JEE Chemistry States Of Matter MCQs Set 07

Multiple Choice Questions (MCQs) for JEE Chemistry: States Of Matter

Review structured MCQ sets for JEE Chemistry States Of Matter. Built according to official JEE guidelines, these downloadable questions support daily revision and core concept reinforcement.

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Multiple Choice Questions

Question 1: There are \( 6.02\times10^{22} \) molecules each of \( \ce{N2} \), \( \ce{O2} \) and \( \ce{H2} \) mixed together. The mass of the mixture in grams is
(a) 6.2
(b) 4.12
(c) 3.09
(d) 7
Show Answer & Explanation

Answer: (a) 6.2

Explanation:
1. Each gas is \( 6.02\times10^{22}/6.02\times10^{23}=0.1 \) mol.
2. Mass = 0.1(28 + 32 + 2) = 6.2 g.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 2: The volume of 4.4 g of \( \ce{CO2} \) at NTP is
(a) 22.4 L
(b) 44.8 L
(c) 2.24 L
(d) 4.48 L
Show Answer & Explanation

Answer: (c) 2.24 L

Explanation:
1. 4.4 g is 4.4/44 = 0.1 mol.
2. Volume = 0.1 \( \times \) 22.4 = 2.24 L.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 3: The energy of an ideal gas depends only on its
(a) pressure
(b) volume
(c) number of moles
(d) temperature
Show Answer & Explanation

Answer: (d) temperature

Explanation:
1. For an ideal gas there are no intermolecular forces, so only kinetic energy exists.
2. The kinetic energy per mole is \( \frac32RT \), which depends only on temperature.

Teacher's Note:
1. Joule's experiment on free expansion showed that the internal energy of an ideal gas does not depend on volume.

Question 4: A bottle of cold drink contains 200 mL of liquid in which \( \ce{CO2} \) is 0.1 molar. If \( \ce{CO2} \) behaves like an ideal gas, the volume of the dissolved \( \ce{CO2} \) at STP is
(a) 0.224 litre
(b) 0.448 litre
(c) 22.4 litre
(d) 2.24 litre
Show Answer & Explanation

Answer: (b) 0.448 litre

Explanation:
1. Moles of \( \ce{CO2} \) = 0.1 \( \times \) 0.2 = 0.02 mol.
2. Volume at STP = 0.02 \( \times \) 22.4 = 0.448 L.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 5: The vapour density of a gas is 11.2. The volume occupied by 11.2 g of this gas at NTP is
(a) 1 L
(b) 11.2 L
(c) 22.4 L
(d) 20 L
Show Answer & Explanation

Answer: (b) 11.2 L

Explanation:
1. Molar mass = 2 \( \times \) vapour density = 22.4 g/mol.
2. 11.2 g is 0.5 mol, which occupies 0.5 \( \times \) 22.4 = 11.2 L.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 6: A pre-weighed vessel was filled with oxygen at NTP and weighed. It was then evacuated, filled with \( \ce{SO2} \) at the same temperature and pressure, and weighed again. The weight of oxygen will be
(a) the same as that of \( \ce{SO2} \)
(b) 1/2 that of \( \ce{SO2} \)
(c) twice that of \( \ce{SO2} \)
(d) one fourth that of \( \ce{SO2} \)
Show Answer & Explanation

Answer: (b) 1/2 that of \( \ce{SO2} \)

Explanation:
1. Equal volumes of gases at the same T and P contain equal moles (Avogadro).
2. Mass ratio = molar mass ratio = 32 : 64 = 1 : 2.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 7: Five grams each of HF, HCl, HBr and HI are taken at the same temperature (87 C) and pressure (750 mm). Which has the least volume?
(a) \( \ce{HF} \)
(b) \( \ce{HCl} \)
(c) \( \ce{HBr} \)
(d) \( \ce{HI} \)
Show Answer & Explanation

Answer: (d) \( \ce{HI} \)

Explanation:
1. At the same T and P, volume is proportional to moles = mass/molar mass.
2. For equal masses, the gas with the largest molar mass has the smallest volume; HI (128) has the largest.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 8: At what pressure will a quantity of gas occupy 60 mL if it occupies 100 mL at 720 mm Hg (temperature constant)?
(a) 700 mm
(b) 800 mm
(c) 100 mm
(d) 1200 mm
Show Answer & Explanation

Answer: (d) 1200 mm

Explanation:
1. Boyle's law: \( P_1V_1=P_2V_2 \).
2. 720 \( \times \) 100 = \( P_2 \) \( \times \) 60, so \( P_2 \) = 1200 mm.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 9: At constant temperature and pressure, which gas will diffuse first, \( \ce{H2} \) or \( \ce{O2} \)?
(a) Hydrogen
(b) Oxygen
(c) Both will diffuse in the same time
(d) None of the above
Show Answer & Explanation

Answer: (a) Hydrogen

Explanation:
1. Rate of diffusion is inversely proportional to the square root of molar mass (Graham's law).
2. Hydrogen is lighter, so it diffuses faster.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 10: When a jar containing a gaseous mixture of equal volumes of \( \ce{CO2} \) and \( \ce{H2} \) is placed in a solution of sodium hydroxide, the solution level will
(a) rise
(b) fall
(c) remain constant
(d) become zero
Show Answer & Explanation

Answer: (a) rise

Explanation:
1. \( \ce{CO2} \) is absorbed by NaOH to form sodium carbonate, so the gas pressure inside falls.
2. The solution is pushed up and the level rises; \( \ce{H2} \) is not absorbed.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 11: At STP, 1 g of \( \ce{CaCO3} \) on decomposition gives \( \ce{CO2} \) of volume
(a) 22.4 litre
(b) 2.24 litre
(c) 0.224 litre
(d) 11.2 litre
Show Answer & Explanation

Answer: (c) 0.224 litre

Explanation:
1. \( \ce{CaCO3 -> CaO + CO2} \).
2. 1 g = 0.01 mol, so \( \ce{CO2} \) = 0.01 mol = 0.01 \( \times \) 22.4 = 0.224 L.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 12: At NTP, the density of a gas whose molecular weight is 45 is
(a) 44.8 g/litre
(b) 11.4 g/litre
(c) 2 g/litre
(d) 3 g/litre
Show Answer & Explanation

Answer: (c) 2 g/litre

Explanation:
1. Density = M/22.4 = 45/22.4 = 2.0 g/L.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 13: The ratio of the rates of diffusion of oxygen and hydrogen is
(a) 1 : 4
(b) 4 : 1
(c) 1 : 8
(d) 8 : 1
Show Answer & Explanation

Answer: (a) 1 : 4

Explanation:
1. Graham's law: \( r_{O_2}/r_{H_2}=\sqrt{2/32}=1/4 \).
2. The ratio is 1 : 4.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 14: The maximum number of molecules is present in
(a) 0.5 g of \( \ce{H2} \) gas
(b) 10 g of \( \ce{O2} \) gas
(c) 15 L of \( \ce{H2} \) gas at STP
(d) 5 L of \( \ce{N2} \) gas at STP
Show Answer & Explanation

Answer: (c) 15 L of \( \ce{H2} \) gas at STP

Explanation:
1. 0.5 g \( \ce{H2} \) = 0.25 mol; 10 g \( \ce{O2} \) = 0.3125 mol.
2. 15 L \( \ce{H2} \) = 15/22.4 = 0.67 mol; 5 L \( \ce{N2} \) = 0.22 mol.
3. The largest number of moles is in 15 L of hydrogen.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 15: At what pressure will a gas that occupies 100 mL at 720 mm occupy 84 mL (constant temperature)?
(a) 736.18 mm
(b) 820.20 mm
(c) 784.15 mm
(d) 857.14 mm
Show Answer & Explanation

Answer: (d) 857.14 mm

Explanation:
1. 720 \( \times \) 100 = \( P_2\times84 \).
2. P2 = 857.14 mm.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 16: Containers A and B hold the same gas. The pressure, volume and temperature of A are all twice those of B. The ratio of the number of molecules of A to B is
(a) 1 : 2
(b) 2
(c) 1 : 4
(d) 4
Show Answer & Explanation

Answer: (b) 2

Explanation:
1. \( n=PV/RT \).
2. For A: \( n_A\propto\frac{2P\times2V}{2T}=\frac{2PV}{T} \), for B: \( n_B\propto PV/T \).
3. So \( n_A/n_B=2 \).

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 17: A mixture of \( \ce{NO2} \) and \( \ce{N2O4} \) has a vapour density of 38.3 at 300 K. The number of moles of \( \ce{NO2} \) in 100 g of the mixture is
(a) 0.043
(b) 4.4
(c) 3.4
(d) 0.437
Show Answer & Explanation

Answer: (d) 0.437

Explanation:
1. Average molar mass = 2 \( \times \) 38.3 = 76.6.
2. Let mole fraction of \( \ce{NO2} \) be x: 46x + 92(1 - x) = 76.6, so x = 0.335.
3. Total moles in 100 g = 100/76.6 = 1.306, so \( \ce{NO2} \) = 0.335 \( \times \) 1.306 = 0.437 mol.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 18: The pressure and temperature of 4 \( dm^3 \) of carbon dioxide are both doubled. The volume of the gas becomes
(a) \( 2\ dm^3 \)
(b) \( 3\ dm^3 \)
(c) \( 4\ dm^3 \)
(d) \( 8\ dm^3 \)
Show Answer & Explanation

Answer: (c) \( 4\ dm^3 \)

Explanation:
1. \( V\propto T/P \).
2. Both T and P doubled, so V is unchanged at 4 \( dm^3 \).

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 19: If the absolute temperature of an ideal gas is doubled and the pressure is halved, the volume of the gas will
(a) remain unchanged
(b) be doubled
(c) become four times
(d) be halved
Show Answer & Explanation

Answer: (c) become four times

Explanation:
1. \( V\propto T/P \) = 2/(1/2) = 4, so the volume becomes four times.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 20: In the reaction \( \ce{C(s) + H2O(l) -> CO(g) + H2(g)} \), the volume of gases produced at STP from 48.0 g of carbon is
(a) 179.2 L
(b) 89.6 L
(c) 44.8 L
(d) 22.4 L
Show Answer & Explanation

Answer: (a) 179.2 L

Explanation:
1. 48 g carbon = 4 mol.
2. 4 mol C gives 4 mol CO + 4 mol \( \ce{H2} \) = 8 mol of gas.
3. Volume = 8 \( \times \) 22.4 = 179.2 L.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 21: A closed vessel contains equal numbers of nitrogen and oxygen molecules at a pressure of P mm. If nitrogen is removed from the system, the pressure will be
(a) \( P \)
(b) \( 2P \)
(c) \( P/2 \)
(d) \( P^2 \)
Show Answer & Explanation

Answer: (c) \( P/2 \)

Explanation:
1. Pressure is proportional to the number of molecules.
2. Removing half the molecules halves the pressure: P/2.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 22: 50 mL of hydrogen diffuses out through a small hole from a vessel in 20 minutes. The time needed for 40 mL of oxygen to diffuse out is
(a) 12 min
(b) 64 min
(c) 8 min
(d) 32 min
Show Answer & Explanation

Answer: (b) 64 min

Explanation:
1. Rate of \( \ce{H2} \) = 50/20 = 2.5 mL/min.
2. Rate of \( \ce{O2} \) = 2.5 \( \times\sqrt{2/32} \) = 0.625 mL/min.
3. Time = 40/0.625 = 64 min.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 23: Under what conditions will a pure sample of an ideal gas exhibit a pressure of 1 atm and also a concentration of 1 mol per litre? (R = 0.082 L atm \( mol^{-1}K^{-1} \))
(a) at STP
(b) when V = 22.4 L
(c) when T = 12 K
(d) impossible under any conditions
Show Answer & Explanation

Answer: (c) when T = 12 K

Explanation:
1. \( P=\frac nV RT \), so T = P/(CR) = 1/(1 \( \times \) 0.082) = 12.2 K.
2. So the temperature must be about 12 K.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 24: Equal masses of two gases of molecular weights 4 and 40 are mixed. The pressure of the mixture is 1.1 atm. The partial pressure of the light gas is
(a) 0.55 atm
(b) 0.11 atm
(c) 1 atm
(d) 0.12 atm
Show Answer & Explanation

Answer: (c) 1 atm

Explanation:
1. For a mass m: moles of light gas = m/4, heavy gas = m/40; ratio 10 : 1.
2. Mole fraction of light gas = 10/11.
3. Partial pressure = 1.1 \( \times \) 10/11 = 1 atm.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 25: A gas has the formula \( (CO)_x \). If its vapour density is 70, the value of x is
(a) 2.5
(b) 3.0
(c) 5.0
(d) 6.0
Show Answer & Explanation

Answer: (c) 5.0

Explanation:
1. Molar mass = 2 \( \times \) 70 = 140.
2. Mass of CO = 28, so x = 140/28 = 5.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

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FAQs

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