) van der Waals’ forces, (III) Hydrogen bonding, (IV) Dipole-dipole interaction, which represents the correct order of increasing stability?
View answer and explanation
Correct option: B
(II)<(III)<(IV)<(I)\left( {\mathrm{II}} \right) < \left( {\mathrm{III}} \right) < \left( {\mathrm{IV}} \right) < \left( I \right)
Option B is correct because it accurately represents the increasing order of stability among the given intermolecular forces: van der Waals < Hydrogen bonding < Dipole-dipole < Covalent bond.
355 Medium
355 of 394
Analyzing +4 | -1
Assertion (A): For a fixed amount of gas, the product PV is always constant.

Reason (R): Gases deviate more from ideal behavior at high pressure and low temperatures.
View answer and explanation
Correct option: D
Assertion is incorrect, reason is correct
Option D is correct. Assertion is incorrect because PV is constant only for ideal gases at constant temperature, and the reason is correct as real gases deviate at high pressure and low temperature.
356 Hard
356 of 394
Remembering +4 | -1
The relationship between Pc,Vc{P_c},{V_c} and Tc{T_c} is:
View answer and explanation
Correct option: D
PcVc=38RTc{P_c}{V_c} = \frac{3}{8}R{T_c}
Option D is correct because the relationship between critical pressure, critical volume, and critical temperature is given by PcVc=38RTc{P_c}{V_c} = \frac{3}{8}R{T_c}.
357 Medium
357 of 394
Understanding +4 | -1
Boyle’s law is applicable in:
View answer and explanation
Correct option: C
Isothermal process
Option C (Isothermal process) is correct because Boyle's Law describes the relationship between pressure and volume at a constant temperature.
358 Medium
358 of 394
Analyzing +4 | -1
The ratio a/ba/b (the terms used in van der Waals’ equation) has the unit:
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because the units of a/b can be expressed in atm litre mol⁻¹, atm dm³mol⁻¹, and dyne cm mol⁻¹.
359 Easy
359 of 394
Remembering +4 | -1
Vapour pressure of a liquid depends upon:-
View answer and explanation
Correct option: A
Temperature of liquid
Option A (Temperature of liquid) is correct because the vapor pressure of a liquid is directly dependent on its temperature.
360 Medium
360 of 394
Application +4 | -1
A monoatomic ideal gas undergoes a process in which the ratio of PP to VV at any instant is constant and equal to unity. The molar heat capacity of the gas is
View answer and explanation
Correct option: A
4R2\frac{{4R}}{2}
Option A (4R/2) is correct because the molar heat capacity is calculated using the given condition P/V = 1 and the ideal gas law.
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