StreakPeaked· Practice

ExamsJEE MainChemistry › States of Matter

JEE Main Chemistry: States of Matter questions with solutions

112 questions with worked solutions.

Questions

Q1. A sample of gas has a volume of 300 cc at 27°C under a pressure of 620 mm. What will its volume be at 47°C and 640 mm pressure?

  1. 260 cc
  2. 310 cc
  3. 390 cc
  4. 450 cc

Answer: 310 cc

The correct option is derived from applying the combined gas law, which states that the volume of a gas is directly proportional to its temperature and inversely proportional to its pressure. By substituting the initial and final conditions into the equation, the calculations yield a final volume of 310 cc.

Q2. When the temperature of neon is increased from 20°C to 40°C, by what factor does the mean kinetic energy of its atoms change?

  1. 313/293
  2. √313/293
  3. 1/2
  4. 2

Answer: 313/293

Mean kinetic energy = (3/2)kT scales linearly with absolute T, so the factor is 313/293 (40C=313K, 20C=293K), not the square root of it.

Q3. Which gas is expected to show the greatest departure from ideal-gas behavior?

  1. N₂(g)
  2. CH₄(g)
  3. NH₃(g)
  4. H₂(g)

Answer: NH₃(g)

Ammonia (NH₃) exhibits strong hydrogen bonding and has a relatively high molecular weight compared to the other gases listed, leading to significant intermolecular forces that cause it to deviate more from ideal gas behavior.

Q4. Equal amounts (in moles) of hydrogen gas and oxygen gas are kept in a vessel that has a tiny pinhole, allowing both gases to effuse out. By the time one-half of the hydrogen has escaped, what fraction of the oxygen has escaped?

  1. 1/8
  2. 1/4
  3. 3/8
  4. 1/2

Answer: 1/8

Rate ratio = sqrt(32/2) = 4, so when half (0.5) of H2 has escaped, the fraction of O2 escaped is 0.5/4 = 1/8, not 1/4.

Q5. Which of the following statements is incorrect regarding how raising the temperature affects the distribution of molecular speeds in a gas?

  1. The total area under the speed distribution curve stays the same as it was at the lower temperature
  2. The distribution curve spreads out more widely
  3. The proportion of molecules having the most probable speed increases
  4. The most probable speed shifts to a higher value

Answer: The proportion of molecules having the most probable speed increases

As temperature increases, the distribution of molecular speeds broadens, leading to a decrease in the proportion of molecules at the most probable speed, rather than an increase. This is because higher temperatures result in a greater variety of speeds among the molecules.

Q6. As stated by the kinetic theory of gases, how does an ideal gas molecule move during the interval between two consecutive collisions?

  1. Along a zigzag trajectory
  2. Along a straight-line trajectory
  3. With increasing speed due to acceleration
  4. Along a circular trajectory

Answer: Along a straight-line trajectory

According to the kinetic theory of gases, ideal gas molecules move in straight lines between collisions, as they are assumed to have no interactions except during these brief collisions.

Q7. For 1 mole of an ideal gas maintained at a fixed temperature T, what is the nature of the graph between log(P) and log(V), where P denotes pressure and V denotes volume?

  1. A straight line parallel to the x-axis
  2. A straight line having a negative slope
  3. A curve that begins from the origin
  4. A straight line that passes through the origin

Answer: A straight line having a negative slope

The relationship between pressure and volume for an ideal gas at constant temperature is described by Boyle's Law, which states that pressure is inversely proportional to volume. When both pressure and volume are expressed in logarithmic form, this inverse relationship results in a straight line with a negative slope on a log-log graph.

Q8. In the gaseous state, let the most probable speed be represented by C*, the mean speed by C̅, and the root mean square speed by C. For a very large number of molecules, what is the correct ratio among these speeds?

  1. C*: C̅: C = 1.225: 1.128: 1
  2. C*: C̅: C = 1.128: 1.225: 1
  3. C*: C̅: C = 1: 1.128: 1.225
  4. C*: C̅: C = 1: 1.225: 1.128

Answer: C*: C̅: C = 1: 1.128: 1.225

C* : Cbar : Crms = sqrt2 : sqrt(8/pi) : sqrt3 = 1 : 1.128 : 1.225 (most probable smallest, rms largest). The stored ratio is reversed.

Q9. When concentrated hydrochloric acid is left exposed to the atmosphere, it may give off a white fuming cloud. This happens because

  1. oxygen present in air reacts with the released HCl gas, producing a cloud of chlorine gas
  2. HCl gas has a strong attraction for moisture in air, causing tiny droplets of acid solution to form and look like white smoke
  3. because of its strong attraction for water, concentrated hydrochloric acid draws in atmospheric moisture, which condenses into water droplets and forms the cloud
  4. concentrated hydrochloric acid continuously releases strongly smelling HCl gas

Answer: HCl gas has a strong attraction for moisture in air, causing tiny droplets of acid solution to form and look like white smoke

Concentrated HCl gives off HCl gas, which is strongly hygroscopic. The gas attracts water vapour from the air to form tiny droplets of hydrochloric acid that appear as white fumes.

Q10. Which of the following is employed to generate and maintain strong superconducting magnets, which are an essential component of an NMR spectrometer?

  1. Argon
  2. Neon
  3. Radon
  4. Helium

Answer: Helium

Helium is used in superconducting magnets because it remains in a liquid state at very low temperatures, which is necessary to achieve and maintain the superconducting state required for optimal performance in NMR spectrometers.

Q11. What is the numerical value of the universal gas constant R?

  1. 0.082 litre atm
  2. 0.987 cal mol⁻¹ K⁻¹
  3. 8.3 J mol⁻¹ K⁻¹
  4. 83 erg mol⁻¹ K⁻¹

Answer: 8.3 J mol⁻¹ K⁻¹

The universal gas constant R is defined as 8.314 J mol⁻¹ K⁻¹, which is commonly rounded to 8.3 J mol⁻¹ K⁻¹ for simplicity in calculations, making option C the correct choice.

Q12. The kinetic theory of gases is able to explain which of the following gas laws?

  1. Boyle's law alone
  2. Charles' law alone
  3. Avogadro's law alone
  4. All of these laws

Answer: All of these laws

The kinetic theory of gases provides a molecular-level understanding of gas behavior, which encompasses the relationships described by Boyle's law, Charles' law, and Avogadro's law, thus explaining all of these laws collectively.

Q13. As stated by the kinetic theory of gases, how does a molecule of an ideal gas move during the interval between two consecutive collisions?

  1. Along a curved route
  2. Along a straight-line path
  3. With increasing speed due to acceleration
  4. Along a circular path

Answer: Along a straight-line path

According to the kinetic theory of gases, molecules of an ideal gas move in straight lines between collisions, as they are considered to have no interactions except during these brief collisions.

Q14. When the temperature is increased from 20°C to 40°C, by what factor does the average kinetic energy of neon atoms change?

  1. 313/293
  2. √(313/293)
  3. 1/2
  4. 2

Answer: 313/293

The average kinetic energy of gas particles is directly proportional to the absolute temperature in Kelvin. By converting the given temperatures to Kelvin (293 K and 313 K), the ratio of the average kinetic energies is calculated as 313/293.

Q15. In the van der Waals equation of state for a gas, the constant 'b' represents the measure of

  1. the space occupied by the gas molecules
  2. attractive forces between molecules
  3. repulsive forces between molecules
  4. collisions between molecules per unit volume

Answer: the space occupied by the gas molecules

The constant 'b' in the van der Waals equation accounts for the volume occupied by the gas molecules themselves, reflecting the finite size of the particles and thus adjusting the ideal gas behavior to more accurately represent real gases.

Q16. Which of the following statements is incorrect regarding how raising the temperature affects the distribution of molecular speeds in a gas?

  1. The total area under the speed distribution curve stays unchanged compared with that at the lower temperature
  2. The speed distribution spreads out more
  3. The proportion of molecules having the most probable speed increases
  4. The most probable speed shifts to a higher value

Answer: The proportion of molecules having the most probable speed increases

As temperature increases, the distribution of molecular speeds broadens, meaning that while the most probable speed increases, the proportion of molecules at that speed actually decreases. Therefore, option C is incorrect.

Q17. A volume of 10⁻⁴ dm³ of liquid water is placed in a 1.0 dm³ container and kept at 300 K. At equilibrium, how many moles of water will be present in the vapour phase? (Given: vapour pressure of H2O at 300 K = 3170 Pa; R = 8.314 J K⁻¹ mol⁻¹)

  1. 5.56 × 10⁻³ mol
  2. 1.53 × 10⁻² mol
  3. 4.46 × 10⁻² mol
  4. 1.27 × 10⁻³ mol

Answer: 1.27 × 10⁻³ mol

The correct option is derived from applying the ideal gas law to the vapor pressure of water at the given temperature, which allows us to calculate the number of moles of water vapor that can exist in the container at equilibrium. By using the vapor pressure and the volume of the container, we find that the moles of water in the vapor phase is 1.27 × 10⁻³ mol.

Q18. If r, P, and M denote the diffusion rate, pressure, and molecular mass respectively, then the ratio of diffusion rates of two gases A and B, rA/rB, is expressed as:

  1. (P_A/P_B)(M_B/M_A)^(1/2)
  2. (P_A/P_B)^(1/2)(M_B/M_A)
  3. (P_A/P_B)(M_A/M_B)^(1/2)
  4. (P_A/P_B)^(1/2)(M_A/M_B)

Answer: (P_A/P_B)(M_B/M_A)^(1/2)

The correct option is derived from Graham's law of effusion, which states that the rate of diffusion of a gas is inversely proportional to the square root of its molecular mass and directly proportional to its pressure. Therefore, the ratio of diffusion rates incorporates both the pressure ratio and the inverse square root of the molecular mass ratio.

Q19. For a gas, its molecular speed varies with temperature as:

  1. inversely proportional to absolute temperature.
  2. directly proportional to the square of temperature.
  3. directly proportional to the square root of temperature.
  4. inversely proportional to the square root of temperature.

Answer: directly proportional to the square root of temperature.

Molecular speeds (rms, average, most probable) all scale as v ~ sqrt(3RT/M), i.e. directly proportional to the square root of absolute temperature.

Q20. The compressibility factor for a real gas at high pressure is:

  1. 1 + RT/Pb
  2. 1
  3. 1 + Pb/RT
  4. 1 - Pb/RT

Answer: 1 + Pb/RT

At high pressure the molecular-volume term of the van der Waals equation dominates, giving Z = PV/RT = 1 + Pb/RT, i.e. Z > 1.

Q21. For gaseous state, if most probable speed is denoted by C*, average speed by C̄ and mean square speed by C, then for a large number of molecules the ratios of these speeds are:

  1. C*: C̄: C = 1.225: 1.128: 1
  2. C*: C̄: C = 1.128: 1.225: 1
  3. C*: C̄: C = 1: 1.128: 1.225
  4. C*: C̄: C = 1.225: 1: 1.128

Answer: C*: C̄: C = 1: 1.128: 1.225

The correct option reflects the established relationships between the most probable speed, average speed, and mean square speed of gas molecules, where the most probable speed is the lowest, followed by the average speed, and the mean square speed is the highest, consistent with kinetic theory.

Q22. If Z is a compressibility factor, van der Waals equation at low pressure can be written as:

  1. Z = 1 + RT/Pb
  2. Z = 1 - a/VRT
  3. Z = 1 - Pb/RT
  4. Z = 1 + Pb/RT

Answer: Z = 1 - a/VRT

At low pressure the volume is large so b is negligible and the a/V^2 term dominates the deviation, giving Z = 1 - a/(VRT).

Q23. The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is:

  1. London force
  2. hydrogen bond
  3. ion - ion interaction
  4. ion - dipole interaction

Answer: hydrogen bond

Ion-ion energy varies as 1/r, ion-dipole as 1/r^2, London dispersion as 1/r^6, while dipole-dipole (and hydrogen bonding, which is a strong dipole-dipole interaction) varies as 1/r^3. So the interaction with inverse-cube distance dependence is the hydrogen bond.

Q24. Two closed bulbs of equal volume (V) containing an ideal gas initially at pressure p1 and temperature T1 are connected through a narrow tube of negligible volume. The temperature of one of the bulbs is then raised to T2. The final pressure p_f is:

  1. 2p1 (T2/(T1 + T2))
  2. 2p1 (T1T2/(T1 + T2))
  3. p1 (T1T2/(T1 + T2))
  4. 2p1 (T1/(T1 + T2))

Answer: 2p1 (T2/(T1 + T2))

Total moles are conserved: 2 p1 V/(R T1) = pf V/(R T1) + pf V/(R T2). Solving gives pf = 2 p1 T2/(T1 + T2).

Q25. 0.5 moles of gas A and x moles of gas B exert a pressure of 200 Pa in a container of volume 10 m³ at 1000 K. Given R is the gas constant in JK⁻¹ mol⁻¹, x is:

  1. 2R/(4 + R)
  2. 2R/(4 - R)
  3. (4 + R)/(2R)
  4. (4 - R)/(2R)

Answer: (4 - R)/(2R)

PV=nRT gives total moles = (200*10)/(R*1000) = 2/R. With 0.5 + x = 2/R, x = 2/R - 0.5 = (4 - R)/(2R).

Q26. Consider the van der Waals constants, a and b, for the following gases. Gas: Ar, Ne, Kr, Xe a/(atm dm⁶ mol⁻²): 1.3, 0.2, 5.1, 4.1 b/(10⁻² dm³ mol⁻¹): 3.2, 1.7, 1.5, 0.9 Which gas is expected to have the highest critical temperature?

  1. Kr
  2. Ne
  3. Xe
  4. Ar

Answer: Xe

Critical temperature Tc = 8a/(27Rb), so it scales with a/b. Computing a/b: Ar=0.41, Ne=0.12, Kr=3.4, Xe=4.56. Xe has the largest a/b and thus the highest critical temperature.

Q27. A pressure cooker shortens the time needed to cook food because

  1. the boiling point of water used in cooking gets raised
  2. the greater pressure inside the cooker physically crushes the food
  3. cooking is a chemical process that is accelerated by a higher temperature
  4. heat spreads more uniformly throughout the cooker

Answer: the boiling point of water used in cooking gets raised

Inside a sealed cooker the increased pressure raises the boiling point of water above 100 C, so cooking proceeds at a higher temperature and finishes faster.

Q28. Concentrated hydrochloric acid when kept in open air, sometimes produces a cloud of white fumes. The explanation for it is that

  1. oxygen in air reacts with the emitted HCl gas to form a cloud of chlorine gas
  2. strong affinity of HCl gas for moisture in air results in forming of droplets of liquid solution which appears like a cloudy smoke.
  3. due to strong affinity for water, concentrated hydrochloric acid pulls moisture of air towards itself. This moisture forms droplets of water and hence the cloud.
  4. concentrated hydrochloric acid emits strongly smelling HCl gas all the time.

Answer: strong affinity of HCl gas for moisture in air results in forming of droplets of liquid solution which appears like a cloudy smoke.

Concentrated HCl is volatile; the escaping HCl gas has a strong affinity for water vapour in the air and condenses it into tiny droplets of hydrochloric acid solution, which appear as white fumes. No oxidation to chlorine occurs.

Q29. For gaseous state, if most probable speed is denoted by C*, average speed by C̄ and mean square speed by C, then for a large number of molecules the ratios of these speeds are -

  1. C*: C̄: C = 1: 1.128: 1.225
  2. C*: C̄: C = 1: 1.225: 1.128
  3. C*: C̄: C = 1.225: 1.128: 1
  4. C*: C̄: C = 1.128: 1.225: 1

Answer: C*: C̄: C = 1: 1.128: 1.225

The correct option reflects the established relationships between the most probable speed, average speed, and mean square speed of gas molecules, where the most probable speed is the lowest, followed by the average speed, and the mean square speed is the highest, consistent with kinetic theory.

Q30. If Z is a compressibility factor, Vander Waals equation at low pressure can be written as:

  1. Z = 1 − a / VRT
  2. Z = 1 − Pb / RT
  3. Z = 1 + Pb / RT
  4. Z = 1 + RT / Pb

Answer: Z = 1 − a / VRT

The correct option reflects how the compressibility factor Z deviates from ideal gas behavior due to intermolecular forces, specifically accounting for the attractive forces represented by the parameter 'a' in the Van der Waals equation, which becomes significant at low pressures.

Q31. The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is:

  1. Ion - ion interaction
  2. Ion - dipole interaction
  3. London force
  4. Hydrogen bond

Answer: Ion - dipole interaction

Ion-dipole interactions arise from the electrostatic attraction between an ion and a polar molecule, and their strength decreases with the cube of the distance between them, making this relationship distinct from other types of intermolecular forces.

Q32. A mixture of one mole each of H2, He and O2 each are enclosed in a cylinder of volume V at temperature T. If the partial pressure of H2 is 2 atm, the total pressure of the gases in the cylinder is -

  1. 14 atm
  2. 22 atm
  3. 38 atm
  4. 6 atm

Answer: 6 atm

With one mole each, all partial pressures are equal; since p(H2) = 2 atm, total = 3 x 2 = 6 atm.

Q33. An empty LPG cylinder weighs 14.8 kg. When full, it weighs 29.0 kg and shows a pressure of 3.47 atm. In the course of use at ambient temperature, the mass of the cylinder is reduced to 23.0 kg. The final pressure inside of the cylinder is ____ atm. (Nearest integer) (Assume LPG to be an ideal gas)

  1. 1
  2. 2
  3. 3
  4. 4

Answer: 2

The pressure of an ideal gas is directly proportional to its mass when temperature and volume are constant. Since the mass of the LPG in the cylinder decreases from 14.2 kg (29.0 kg - 14.8 kg) to 8.2 kg (23.0 kg - 14.8 kg), the pressure also decreases proportionally, leading to a final pressure of approximately 2 atm.

Q34. Two small drops of mercury each of radius R coalesce to form a single large drop. The ratio of total surface energy before and after the change is:

  1. 2^(1/3): 1
  2. 1: 2^(1/3)
  3. 2: 1
  4. 1: 2

Answer: 2^(1/3): 1

Volume conservation gives R' = 2^(1/3) R. Surface energy is proportional to area (r^2). Before/after = (2*4piR^2)/(4pi*2^(2/3)R^2) = 2/2^(2/3) = 2^(1/3). So the ratio is 2^(1/3):1.

Q35. The water having more dissolved O2 is:

  1. boiling water
  2. water at 80°C
  3. polluted water
  4. water at 4°C

Answer: water at 4°C

Dissolved oxygen increases as water temperature decreases, so water at 4 degC holds more dissolved O2 than warm, boiling, or polluted water.

Q36. The interaction energy of London forces between two particles is proportional to r−x, where r is the distance between the particles. The value of x is: (1) 3 (2) −3 (3) −6 (4) 6

  1. 3
  2. −3
  3. −6
  4. 6

Answer: 6

The instantaneous-dipole induced-dipole (London) interaction energy is proportional to 1/r^6 = r^-6, so the exponent x in r^-x is 6.

Q37. The unit of the van der Waals gas equation parameter 'a' in (P + a n²/V²)(V - nb) = nRT is:

  1. kg m s⁻²
  2. dm³ mol⁻¹
  3. kg m s⁻¹
  4. atm dm⁶ mol⁻²

Answer: atm dm⁶ mol⁻²

The parameter 'a' in the van der Waals equation accounts for the attractive forces between gas molecules, and its units must reflect pressure multiplied by volume squared per mole squared. Therefore, the correct unit is atm dm⁶ mol⁻², which aligns with the dimensional analysis of the equation.

Q38. A rigid nitrogen tank stored inside a laboratory has a pressure of 30 atm at 06:00 am when the temperature is 27°C. At 03:00 pm, when the temperature is 45°C, the pressure in the tank will be ______ atm. [nearest integer]

  1. 32.00
  2. 32.00
  3. 32.00
  4. 32.00

Answer: 32.00

The pressure of a gas is directly proportional to its temperature when the volume is constant, according to Gay-Lussac's law. As the temperature increases from 27°C to 45°C, the pressure also increases, resulting in a calculated pressure of approximately 32 atm.

Q39. Decreasing order of the hydrogen bonding in following forms of water is correctly represented by A. Liquid water B. Ice C. Impure water Choose the correct answer from the options given below:

  1. A > B > C
  2. B > A > C
  3. A = B > C
  4. C > B > A

Answer: B > A > C

Ice has a structured lattice that maximizes hydrogen bonding, making it the strongest in terms of hydrogen bonds. Liquid water has a high degree of hydrogen bonding but is less structured than ice, while impure water has fewer hydrogen bonds due to the presence of solutes, resulting in the weakest hydrogen bonding.

Q40. Arrange the following gases in increasing order of van der Waals constant ‘a’: (A) Ar (B) CH4 (C) H2O (D) C6H6 Choose the correct option from the following.

  1. D, C, B and A
  2. B, C, D and A
  3. C, D, B and A
  4. A, B, C and D

Answer: A, B, C and D

The van der Waals constant 'a' reflects the strength of intermolecular forces; noble gases like Ar have weak interactions, while polar molecules like H2O have stronger forces, leading to a higher 'a' value. Thus, the order from lowest to highest 'a' is Ar, CH4, H2O, and C6H6.

Q41. A big drop is formed by coalescing 1000 small identical drops of water. If E1 be the total surface energy of 1000 small drops of water and E2 be the surface energy of single big drop of water, then E1: E2 is x: 1 where x = _____.

  1. 10
  2. 100
  3. 1
  4. 1000

Answer: 10

The surface energy of a drop is proportional to its surface area, which scales with the square of the radius. When 1000 small drops coalesce into one larger drop, the total surface area decreases, leading to a reduction in surface energy. Specifically, the ratio of the surface energy of the small drops to that of the big drop is 10:1, as the surface area of the big drop is 10 times that of each small drop.

Q42. If the collision frequency of hydrogen molecules in a closed chamber at 27°C is Z, then the collision frequency of the same system at 127° C is:

  1. √3/2 Z
  2. 4/3 Z
  3. 2/√3 Z
  4. 3/4 Z

Answer: 2/√3 Z

The collision frequency of gas molecules is proportional to the square root of the temperature in Kelvin. By converting the temperatures from Celsius to Kelvin and applying the formula, we find that the collision frequency at 127°C is 2/√3 times that at 27°C.

Q43. Ice and water are placed in a closed container at a pressure of 1 atm and temperature 273.15 K. If pressure of the system is increased 2 times, keeping temperature constant, then identify correct observation from following: (1) Volume of system increases. (2) Liquid phase disappears completely. (3) The amount of ice decreases. (4) The solid phase (ice) disappears completely.

  1. (1), (2) and (3)
  2. (2) and (4)
  3. (1) and (4)
  4. (4)

Answer: (4)

Increasing the pressure while keeping the temperature constant will favor the liquid phase over the solid phase in a system containing ice and water. At 273.15 K, the solid phase (ice) will disappear completely as the pressure increases, leading to a complete transition to the liquid phase.

Q44. Given below are two statements: Statement I: When a system containing ice in equilibrium with water (liquid) is heated, heat is absorbed by the system and there is no change in the temperature of the system until whole ice gets melted. Statement II: At melting point of ice, there is absorption of heat in order to overcome intermolecular forces of attraction within the molecules of water in ice and kinetic energy of molecules is not increased at melting point. In the light of the above statements choose the correct answer from the options given below. (1) Both Statement I and Statement II are true (2) Statement I is true but Statement II is false (3) Both Statement I and Statement II is false (4) Statement I is false but Statement II is true

  1. Both Statement I and Statement II are true
  2. Statement I is true but Statement II is false
  3. Both Statement I and Statement II is false
  4. Statement I is false but Statement II is true

Answer: Both Statement I and Statement II are true

Both statements accurately describe the process of melting ice; heat is absorbed without a temperature change until all ice melts, and during melting, energy is used to overcome intermolecular forces, not to increase kinetic energy.

Q45. Given below are two statements: Statement-I: Sublimation is used for the separation and purification of compounds with low melting point. Statement-II: The boiling point of a liquid increases as the external pressure is reduced. In the light of the above statements, choose the correct answer from the options given below:

  1. Statement-I is false but Statement-II is true.
  2. Statement-I is true but Statement-II is false.
  3. Both Statement-I and Statement-II are true.
  4. Both Statement-I and Statement-II are false.

Answer: Both Statement-I and Statement-II are false.

Sublimation is typically used for substances that can transition directly from solid to gas without passing through a liquid phase, often applicable to compounds with low melting points, making Statement-I misleading. Additionally, the boiling point of a liquid actually decreases when external pressure is reduced, contradicting Statement-II.

Q46. A pressure cooker reduces cooking time for food because

  1. boiling point of water involved in cooking is increased
  2. the higher pressure inside the cooker crushes the food material
  3. cooking involves chemical changes helped by a rise in temperature
  4. heat is more evenly distributed in the cooking space

Answer: boiling point of water involved in cooking is increased

Inside a pressure cooker the increased pressure raises the boiling point of water above 100 C, so food cooks at a higher temperature and therefore faster. The crushing/even-distribution options are not the mechanism.

Q47. Concentrated hydrochloric acid when kept in open air sometimes produces a cloud of white fumes. The explanation for it is that

  1. it reacts with moisture in air to form HCl fumes
  2. it oxidizes to chlorine gas
  3. it decomposes to hydrogen and chlorine
  4. it absorbs carbon dioxide from air

Answer: it reacts with moisture in air to form HCl fumes

Concentrated hydrochloric acid releases hydrogen chloride gas, which readily reacts with moisture in the air to form hydrochloric acid mist, resulting in the visible white fumes.

Q48. Which one of the following statements is NOT true about the effect of an increase in temperature on the distribution of molecular speeds in a gas?

  1. The most probable speed increases
  2. The distribution curve becomes broader
  3. The area under the curve remains the same
  4. The peak of the distribution curve shifts to lower speeds

Answer: The peak of the distribution curve shifts to lower speeds

Raising temperature increases the most probable speed and broadens/flattens the curve while the area (total fraction) stays 1. The peak shifts to HIGHER speeds, so the statement that it shifts to lower speeds is the false (not true) one.

Q49. Equal masses of methane and oxygen are mixed in an empty container at 25°C. The fraction of the total pressure exerted by oxygen is

  1. 2/3
  2. 1/3
  3. 1/2
  4. 1/3 × 273/298

Answer: 1/3

For equal mass m, moles of CH4 = m/16 and O2 = m/32. Mole fraction of O2 = (1/32)/(1/16+1/32) = (1/32)/(3/32) = 1/3, so oxygen exerts 1/3 of the total pressure.

Q50. 'a' and 'b' are van der Waals' constants for gases. Chlorine is more easily liquefied than ethane because

  1. a and b for Cl2 < a for C2H6 but b for Cl2 < b for C2H6
  2. a for Cl2 < a for C2H6 but b for Cl2 > b for C2H6
  3. a for Cl2 > a for C2H6 but b for Cl2 < b for C2H6
  4. a and b for Cl2 > a and b for C2H6

Answer: a for Cl2 > a for C2H6 but b for Cl2 < b for C2H6

Chlorine has a higher 'a' constant, indicating stronger intermolecular attractions compared to ethane, which facilitates its liquefaction. Meanwhile, a lower 'b' constant for chlorine suggests it occupies less volume per mole, further aiding in its ability to be liquefied.

⚔️ Practice JEE Main Chemistry free + battle 1v1 →