States of Matter — NEET Chemistry MCQs with answers

States of Matter is a Chemistry chapter in the NEET (UG) syllabus. NEET720 has 437 reviewed practice questions on it, each with a quick answer and a step-by-step explanation. The 8 questions below are free and fixed, so you can bookmark this page; the full chapter, plus mistake tracking and spaced revision, is in the app.

Topics covered

Conceptual · States of Matter (Gaseous State) · Ideal gas equation · Boyle's law · Charles's law · Gay-Lussac's law · Dalton's law of partial pressures · Graham's law of diffusion · Kinetic theory of gases · Root mean square velocity · Real gases and van der Waals equation · Van der Waals equation · Combined gas law · Molar mass from ideal gas equation · Compressibility factor · Avogadro's law · Assertion-reason on gas laws · Kinetic energy and temperature · Statement-based on gas laws · Graph interpretation - PV vs P · Number of moles from PVT data · Most probable, average, and rms speed · Effect of temperature on gas volume graph · Vapour pressure and boiling · Critical constants · Numerical on Graham's law with time · Standard temperature and pressure · Surface tension · Viscosity · Numericals on Dalton's law with mass · Liquefaction of gases · Ideal gas equation with unknown gas · Effusion vs diffusion · Numericals combining Boyle's and Charles's law · Molar volume conceptual · Numerical on mixture of gases and total moles · Boyle temperature · Numerical - moles from two states · Assertion-reason on real gas deviation · Barometric/atmospheric pressure units

8 free States of Matter practice questions with answers

Choose an answer in your head before opening it. Each explanation says why the correct option is right and, where relevant, why the tempting wrong option is wrong.

1. Question 1 · hard · Graham's Law of Diffusion

   A gas X takes 40 s and O2 takes 20 s to diffuse through the same volume under identical conditions. The molar mass of X is: (M(O2)=32)

   • A.128 g/mol
   • B.64 g/mol
   • C.16 g/mol
   • D.8 g/mol
   Show answer and explanation

   Answer: A. 128 g/mol

   r(X)/r(O2)=t(O2)/t(X)=20/40=0.5=sqrt(32/Mx); Mx=128

   r(X)/r(O2) = t(O2)/t(X) = 20/40 = 0.5. By Graham's law, 0.5 = sqrt(32/Mx), so 0.25 = 32/Mx, giving Mx = 32/0.25 = 128 g/mol.

   Common mistake: Not squaring the rate ratio before solving for molar mass

2. Question 2 · medium · Dalton's Law of Partial Pressures

   A mixture of 2 mol gas A and 3 mol gas B has a total pressure of 2 atm. The partial pressure of gas B is:

   • A.1.2 atm
   • B.0.8 atm
   • C.1.5 atm
   • D.0.6 atm
   Show answer and explanation

   Answer: A. 1.2 atm

   mole fraction B = 3/5; PB = (3/5)x2 = 1.2 atm

   Total moles = 2+3 = 5. Mole fraction of B = 3/5 = 0.6. Partial pressure of B = mole fraction x total pressure = 0.6x2 = 1.2 atm.

   Common mistake: Using the mole fraction of the wrong component

3. Question 3 · hard · Graham's Law of Diffusion

   He gas effuses through a pinhole in 10 min. Under identical conditions, the time taken by CO2 gas to effuse the same volume is: (M(He)=4, M(CO2)=44)

   • A.33.2 min
   • B.3.02 min
   • C.110 min
   • D.20 min
   Show answer and explanation

   Answer: A. 33.2 min

   t2=t1xsqrt(M2/M1)=10xsqrt(44/4)=10x3.317=33.2 min

   Since rate is inversely proportional to time, t(CO2)/t(He) = sqrt(M(CO2)/M(He)). t(CO2) = 10 x sqrt(44/4) = 10 x sqrt(11) = 10 x 3.3166 = 33.17 min.

   Common mistake: Not taking the square root of the molar mass ratio

4. Question 4 · hard · Graham's Law

   A gas X (M=64) takes 40 s to effuse a certain volume. Under identical conditions, gas Y (M=16) takes how long to effuse the same volume?

   • A.20 s
   • B.80 s
   • C.10 s
   • D.160 s
   Show answer and explanation

   Answer: A. 20 s

   t(Y)/t(X) = sqrt(M(Y)/M(X)) = sqrt(16/64) = 0.5; t(Y) = 0.5x40 = 20 s

   Since rate is inversely proportional to time, t2/t1 = sqrt(M2/M1). t(Y) = t(X) x sqrt(M(Y)/M(X)) = 40 x sqrt(16/64) = 40 x 0.5 = 20 s.

   Common mistake: Not converting rate relationship into time relationship correctly

5. Question 5 · hard · Van der Waals Equation

   In the van der Waals equation (P + an²/V²)(V − nb) = nRT, the term an²/V² accounts for:

   • A.The reduction in pressure due to intermolecular attractive forces
   • B.The finite volume occupied by gas molecules
   • C.The increase in pressure due to molecular collisions with walls
   • D.The kinetic energy of gas molecules
   Show answer and explanation

   Answer: A. The reduction in pressure due to intermolecular attractive forces

   an²/V² corrects the measured pressure upward to account for attraction-caused pressure loss; b corrects volume for molecular size.

   Real gas molecules experience intermolecular attraction, which reduces the pressure they exert on the walls compared to an ideal gas. The term an²/V² is added to the observed pressure to correct for this reduction. The term nb (subtracted from V) corrects for the finite volume of molecules, not an²/V².

   Common mistake: Confusing pressure correction (a) with volume correction (b)

6. Question 6 · hard · Kinetic Theory

   For a gas at a given temperature, the correct order of molecular speeds is:

   • A.most probable speed < average speed < rms speed
   • B.rms speed < average speed < most probable speed
   • C.average speed < most probable speed < rms speed
   • D.All three speeds are equal at any temperature
   Show answer and explanation

   Answer: A. most probable speed < average speed < rms speed

   From Maxwell distribution: vmp : vavg : vrms = 1 : 1.128 : 1.225, so vmp < vavg < vrms.

   From the Maxwell-Boltzmann speed distribution, the ratio vmp : vavg : vrms = √(2RT/M) : √(8RT/πM) : √(3RT/M) ≈ 1 : 1.128 : 1.225. Hence most probable speed is smallest and rms speed is largest.

   Common mistake: Reversing or confusing the order of the three speeds

7. Question 7 · easy · Kinetic Theory

   At the same temperature, the average kinetic energy of 1 mole of He and 1 mole of O2 gas is:

   • A.Equal for both gases, since KE depends only on temperature
   • B.Greater for O2 because it has higher molar mass
   • C.Greater for He because it is lighter and moves faster
   • D.Cannot be determined without knowing pressure
   Show answer and explanation

   Answer: A. Equal for both gases, since KE depends only on temperature

   Average KE per mole = (3/2)RT, independent of the nature/molar mass of the gas.

   According to kinetic theory, average kinetic energy per mole = (3/2)RT, which depends only on absolute temperature, not on the identity or molar mass of the gas. So at the same T, He and O2 have equal average KE (though different rms speeds).

   Common mistake: Confusing kinetic energy with molecular speed dependence on mass

8. Question 8 · hard · Ideal Gas Equation

   Two flasks of equal volume (5 L each) contain gas A at 2 atm and gas B at 3 atm, both at the same temperature. If both gases are mixed into one 5 L flask at the same temperature, the resultant pressure is:

   • A.5 atm
   • B.2.5 atm
   • C.6 atm
   • D.1 atm
   Show answer and explanation

   Answer: A. 5 atm

   Since final volume equals each original volume and T constant, pressures simply add: 2+3=5 atm (Amagat/Dalton-type combination since n∝P at same V,T).

   Since P∝n at fixed V and T, combining moles from two flasks of the same volume V into one flask of volume V at the same T means total pressure = sum of original pressures = 2+3 = 5 atm.

   Common mistake: Averaging pressures instead of adding when final volume equals each original volume

Questions about States of Matter for NEET

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Questions are original NEET720 compositions reviewed for correctness, syllabus fit and option quality. Counts update as the bank grows (437 active practice questions in this chapter today).