JEE Chemistry States Of Matter MCQs Set 22

Answer: (c) C* : C̄ : C = 1 : 1.128 : 1.225

Explanation:
1. \( C\cdot \) = √(2RT/M), C̄ = √(8RT/πM), \( C \) = √(3RT/M).
2. Ratio = √2 : √(8/π) : \( \sqrt{3} = 1.414 \) : 1.596 : \( 1.732 = 1 \) : 1.128 : 1.225.
3. So \( C\cdot \) < C̄ < \( C \) (rms).

Teacher's Note:
1. The order is most probable < average < rms.
2. Divide all three speeds by √(2RT/M) to get the ratio.

Answer: (c) 1 + pb/RT

Explanation:
1. At high pressure \( p \) ≫ \( a/V^{2} \), so \( p(V - b) \) = RT.
2. pV = RT + pb, so \( Z \) = pV/RT = 1 + pb/RT.

Teacher's Note:
1. Repulsive (volume) effects dominate at high pressure, so \( Z \) is greater than 1.
2. At low pressure the attraction term dominates instead.

Answer: (d) a for Cl₂ > a for C₂H₆ but b for Cl₂ < b for C₂H₆

Explanation:
1. a measures the intermolecular forces of attraction; \( b \) measures the finite molecular size.
2. Easier liquefaction needs stronger attraction (larger \( a \)) and a smaller \( b \).
3. For Cl₂: \( a = 6.49 \) atm \( L^{2} \) mol⁻², \( b = 0.056 L \) mol⁻¹; for \( C_{2}H_{6} \): \( a = 5.56, b = 0.064 \).

Teacher's Note:
1. Critical temperature depends on \( a/b \).
2. A large a and small \( b \) raise the critical temperature.

Answer: (d) A₂B₅

Explanation:
1. Atoms of \( A \) per cell = \( 8 \times 1/8 = 1 \).
2. Atoms of \( B \) per cell = \( (6 - 1) \times 1/2 = 5/2 \).
3. Formula = \( A_{1}B_{5}/_{2} = A_{2}B_{5} \).

Teacher's Note:
1. Each face atom contributes 1/2.
2. Clear the fraction by doubling the subscripts.

Answer: (d) directly proportional to the square root of temperature

Explanation:
1. Average, rms and most probable speeds are proportional to \( \sqrt{T/M} \).
2. So molecular velocity is directly proportional to the square root of absolute temperature.

Teacher's Note:
1. All the three kinds of speed have the same \( T \) dependence.
2. The molar mass \( M \) is constant for one gas.

Answer: (b) \( (p_A/p_B)(M_B/M_A)^{1/2} \)

Explanation:
1. Rate of diffusion \( r ∝ p/\sqrt{M} \).
2. \( r_A/r_B = (p_A/p_B) \times \sqrt{M_B/M_A} \).

Teacher's Note:
1. Diffusion rate is directly proportional to pressure and inversely proportional to \( \sqrt{M} \).
2. Graham's law is the special case \( p_A = p_B \).

Answer: (d) 1.27 × 10⁻³ mol

Explanation:
1. Moles in vapour at saturation = pV/RT = \( 3170 \times 10^{-3} / (8.314 \times 300) = 1.27 \times 10^{-3} \) mol.
2. The water added (\( 0.1 g = 5.56 \times 10^{-3} \) mol) is more than this, so the vapour is saturated and the pressure equals the vapour pressure.

Teacher's Note:
1. Check whether all the liquid can evaporate before using the vapour pressure.
2. \( V \) must be in \( m^{3} \) when \( p \) is in Pa.

Answer: (d) 144 pm

Explanation:
1. In an fcc ionic lattice the ions touch along the edge: \( 2(r^{+} + r^{-}) = a \).
2. \( 2(110 + r^{-}) = 508 \), so \( r^{-} = 144 \) pm.

Teacher's Note:
1. Anions form the fcc lattice and cations fit in octahedral voids along the edge.
2. The edge contains two radii of each ion.

Answer: (b) 26% and 32%

Explanation:
1. Packing fraction of ccp = \( \pi/(3\sqrt{2}) = 0.74 \), so free space = 26%.
2. Packing fraction of bcc = \( \pi\sqrt{3}/8 = 0.68 \), so free space = 32%.

Teacher's Note:
1. ccp is the most efficient of the cubic packings.
2. Free space = 100 − packing efficiency.

Answer: (a) X₄Y₃

Explanation:
1. For \( 100 Y \) atoms there are 200 tetrahedral voids.
2. \( X \) atoms = \( 2/3 \times 200 = 400/3 \).
3. \( X \) : \( Y = (400/3) \) : \( 100 = 4 \) : 3, so the formula is \( X_{4}Y_{3} \).

Teacher's Note:
1. ccp has twice as many tetrahedral voids as atoms.
2. Reduce the ratio to whole numbers.

Answer: (d) 16πr³/3

Explanation:
1. An fcc cell has \( 8 \times 1/8 + 6 \times 1/2 = 4 \) atoms.
2. Volume = \( 4 \times (4/3)\pir^{3} = 16\pir^{3}/3 \).

Teacher's Note:
1. Multiply the number of atoms per cell by the volume of one sphere.
2. The packing fraction follows by dividing by \( a^{3} = (2\sqrt{2} r)^{3} \).

Answer: (c) The fraction of the molecules with the most probable speed increases

Explanation:
1. The total number of molecules is fixed, so the area under the curve stays the same.
2. On heating, the curve flattens and broadens and its peak shifts to a higher speed.
3. The peak becomes lower, so the fraction of molecules with the most probable speed decreases, not increases.

Teacher's Note:
1. Higher temperature spreads molecules over a wider range of speeds.
2. The area under a normalised distribution curve is constant.

Answer: (b) AB₃

Explanation:
1. A: \( 8 \times 1/8 = 1 \); \( B \): \( 6 \times 1/2 = 3 \).
2. Formula = AB₃.

Teacher's Note:
1. Corner ions are shared by 8 cells and face ions by 2 cells.
2. This is the structure type of ReO₃-like compounds.

Answer: (c) volume occupied by the molecules

Explanation:
1. The term \( (V - b) \) corrects the volume for the finite size of the molecules.
2. So \( b \) is a measure of the volume occupied by the molecules (about four times their actual volume).

Teacher's Note:
1. a corrects for attraction; \( b \) corrects for molecular volume.
2. Both constants are specific to each gas.

Answer: (c) in a straight line path

Explanation:
1. Ideal gas molecules exert no forces on each other except during collisions.
2. With no net force, a molecule moves with uniform velocity in a straight line between collisions.

Teacher's Note:
1. No force means no acceleration (Newton's first law).
2. A real gas would curve slightly because of attractions.

Answer: (a) 2.57 × 10²¹

Explanation:
1. Molar mass of NaCl = 58.5 g/mol; \( 1.00 g = 1/58.5 \) mol of formula units = \( 6.02 \times 10^{23}/58.5 = 1.03 \times 10^{22} \) units.
2. Each fcc unit cell of NaCl contains 4 formula units.
3. Number of unit cells = \( 1.03 \times 10^{22}/4 = 2.57 \times 10^{21} \).

Teacher's Note:
1. NaCl has the rock salt structure with \( Z = 4 \).
2. Find formula units first, then divide by 4.

Answer: (c) p/RT

Explanation:
1. From pV = nRT, \( n/V \) = p/RT.

Teacher's Note:
1. Moles per litre is the concentration \( n/V \).
2. Rearrange the ideal gas equation.

Answer: (c) 2, 4

Explanation:
1. bcc: \( 8 \times 1/8 + 1 = 2 \) atoms.
2. fcc: \( 8 \times 1/8 + 6 \times 1/2 = 4 \) atoms.

Teacher's Note:
1. \( Z = 2 \) for bcc and \( Z = 4 \) for fcc.
2. Corner atoms count 1/8 and face atoms 1/2.