JEE Chemistry States Of Matter MCQs Set 25

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Multiple Choice Questions

Question 1: The approximate energy required to break an ionic crystal of the type A+B- into its ions is in the range of:
(a) 10 to 100 kJ/mol
(b) 50 to 150 kJ/mol
(c) 500 to 1000 kJ/mol
(d) 2 to 50 kJ/mol
Show Answer & Explanation

Answer: (c) 500 to 1000 kJ/mol

Explanation:
1. Ionic bonds are very strong and the lattice enthalpy of typical crystals such as NaCl is several hundred kJ/mol.

Teacher's Note:
1. Lattice energies are large.

Question 2: Ion-dipole attractions are present in:
(a) Pure water
(b) Sodium chloride dissolved in water
(c) Benzene
(d) All of these
Show Answer & Explanation

Answer: (b) Sodium chloride dissolved in water

Explanation:
1. Ions of NaCl are hydrated by polar water molecules, which gives ion-dipole attraction.

Teacher's Note:
1. An ion plus a polar molecule means ion-dipole.

Question 3: F2 is a gas but I2 is a solid because:
(a) Larger London forces are present in I2 than in F2
(b) Fewer London forces are present in I2 than in F2
(c) F2 and I2 have the same extent of London forces
(d) I2 has a low bond dissociation energy
Show Answer & Explanation

Answer: (a) Larger London forces are present in I2 than in F2

Explanation:
1. London dispersion forces increase with the number of electrons and polarisability, so they are far stronger in I2.

Teacher's Note:
1. Bigger molecules have stronger dispersion forces.

Question 4: Hydration of ions is an example of:
(a) Ion-dipole interaction
(b) Dipole-dipole interaction
(c) Dipole-induced dipole interaction
(d) Dispersion forces
Show Answer & Explanation

Answer: (a) Ion-dipole interaction

Explanation:
1. Water is polar, and its partial charges are attracted to the ion.

Teacher's Note:
1. Ion-dipole forces decide solubility of salts.

Question 5: The intermolecular forces present in the noble gases are:
(a) Ion-ion
(b) Ion-dipole
(c) Dipole-dipole
(d) Dispersion (London) forces
Show Answer & Explanation

Answer: (d) Dispersion (London) forces

Explanation:
1. Noble gas atoms are non-polar, so only temporary induced dipoles give attraction.

Teacher's Note:
1. Weak forces explain low boiling points.

Question 6: For a fixed mass of a gas at constant temperature, if the volume becomes three times, the pressure becomes:
(a) 3P
(b) P/3
(c) 9P
(d) P/9
Show Answer & Explanation

Answer: (b) P/3

Explanation:
1. Boyle's law gives PV = constant, so P becomes P/3.

Teacher's Note:
1. Pressure is inversely proportional to volume.

Question 7: The pressure and absolute temperature of a certain mass of a gas are both doubled. The new volume is:
(a) V/2
(b) 2V
(c) V
(d) 4V
Show Answer & Explanation

Answer: (c) V

Explanation:
1. V is proportional to T/P, and the factor 2/2 = 1.

Teacher's Note:
1. Volume depends on the ratio T/P.

Question 8: The volume of a gas at 0 degrees C is doubled at constant pressure. The new temperature is:
(a) 273 K
(b) 273 degrees C
(c) 546 degrees C
(d) 127 degrees C
Show Answer & Explanation

Answer: (b) 273 degrees C

Explanation:
1. By Charles's law, V is proportional to T. Doubling V at 273 K gives 546 K, which is 273 degrees C.

Teacher's Note:
1. Always convert to kelvin.

Question 9: A balloon is filled with hydrogen at room temperature. It bursts if the pressure exceeds 0.2 bar. If at 1 bar pressure the gas occupies 2.27 L, the volume up to which the balloon can be expanded is:
(a) 12 L
(b) 11.35 L
(c) 12.35 L
(d) 13 L
Show Answer & Explanation

Answer: (b) 11.35 L

Explanation:
1. P1V1 = P2V2, so V2 = (1 x 2.27)/0.2 = 11.35 L.

Teacher's Note:
1. Boyle's law at constant temperature.

Question 10: Equal amounts of H2 and CH4 are in two vessels of equal volume, the first at 300 K and the second at 600 K. The ratio of pressures P(first) : P(second) is:
(a) 2 : 1
(b) 1 : 1
(c) 1 : 2
(d) 1 : 4
Show Answer & Explanation

Answer: (c) 1 : 2

Explanation:
1. P = nRT/V and n, V are equal, so P is proportional to T. The ratio is 300 : 600 = 1 : 2.

Teacher's Note:
1. At constant n and V, P is proportional to T.

Question 11: A steel vessel of capacity 22.4 L contains 2 g of H2, 8 g of O2 and 22 g of CO2 at 0 degrees C. The total pressure of the mixture is:
(a) 3.5 atm
(b) 5.25 atm
(c) 1.75 atm
(d) 4.0 atm
Show Answer & Explanation

Answer: (c) 1.75 atm

Explanation:
1. n = 1 + 0.25 + 0.5 = 1.75 mol.
2. At 273 K in 22.4 L, one mole exerts 1 atm, so P = 1.75 atm.

Teacher's Note:
1. Add moles and use PV = nRT.

Question 12: The total mass of a gas mixture containing 11.2 L of CO2, 5.6 L of O2, 22.4 L of N2 and 44.8 L of He at 0 degrees C and 760 mm Hg is:
(a) 44 g
(b) 22 g
(c) 33 g
(d) 66 g
Show Answer & Explanation

Answer: (d) 66 g

Explanation:
1. Moles: CO2 0.5 (22 g), O2 0.25 (8 g), N2 1 (28 g), He 2 (8 g).
2. Total mass = 22 + 8 + 28 + 8 = 66 g.

Teacher's Note:
1. Convert litres at STP to moles first.

Question 13: An open vessel at 27 degrees C is heated until 3/5 of the air in it has been expelled. The final temperature (in degrees C) is:
(a) 750
(b) 300
(c) 600
(d) 477
Show Answer & Explanation

Answer: (d) 477

Explanation:
1. At constant P and V, n is proportional to 1/T. Remaining air is 2/5, so T = 300 x 5/2 = 750 K = 477 degrees C.

Teacher's Note:
1. Remaining fraction is inversely proportional to T.

Question 14: The density of CO2 gas at 27 degrees C and 1 atm is about (g/L):
(a) 1.78
(b) 1.52
(c) 1.96
(d) 1.20
Show Answer & Explanation

Answer: (a) 1.78

Explanation:
1. d = PM/RT = (1 x 44)/(0.0821 x 300) = 1.79, which is about 1.78 g/L.

Teacher's Note:
1. Use d = PM/RT.

Question 15: Four one-litre flasks are filled with H2, He, O2 and O3 at the same temperature and pressure. The ratio of the masses of the gases is:
(a) 1 : 1 : 1 : 1
(b) 2 : 1 : 2 : 3
(c) 1 : 2 : 16 : 24
(d) 2 : 1 : 16 : 24
Show Answer & Explanation

Answer: (c) 1 : 2 : 16 : 24

Explanation:
1. Equal volumes contain equal moles, so the masses are in the ratio of molar masses 2 : 4 : 32 : 48 = 1 : 2 : 16 : 24.

Teacher's Note:
1. Avogadro's law.

Question 16: A gas occupies 600 mL at 25 degrees C and 760 mm Hg. At a height where the temperature is 10 degrees C, its volume is 640 mL. The pressure there is:
(a) 760 mm Hg
(b) 690 mm Hg
(c) 800 mm Hg
(d) 676.6 mm Hg
Show Answer & Explanation

Answer: (d) 676.6 mm Hg

Explanation:
1. P2 = P1V1T2/(T1V2) = 760 x 600 x 283/(298 x 640) = 676.6 mm Hg.

Teacher's Note:
1. Combined gas law with T in kelvin.

Question 17: The total number of electrons in 1.4 g of dinitrogen gas is:
(a) \( 4.2164 x 10^{23} \)
(b) \( 4.2154 x 10^{23} \)
(c) \( 4.2144 x 10^{23} \)
(d) \( 4.2174 x 10^{23} \)
Show Answer & Explanation

Answer: (b) \( 4.2154 x 10^{23} \)

Explanation:
1. 1.4 g = 0.05 mol, which is \( 0.05 x 6.022 x 10^{23} \) molecules. Each N2 has 14 electrons.
2. Electrons = \( 0.05 x 6.022 x 10^{23} x 14 = 4.2154 x 10^{23} \).

Teacher's Note:
1. Multiply by 14 electrons per N2.

Question 18: The total pressure of a mixture of 8 g of dioxygen and 4 g of dihydrogen in a vessel of 1 dm3 at 27 degrees C is (R = 0.083 bar dm3 K-1 mol-1):
(a) 56.0 bar
(b) 28.0 bar
(c) 112.0 bar
(d) 14.0 bar
Show Answer & Explanation

Answer: (a) 56.0 bar

Explanation:
1. n = 0.25 + 2 = 2.25 mol.
2. P = nRT/V = 2.25 x 0.083 x 300 / 1 = 56.0 bar.

Teacher's Note:
1. Use the R value given in bar units.

Question 19: Density of a gas is 5.46 g/dm3 at 27 degrees C and 2 bar. Its density at STP (273 K, 1 bar) is:
(a) 2 g/dm3
(b) 5 g/dm3
(c) 4 g/dm3
(d) 3 g/dm3
Show Answer & Explanation

Answer: (d) 3 g/dm3

Explanation:
1. Density is proportional to P/T.
2. d(STP) = 5.46 x (1/2) x (300/273) = 3.0 g/dm3.

Teacher's Note:
1. d = PM/RT.

Question 20: 34.05 mL of phosphorus vapour weighs 0.0625 g at 546 degrees C and 0.1 bar. The molar mass of phosphorus (R = 0.083 bar L K-1 mol-1) is:
(a) 124.8 g/mol
(b) 1247.7 g/mol
(c) 12.5 g/mol
(d) 12477 g/mol
Show Answer & Explanation

Answer: (b) 1247.7 g/mol

Explanation:
1. n = PV/RT = \( (0.1 x 0.03405)/(0.083 x 819) = 5.01 x 10^{-5} \) mol.
2. \( M = 0.0625/5.01 x 10^{-5} = 1247.7 \) g/mol.

Teacher's Note:
1. Convert mL to litres.

Question 21: The pressure of 1 g of an ideal gas A at 27 degrees C is 2 bar. When 2 g of another ideal gas B is added to the same flask at the same temperature, the total pressure is 3 bar. The relation between their molar masses is:
(a) MB = 4 MA
(b) MB = 3 MA
(c) MB = 2 MA
(d) MB = MA
Show Answer & Explanation

Answer: (a) MB = 4 MA

Explanation:
1. B contributes 1 bar and A contributes 2 bar, so n(B) = n(A)/2.
2. 2/MB = (1/MA)/2, so MB = 4 MA.

Teacher's Note:
1. At fixed V and T, pressure is proportional to moles.

Question 22: The mass x mg that exerts a pressure of 1.00 Pa when resting on an area of 1.00 cm2 is about (g = 9.81 m/s2):
(a) 10.300 mg
(b) 10.308 mg
(c) 1030 mg
(d) 10.2 mg
Show Answer & Explanation

Answer: (d) 10.2 mg

Explanation:
1. \( F = P x A = 1 x 10^{-4} N \), so \( m = 10^{-4}/9.81 \) kg = \( 1.02 x 10^{-5} \) kg = 10.2 mg.

Teacher's Note:
1. Pressure = weight/area.

Question 23: Hydrogen diffuses six times faster than a gas X. The molar mass of X is:
(a) 36
(b) 72
(c) 28
(d) 48
Show Answer & Explanation

Answer: (b) 72

Explanation:
1. \( r(H2)/r(X) = \sqrt{M(X)/2} = 6 \), so M(X) = 2 x 36 = 72.

Teacher's Note:
1. Graham's law.

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