solid state 12th cet mht
Solid State MCQ Practice
Chemistry | MHT-CET Previous year questions with detailed solutions
MHT-CET 2015
Q1. Select a ferromagnetic material from the following.
Correct Answer
(B) Chromium(IV) oxide
Detailed Explanation:
- Ferromagnetism: These are substances that show permanent magnetism even when an external magnetic field is removed. This happens because the magnetic moments (domains) align spontaneously in the same direction.
- While common metals like Iron (Fe), Cobalt (Co), and Nickel (Ni) are ferromagnetic, Chromium(IV) oxide (CrO2) is a key example of a ferromagnetic metallic oxide.
- Due to its high coercivity and remanence, it was the standard material used for high-fidelity magnetic recording tapes.
MHT-CET 2016
Q2. Which among the following solids is a non-polar solid?
Correct Answer
(D) Carbon dioxide (CO2)
Detailed Explanation:
- In solid CO2 (dry ice), the molecule is linear (O=C=O).
- The individual C-O bond dipoles are equal and opposite, cancelling each other out, resulting in a net dipole moment of zero.
- HCl, SO2, and H2O are all polar molecules due to their geometry and electronegativity differences.
MHT-CET 2016
Q3. Which metal crystallizes in a simple cubic structure?
Correct Answer
(A) Polonium
Detailed Explanation:
- The simple cubic structure is very rare for metals because it is an inefficient way to pack atoms (only 52.4% space occupied).
- Polonium is the only metal known to crystallize in this form under standard conditions.
- Copper and Nickel are FCC (Face-Centered Cubic), and Iron is typically BCC (Body-Centered Cubic) at room temperature.
MHT-CET 2016
Q4. In face-centred cubic unit cell, what is the volume occupied?
Correct Answer
(C) \(\frac{16}{3}\pi r^3\)
Detailed Explanation & Calculation:
- In an FCC unit cell, the total number of atoms (Z) is 4.
- The volume of one atom (treated as a sphere) is \(\frac{4}{3}\pi r^3\).
- Total volume occupied = \(Z \times \text{Volume of one sphere}\)
- = \(4 \times \frac{4}{3}\pi r^3 = \frac{16}{3}\pi r^3\)
MHT-CET 2019
Q5. How many total spheres of constituent particles are present in bcc type of unit cell?
Correct Answer
(A) 2
Detailed Explanation & Calculation:
- Corners: 8 corners × \(\frac{1}{8}\) (contribution per corner) = 1 atom
- Body center: 1 body center × 1 (full contribution) = 1 atom
- Total = 1 + 1 = 2 atoms
MHT-CET 2019
Q6. Which among the following defects is observed in Brass?
Correct Answer
(B) Substitution impurity
Detailed Explanation:
- Brass is an alloy consisting of Copper (Cu) and Zinc (Zn).
- In this alloy, Zinc atoms occupy the lattice sites normally held by Copper atoms.
- Since one type of atom is substituted by another of similar size, it is a substitution impurity defect.
- This type of defect occurs when atoms of the original metal are replaced by atoms of a different element.
MHT-CET 2020
Q7. What is the number of atoms present per unit cell of aluminium having edge length 4 Å? (If density of Al = 2.7 g cm⁻³, At. Mass of Al = 27)
Correct Answer
(C) 4
Detailed Explanation & Calculation:
- Using the density formula: \(\rho = \frac{Z \times M}{a^3 \times N_A}\)
- Rearranging: \(Z = \frac{\rho \times a^3 \times N_A}{M}\)
- Given values:
- \(\rho = 2.7 \text{ g/cm}^3\)
- \(a = 4 \text{ Å} = 4 \times 10^{-8} \text{ cm}\)
- \(a^3 = 64 \times 10^{-24} \text{ cm}^3\)
- \(M = 27 \text{ g/mol}\)
- \(N_A = 6.022 \times 10^{23}\)
- \(Z = \frac{2.7 \times 64 \times 10^{-24} \times 6.022 \times 10^{23}}{27}\)
- \(Z = \frac{2.7 \times 64 \times 6.022 \times 10^{-1}}{27}\)
- \(Z = \frac{1040.6 \times 0.1}{27} = \frac{104.06}{27} \approx 3.85 \approx \mathbf{4}\)
- Note: Aluminium is an FCC metal, which confirms \(Z = 4\).
MHT-CET 2020
Q8. Lithium crystallises into body centered cubic structure. What is the radius of lithium if edge length of its unit cell is 351 pm?
Correct Answer
(B) 151.98 pm
Detailed Explanation & Calculation:
- For BCC structure, atoms touch along the body diagonal of the cube.
- The relationship between radius (\(r\)) and edge length (\(a\)) is:
- Given: \(a = 351 \text{ pm}\)
- \(\sqrt{3} = 1.732\)
- \(r = \frac{1.732}{4} \times 351\)
- \(r = 0.433 \times 351\)
- \(r = \mathbf{151.98 \text{ pm}}\)
\[4r = \sqrt{3}a \quad \Rightarrow \quad r = \frac{\sqrt{3}}{4}a\]
MHT-CET 2020
Q9. The edge length of fcc type unit cell of copper having atomic radius 127.6 pm is equal to
Correct Answer
(A) 361 pm
Detailed Explanation & Calculation:
- For FCC structure, atoms touch along the face diagonal of the cube.
- The relationship between radius (\(r\)) and edge length (\(a\)) is:
- Given: \(r = 127.6 \text{ pm}\)
- \(2\sqrt{2} = 2 \times 1.414 = 2.828\)
- \(a = 2.828 \times 127.6\)
- \(a = \mathbf{360.8 \text{ pm} \approx 361 \text{ pm}}\)
\[4r = \sqrt{2}a \quad \Rightarrow \quad a = 2\sqrt{2}r\]
MHT-CET
Q10. The mass of a unit cell of a face-centred cubic (FCC) lattice is (given mass of one atom = 6 × 10⁻²³ g)
Correct Answer
(D) 2.4 × 10⁻²² g
Formula: Mass of unit cell = Z × mass of one atom
Given: Mass of one atom = 6 × 10⁻²³ g; for FCC, Z = 4
Calculation: 4 × (6 × 10⁻²³) = 24 × 10⁻²³ g = 2.4 × 10⁻²² g
Given: Mass of one atom = 6 × 10⁻²³ g; for FCC, Z = 4
Calculation: 4 × (6 × 10⁻²³) = 24 × 10⁻²³ g = 2.4 × 10⁻²² g
MHT-CET
Q11. An element having BCC geometry has density 4 g cm⁻³ and edge length 500 pm. Find the atomic mass of the element.
Correct Answer
(B) 150.0
Formula: \(\rho = \frac{Z \times M}{a^3 \times N_A} \implies M = \frac{\rho \times a^3 \times N_A}{Z}\)
Given: BCC (Z = 2), ρ = 4 g cm⁻³, a = 500 pm = 5 × 10⁻⁸ cm
Calculation: \[M = \frac{4 \times (5 \times 10^{-8})^3 \times 6.022 \times 10^{23}}{2}\] \[M = \frac{4 \times 125 \times 10^{-24} \times 6.022 \times 10^{23}}{2} = 250 \times 0.6022 \approx \mathbf{150.0}\]
Given: BCC (Z = 2), ρ = 4 g cm⁻³, a = 500 pm = 5 × 10⁻⁸ cm
Calculation: \[M = \frac{4 \times (5 \times 10^{-8})^3 \times 6.022 \times 10^{23}}{2}\] \[M = \frac{4 \times 125 \times 10^{-24} \times 6.022 \times 10^{23}}{2} = 250 \times 0.6022 \approx \mathbf{150.0}\]
MHT-CET
Q12. Dry ice is an example of which type of solid?
Correct Answer
(D) Molecular solid
Reasoning: Dry ice is solid carbon dioxide (CO₂). It consists of non-polar CO₂ molecules held together by weak London dispersion forces. Therefore, it is classified as a molecular solid.
MHT-CET
Q13. Which of the following is a ferromagnetic substance?
Correct Answer
(A) Gadolinium
Reasoning: Ferromagnetism is the strongest form of magnetism. Common examples include Iron (Fe), Cobalt (Co), Nickel (Ni), and Gadolinium (Gd).
MHT-CET
Q14. If the number of unit cells in a sample of a BCC metal is 12.08 × 10²³, then the total number of atoms present in the sample is
Correct Answer
(B) 2.416 × 10²⁴
Given: Number of unit cells = 12.08 × 10²³; for BCC, Z = 2 atoms per cell
Calculation: Total atoms = 2 × (12.08 × 10²³) = 24.16 × 10²³ = 2.416 × 10²⁴
Calculation: Total atoms = 2 × (12.08 × 10²³) = 24.16 × 10²³ = 2.416 × 10²⁴
MHT-CET
Q15. A simple cubic crystal has edge length 5 Å. The radius of the atom in the crystal is
Correct Answer
(D) 250.0 pm
Formula: For Simple Cubic, a = 2r ⇒ r = a/2
Given: a = 5 Å = 500 pm
Calculation: r = 500 / 2 = 250.0 pm
Given: a = 5 Å = 500 pm
Calculation: r = 500 / 2 = 250.0 pm
MHT-CET
Q16. If the radius of an atom in a simple cubic crystal is 3 nm, then the edge length of the unit cell is
Correct Answer
(B) 6 × 10⁻⁹ m
Formula: a = 2r
Given: r = 3 nm = 3 × 10⁻⁹ m
Calculation: a = 2 × 3 nm = 6 nm = 6 × 10⁻⁹ m
Given: r = 3 nm = 3 × 10⁻⁹ m
Calculation: a = 2 × 3 nm = 6 nm = 6 × 10⁻⁹ m
MHT-CET
Q17. Which of the following is NOT an amorphous solid?
Correct Answer
(B) Camphor
Reasoning: Tar, butter, and rubber are amorphous (lack a regular repeating lattice). Camphor is a crystalline solid with a defined molecular arrangement.
MHT-CET
Q18. A metal having atomic mass 40 g/mol crystallizes in BCC structure. The number of unit cells in 4 g of the metal is
Correct Answer
(C) \(\frac{0.1 \times N_A}{2}\)
Given: Mass = 4 g, Atomic mass (M) = 40
Step 1: Moles = \(\frac{\text{Mass}}{M} = \frac{4}{40} = 0.1\) mole
Step 2: Atoms = \(0.1 \times N_A\)
Step 3: For BCC, 2 atoms make 1 unit cell
Calculation: Unit cells = \(\frac{\text{Atoms}}{2} = \mathbf{\frac{0.1 \times N_A}{2}}\)
Step 1: Moles = \(\frac{\text{Mass}}{M} = \frac{4}{40} = 0.1\) mole
Step 2: Atoms = \(0.1 \times N_A\)
Step 3: For BCC, 2 atoms make 1 unit cell
Calculation: Unit cells = \(\frac{\text{Atoms}}{2} = \mathbf{\frac{0.1 \times N_A}{2}}\)
MHT-CET
Q19. The mass of a unit cell of a FCC crystal is 419 × 10⁻²⁴ g. The mass of one atom is
Correct Answer
(C) 1.048 × 10⁻²² g atom⁻¹
Given: Mass of unit cell = 419 × 10⁻²⁴ g; for FCC, Z = 4
Calculation: Mass of one atom = \(\frac{\text{Mass of unit cell}}{4} = \frac{419 \times 10^{-24}}{4} = 104.75 \times 10^{-24}\) g
Shift decimal: 1.0475 × 10⁻²² ≈ 1.048 × 10⁻²² g atom⁻¹
Calculation: Mass of one atom = \(\frac{\text{Mass of unit cell}}{4} = \frac{419 \times 10^{-24}}{4} = 104.75 \times 10^{-24}\) g
Shift decimal: 1.0475 × 10⁻²² ≈ 1.048 × 10⁻²² g atom⁻¹
MHT-CET
Q20. What is the contribution of an atom at the corner of a cubic unit cell?
Correct Answer
(B) \(\frac{1}{8}\)
Reasoning: In a cubic system, each corner of a unit cell is shared by 8 adjacent unit cells. Therefore, the contribution of an atom at one corner to a single unit cell is \(\frac{1}{8}\).
MHT-CET
Q21. An element having BCC geometry has edge length 5 Å and atomic mass 70 g/mol. Find its density.
Correct Answer
(D) 1.86 g cm⁻³
Formula: \(\rho = \frac{Z \times M}{a^3 \times N_A}\)
Given: BCC (Z = 2), a = 5 Å = 5 × 10⁻⁸ cm, M = 70 g/mol
Calculation: \[\rho = \frac{2 \times 70}{(5 \times 10^{-8})^3 \times 6.022 \times 10^{23}}\] \[\rho = \frac{140}{125 \times 10^{-24} \times 6.022 \times 10^{23}} = \frac{140}{12.5 \times 6.022} \approx \mathbf{1.86\text{ g cm}^{-3}}\]
Given: BCC (Z = 2), a = 5 Å = 5 × 10⁻⁸ cm, M = 70 g/mol
Calculation: \[\rho = \frac{2 \times 70}{(5 \times 10^{-8})^3 \times 6.022 \times 10^{23}}\] \[\rho = \frac{140}{125 \times 10^{-24} \times 6.022 \times 10^{23}} = \frac{140}{12.5 \times 6.022} \approx \mathbf{1.86\text{ g cm}^{-3}}\]
MHT-CET
Q22. In a simple cubic structure, the atomic radius (r) in terms of edge length (a) is
Correct Answer
(B) 0.5 a
Reasoning: In a simple cubic structure, atoms touch along the edge of the cube. Therefore, the edge length a is equal to two radii (2r).
Formula: a = 2r ⇒ r = 0.5 a
Formula: a = 2r ⇒ r = 0.5 a
MHT-CET
Q23. An element crystallizes in BCC structure with density 6 g/cm³ and atomic mass 60 g/mol. The edge length in logarithmic form is
Correct Answer
(C) \((\frac{1}{3} \log 33.2) \times 10^{-8}\text{ cm}\)
Formula: \(a^3 = \frac{Z \times M}{\rho \times N_A}\)
Given: BCC (Z = 2), ρ = 6 g/cm³, M = 60 g/mol
Calculation: \[a^3 = \frac{2 \times 60}{6 \times 6.022 \times 10^{23}} = \frac{20}{6.022 \times 10^{23}} \approx 3.32 \times 10^{-23}\text{ cm}^3\] To match options: \(a^3 = 33.2 \times 10^{-24}\text{ cm}^3\)
Taking cube root: \(a = (33.2)^{1/3} \times 10^{-8}\text{ cm}\)
In log form: \(\log a = \frac{1}{3} \log(33.2) \times 10^{-8}\text{ cm}\)
Given: BCC (Z = 2), ρ = 6 g/cm³, M = 60 g/mol
Calculation: \[a^3 = \frac{2 \times 60}{6 \times 6.022 \times 10^{23}} = \frac{20}{6.022 \times 10^{23}} \approx 3.32 \times 10^{-23}\text{ cm}^3\] To match options: \(a^3 = 33.2 \times 10^{-24}\text{ cm}^3\)
Taking cube root: \(a = (33.2)^{1/3} \times 10^{-8}\text{ cm}\)
In log form: \(\log a = \frac{1}{3} \log(33.2) \times 10^{-8}\text{ cm}\)
MHT-CET
Q24. How many lattice points are present in a face-centred cubic (FCC) unit cell?
Correct Answer
(C) 14
Reasoning: Lattice points are the specific positions where particles are located.
- 8 corners = 8 lattice points
- 6 face centers = 6 lattice points
- Total = 8 + 6 = 14 lattice points
MHT-CET
Q25. How many moles of tetrahedral voids are present in 1 mole of a compound having HCP structure?
Correct Answer
(B) 2.0 mole
Reasoning: In any close-packed structure (HCP or CCP/FCC), if there are N atoms, there are 2N tetrahedral voids.
Given: 1 mole of compound (N = 1 mole)
Calculation: 2 × 1 mole = 2.0 moles
Given: 1 mole of compound (N = 1 mole)
Calculation: 2 × 1 mole = 2.0 moles
MHT-CET
Q26. A BCC crystal has edge length 600 pm. The radius of the atom is
Correct Answer
(A) \(\sqrt{3} \times 150\text{ pm}\)
Formula: \(4r = \sqrt{3}a \implies r = \frac{\sqrt{3}}{4} a\)
Given: a = 600 pm
Calculation: \(r = \frac{\sqrt{3}}{4} \times 600 = \sqrt{3} \times 150\text{ pm}\)
Given: a = 600 pm
Calculation: \(r = \frac{\sqrt{3}}{4} \times 600 = \sqrt{3} \times 150\text{ pm}\)
MHT-CET
Q27. A metal having density 9.8 g/cm³ and edge length 340 pm crystallizes in simple cubic structure. The number of atoms in 20 g of the metal is
Correct Answer
(C) \(5.19 \times 10^{22}\)
Step 1: Volume of 20g metal = \(\frac{\text{Mass}}{\text{Density}} = \frac{20}{9.8} \approx 2.0408\text{ cm}^3\)
Step 2: Volume of one unit cell = a = 340 pm = 3.4 × 10⁻⁸ cm
\(V_{cell} = (3.4 \times 10^{-8})^3 = 39.304 \times 10^{-24}\text{ cm}^3\)
Step 3: Number of unit cells = \(\frac{2.0408}{39.304 \times 10^{-24}} \approx 5.19 \times 10^{22}\)
Step 4: Simple cubic has 1 atom per cell, so atoms = unit cells = 5.19 × 10²²
Step 2: Volume of one unit cell = a = 340 pm = 3.4 × 10⁻⁸ cm
\(V_{cell} = (3.4 \times 10^{-8})^3 = 39.304 \times 10^{-24}\text{ cm}^3\)
Step 3: Number of unit cells = \(\frac{2.0408}{39.304 \times 10^{-24}} \approx 5.19 \times 10^{22}\)
Step 4: Simple cubic has 1 atom per cell, so atoms = unit cells = 5.19 × 10²²
MHT-CET
Q28. What is the coordination number of an atom in Hexagonal Close Packed (HCP) structure?
Correct Answer
(B) 12
Reasoning: In Hexagonal Close Packed (HCP) and Cubic Close Packed (CCP) structures, each sphere is in contact with 12 neighboring spheres (6 in its own layer, 3 above, and 3 below).
Therefore, the coordination number is 12.
Therefore, the coordination number is 12.
MHT-CET
Q29. A metal crystallizes in FCC structure with density 22.4 g cm⁻³ and a³ = 5.6 × 10⁻²³ cm³. Find its molar mass.
Correct Answer
(D) 188.8 g mol⁻¹
Formula: \(M = \frac{\rho \times a^3 \times N_A}{Z}\)
Given: ρ = 22.4 g cm⁻³, Z = 4 (FCC), a³ = 5.6 × 10⁻²³ cm³
Calculation: \[M = \frac{22.4 \times 5.6 \times 10^{-23} \times 6.022 \times 10^{23}}{4}\] \[M = 5.6 \times 5.6 \times 6.022 = 31.36 \times 6.022 \approx \mathbf{188.8\text{ g mol}^{-1}}\]
Given: ρ = 22.4 g cm⁻³, Z = 4 (FCC), a³ = 5.6 × 10⁻²³ cm³
Calculation: \[M = \frac{22.4 \times 5.6 \times 10^{-23} \times 6.022 \times 10^{23}}{4}\] \[M = 5.6 \times 5.6 \times 6.022 = 31.36 \times 6.022 \approx \mathbf{188.8\text{ g mol}^{-1}}\]
MHT-CET 2022
Q30. Which of the following statements is NOT true about polymorphism?
Correct Answer
(C) NaF and MgO are polymorphous
Reasoning: NaF and MgO are actually isomorphous. Isomorphism occurs when two different substances have the same crystal structure and similar chemical formulas (ratio of atoms is 1:1 here). Polymorphism refers to a single substance existing in more than one crystal form (like Carbon existing as Diamond and Graphite).
MHT-CET 2022
Q31. Which among the following is NOT ferromagnetic in nature?
Correct Answer
(A) Zn
Reasoning: Iron (Fe), Cobalt (Co), and Nickel (Ni) are the classic trio of ferromagnetic elements. Zinc (Zn) has a completely filled d-orbital (3d¹⁰), making it diamagnetic.
MHT-CET 2022
Q32. Which of the following is an example of covalent network solid?
Correct Answer
(D) Boron nitride
Reasoning: Boron nitride (BN) forms a giant covalent structure similar to diamond or graphite. Magnesium is metallic, benzoic acid is molecular (held by H-bonds), and NaCl is ionic.
MHT-CET 2022
Q33. What type of unit cell is NOT present in cubic crystal system?
Correct Answer
(C) Base centred
Reasoning: The cubic system only has three Bravais lattices: Simple (Primitive), Body-centred (BCC), and Face-centred (FCC). Base-centred (or end-centred) unit cells do not exist for the cubic system.
MHT-CET 2022
Q34. Which among the following statements is NOT true about molecular solid?
Correct Answer
(C) It is good conductor of heat and electricity
Reasoning: Molecular solids (like ice, dry ice, or naphthalene) consist of discrete molecules held by weak van der Waals forces or hydrogen bonds. Since they have no free electrons or ions, they are insulators (poor conductors).
MHT-CET 2022
Q35. How many types of unit cells are present in tetragonal crystal system?
Correct Answer
(D) 2
Reasoning: The tetragonal system (a = b ≠ c; α = β = γ = 90°) has only two types of Bravais lattices: Primitive (Simple) and Body-centred.
MHT-CET 2022
Q36. Which among the following is an example of molecular solid?
Correct Answer
(B) C₆H₅COOH (Benzoic Acid)
Reasoning: Benzoic acid is composed of individual molecules held together by intermolecular hydrogen bonding. SiO₂ and SiC are covalent network solids, and CaF₂ is an ionic solid.
MHT-CET 2022
Q37. What is the volume of one unit cell of a metal if it exists in simple cubic structure with edge length of unit cell 3 Å?
Correct Answer
(B) 2.7 × 10⁻²³ cm³
Calculation: a = 3 Å = 3 × 10⁻⁸ cm
V = a³ = (3 × 10⁻⁸)³
V = 27 × 10⁻²⁴ cm³ = 2.7 × 10⁻²³ cm³
V = a³ = (3 × 10⁻⁸)³
V = 27 × 10⁻²⁴ cm³ = 2.7 × 10⁻²³ cm³
MHT-CET 2022
Q38. Calculate the mass of bcc unit cell if metal has molar mass 56 g mol⁻¹.
Correct Answer
(A) 1.86 × 10⁻²² g
Calculation:
For BCC, Z = 2
Mass of unit cell = \(\frac{Z \times M}{N_A}\)
Mass = \(\frac{2 \times 56}{6.022 \times 10^{23}} \approx \frac{112}{6 \times 10^{23}} \approx 18.66 \times 10^{-24}\) g
Mass = 1.86 × 10⁻²² g
For BCC, Z = 2
Mass of unit cell = \(\frac{Z \times M}{N_A}\)
Mass = \(\frac{2 \times 56}{6.022 \times 10^{23}} \approx \frac{112}{6 \times 10^{23}} \approx 18.66 \times 10^{-24}\) g
Mass = 1.86 × 10⁻²² g
MHT-CET 2022
Q39. Calculate the density of metal having volume of unit cell 64 × 10⁻²⁴ cm³ and molar mass of metal 192 g mol⁻¹ containing 4 particles in unit cell.
Correct Answer
(B) 19.93 g cm⁻³
Calculation:
\(\rho = \frac{Z \times M}{V \times N_A}\)
\(\rho = \frac{4 \times 192}{(64 \times 10^{-24}) \times (6.022 \times 10^{23})}\)
\(\rho = \frac{768}{64 \times 0.6022} = \frac{12}{0.6022} \approx \mathbf{19.93\text{ g cm}^{-3}}\)
\(\rho = \frac{Z \times M}{V \times N_A}\)
\(\rho = \frac{4 \times 192}{(64 \times 10^{-24}) \times (6.022 \times 10^{23})}\)
\(\rho = \frac{768}{64 \times 0.6022} = \frac{12}{0.6022} \approx \mathbf{19.93\text{ g cm}^{-3}}\)
MHT-CET
Q40. An element crystallizes in BCC structure with density 7.2 g/cm³ and edge length 2.88 × 10⁻⁸ cm. Find the number of atoms in 36 g of the element.
Correct Answer
(B) 4.19 × 10²³
Formula: Number of atoms = \(\frac{\text{Mass} \times Z}{\rho \times a^3}\)
Given: BCC (Z = 2), Mass (m) = 36 g, ρ = 7.2 g/cm³, a = 2.88 × 10⁻⁸ cm
Calculation: \[\text{Atoms} = \frac{36 \times 2}{7.2 \times (2.88 \times 10^{-8})^3}\] \[\text{Atoms} = \frac{72}{7.2 \times 23.887 \times 10^{-24}} = \frac{10}{23.887 \times 10^{-24}} \approx \mathbf{4.19 \times 10^{23}}\]
Given: BCC (Z = 2), Mass (m) = 36 g, ρ = 7.2 g/cm³, a = 2.88 × 10⁻⁸ cm
Calculation: \[\text{Atoms} = \frac{36 \times 2}{7.2 \times (2.88 \times 10^{-8})^3}\] \[\text{Atoms} = \frac{72}{7.2 \times 23.887 \times 10^{-24}} = \frac{10}{23.887 \times 10^{-24}} \approx \mathbf{4.19 \times 10^{23}}\]
MHT-CET
Q41. How many octahedral voids are present in 0.8 mole of a compound having close-packed structure?
Correct Answer
(B) 4.82 × 10²³
Reasoning: In a close-packed structure (HCP or CCP), the number of octahedral voids is equal to the number of atoms (N).
Given: 0.8 mole of compound
Calculation: \[\text{Atoms} = 0.8 \times 6.022 \times 10^{23} = 4.8176 \times 10^{23} \approx \mathbf{4.82 \times 10^{23}}\]
Given: 0.8 mole of compound
Calculation: \[\text{Atoms} = 0.8 \times 6.022 \times 10^{23} = 4.8176 \times 10^{23} \approx \mathbf{4.82 \times 10^{23}}\]
MHT-CET
Q42. In a compound, atoms of A occupy the corners of FCC unit cell and atoms of B occupy the face centers. What is the formula of the compound?
Correct Answer
(C) AB₃
Reasoning:
- Atoms of A (Corners): 8 × \(\frac{1}{8}\) = 1 atom
- Atoms of B (Face centers): 6 × \(\frac{1}{2}\) = 3 atoms
MHT-CET
Q43. Silver crystallizes in a CCP structure. What is its packing efficiency?
Correct Answer
(D) 74%
Reasoning: Silver crystallizes in a CCP (which is FCC) structure. The packing efficiency of an FCC lattice is a constant value calculated as the volume of 4 spheres divided by the volume of the cube.
Value: 74%
Value: 74%
MHT-CET
Q44. How many unit cells are present in one mole of a simple cubic crystal?
Correct Answer
(A) 6.022 × 10²³
Reasoning: One mole contains N_A particles (6.022 × 10²³). In a simple cubic (SC) structure, each unit cell contains exactly 1 particle.
Calculation: Unit cells = \(\frac{\text{Total atoms}}{Z} = \frac{6.022 \times 10^{23}}{1} = \mathbf{6.022 \times 10^{23}}\)
Calculation: Unit cells = \(\frac{\text{Total atoms}}{Z} = \frac{6.022 \times 10^{23}}{1} = \mathbf{6.022 \times 10^{23}}\)
MHT-CET
Q45. A metal crystallizes in BCC structure with density 7.8 g/cm³. If a³ × N_A = 16.2 cm³ mol⁻¹, find its molar mass.
Correct Answer
(A) 63.18 g mol⁻¹
Formula: \(M = \frac{\rho \times (a^3 \times N_A)}{Z}\)
Given: ρ = 7.8 g/cm³, BCC (Z = 2), a³ × N_A = 16.2 cm³ mol⁻¹
Calculation: \[M = \frac{7.8 \times 16.2}{2} = 3.9 \times 16.2 = \mathbf{63.18\text{ g mol}^{-1}}\]
Given: ρ = 7.8 g/cm³, BCC (Z = 2), a³ × N_A = 16.2 cm³ mol⁻¹
Calculation: \[M = \frac{7.8 \times 16.2}{2} = 3.9 \times 16.2 = \mathbf{63.18\text{ g mol}^{-1}}\]
MHT-CET
Q46. The band gaps (E_g) of four solids A, B, C, D are 6.0 eV, 0.0 eV, 2.5 eV, and 5.5 eV respectively. Which one is a good conductor of electricity?
Correct Answer
(B) B
Reasoning: Conductors have overlapping valence and conduction bands (or a band gap of 0.0 eV). Semi-conductors have small gaps (< 3 eV), and insulators have large gaps (> 5 eV).
Analysis: Solid B has E_g = 0.0 eV, making it a metal/conductor.
Analysis: Solid B has E_g = 0.0 eV, making it a metal/conductor.
MHT-CET
Q47. Which combination of elements will produce an n-type semiconductor?
Correct Answer
(B) Si + P
Reasoning: An n-type (negative) semiconductor is formed when a Group 14 element (like Silicon) is doped with a Group 15 element (like Phosphorus or Arsenic) that has 5 valence electrons, providing extra "negative" electrons.
Analysis:
Analysis:
- Si + B (Group 13) → p-type
- Si + P (Group 15) → n-type
- Si + Ga (Group 13) → p-type
MHT-CET
Q48. A metal crystallizes in FCC structure with molar mass 180 g/mol. If a³ × N_A = 36 cm³ mol⁻¹, find its density.
Correct Answer
(A) 20 g cm⁻³
Formula: \(\rho = \frac{Z \times M}{a^3 \times N_A}\)
Given: FCC (Z = 4), M = 180 g/mol, a³ × N_A = 36 cm³ mol⁻¹
Calculation: \(\rho = \frac{4 \times 180}{36} = 4 \times 5 = \mathbf{20\text{ g cm}^{-3}}\)
Given: FCC (Z = 4), M = 180 g/mol, a³ × N_A = 36 cm³ mol⁻¹
Calculation: \(\rho = \frac{4 \times 180}{36} = 4 \times 5 = \mathbf{20\text{ g cm}^{-3}}\)
MHT-CET
Q49. A metal has density 2.7 g cm⁻³, volume of unit cell 6.64 × 10⁻²³ cm³, and molar mass 27 g mol⁻¹. Identify the type of unit cell.
Correct Answer
(C) Face centred cubic
Formula: \(Z = \frac{\rho \times V \times N_A}{M}\)
Given: V = 6.64 × 10⁻²³ cm³, ρ = 2.7 g cm⁻³, M = 27 g mol⁻¹
Calculation: \[Z = \frac{2.7 \times (6.64 \times 10^{-23}) \times (6.022 \times 10^{23})}{27}\] \[Z = 0.1 \times 6.64 \times 6.022 \approx 3.99 \approx \mathbf{4}\] Since Z = 4, it is a Face-centred cubic (FCC) cell.
Given: V = 6.64 × 10⁻²³ cm³, ρ = 2.7 g cm⁻³, M = 27 g mol⁻¹
Calculation: \[Z = \frac{2.7 \times (6.64 \times 10^{-23}) \times (6.022 \times 10^{23})}{27}\] \[Z = 0.1 \times 6.64 \times 6.022 \approx 3.99 \approx \mathbf{4}\] Since Z = 4, it is a Face-centred cubic (FCC) cell.
MHT-CET
Q50. Which of the following statements about metallic solids is NOT correct?
Correct Answer
(B) They are held together by covalent bonds
Reasoning: Metallic solids are composed of positive kernels in a "sea" of mobile electrons. They are held together by metallic bonds, not covalent bonds.
Therefore, statement (B) is incorrect.
Therefore, statement (B) is incorrect.
MHT-CET
Q51. A metal crystallizes in BCC structure with molar mass 56 g mol⁻¹. If ρ × N_A = 4.8 × 10²⁴ g cm⁻³ mol⁻¹, find the volume of unit cell.
Correct Answer
(D) 2.333 × 10⁻²³ cm³
Formula: \(V = \frac{Z \times M}{\rho \times N_A}\)
Given: BCC (Z = 2), M = 56 g mol⁻¹, ρ × N_A = 4.8 × 10²⁴ g cm⁻³ mol⁻¹
Calculation: \[V = \frac{2 \times 56}{4.8 \times 10^{24}} = \frac{112}{4.8 \times 10^{24}}\] \[V = 23.33 \times 10^{-24}\text{ cm}^3 = \mathbf{2.333 \times 10^{-23}\text{ cm}^3}\]
Given: BCC (Z = 2), M = 56 g mol⁻¹, ρ × N_A = 4.8 × 10²⁴ g cm⁻³ mol⁻¹
Calculation: \[V = \frac{2 \times 56}{4.8 \times 10^{24}} = \frac{112}{4.8 \times 10^{24}}\] \[V = 23.33 \times 10^{-24}\text{ cm}^3 = \mathbf{2.333 \times 10^{-23}\text{ cm}^3}\]
MHT-CET
Q52. Find the edge length (in cm) of BCC unit cell if radius of metal atom is 173 pm.
Correct Answer
(B) 4.00 × 10⁻⁸ cm
Formula: For BCC, \(4r = \sqrt{3}a \implies a = \frac{4r}{\sqrt{3}}\)
Given: r = 173 pm
Calculation: \[a = \frac{4 \times 173}{1.732} \approx \frac{692}{1.732} \approx 399.5\text{ pm} \approx 400\text{ pm}\] \[a = \mathbf{4.00 \times 10^{-8}\text{ cm}}\]
Given: r = 173 pm
Calculation: \[a = \frac{4 \times 173}{1.732} \approx \frac{692}{1.732} \approx 399.5\text{ pm} \approx 400\text{ pm}\] \[a = \mathbf{4.00 \times 10^{-8}\text{ cm}}\]
MHT-CET
Q53. How many total Bravais lattices are possible in crystal systems?
Correct Answer
(A) 14
Fact: There are 7 crystal systems which result in exactly 14 possible Bravais lattices.
MHT-CET
Q54. Find the radius of atom in FCC unit cell having edge length 393 pm.
Correct Answer
(D) 138.93 pm
Formula: For FCC, \(4r = \sqrt{2}a \implies r = \frac{a}{2\sqrt{2}}\)
Given: a = 393 pm
Calculation: \[r = \frac{393}{2 \times 1.414} = \frac{393}{2.828} \approx \mathbf{138.93\text{ pm}}\]
Given: a = 393 pm
Calculation: \[r = \frac{393}{2 \times 1.414} = \frac{393}{2.828} \approx \mathbf{138.93\text{ pm}}\]
MHT-CET
Q55. An element has density 20 g cm⁻³, molar mass 190 g mol⁻¹, and a³ × N_A = 38 cm³ mol⁻¹. Find the number of atoms per unit cell.
Correct Answer
(D) 4
Formula: \(Z = \frac{\rho \times (a^3 \times N_A)}{M}\)
Given: ρ = 20 g cm⁻³, M = 190 g mol⁻¹, a³ × N_A = 38 cm³ mol⁻¹
Calculation: \[Z = \frac{20 \times 38}{190} = \frac{760}{190} = \mathbf{4}\]
Given: ρ = 20 g cm⁻³, M = 190 g mol⁻¹, a³ × N_A = 38 cm³ mol⁻¹
Calculation: \[Z = \frac{20 \times 38}{190} = \frac{760}{190} = \mathbf{4}\]
MHT-CET
Q56. Which of the following is NOT a characteristic of crystalline solids?
Correct Answer
(D) They are isotropic
Reasoning: Crystalline solids are anisotropic, meaning their physical properties (like refractive index or electrical resistance) change depending on the direction of measurement. Amorphous solids are isotropic.
Therefore, statement (D) is incorrect for crystalline solids.
Therefore, statement (D) is incorrect for crystalline solids.
MHT-CET 2023
Q57. Which from following is NOT an example of amorphous solid?
Correct Answer
(D) Diamond
Reasoning: Glass, rubber, and plastic are amorphous solids (pseudo-solids) because they lack a long-range repeating order of particles. Diamond is a crystalline solid with a highly ordered, giant covalent network structure.
MHT-CET 2023
Q58. Calculate the molar mass of metal having density 9.3 g cm⁻³ that forms simple cubic unit cell. (a³ × N_A = 22.6 cm³ mol⁻¹)
Correct Answer
(A) 210.2 g mol⁻¹
Calculation:
For Simple Cubic, Z = 1
Formula: \(M = \frac{\rho \times (a^3 \times N_A)}{Z}\)
\(M = \frac{9.3 \times 22.6}{1} = \mathbf{210.18\text{ g mol}^{-1} \approx 210.2\text{ g mol}^{-1}}\)
For Simple Cubic, Z = 1
Formula: \(M = \frac{\rho \times (a^3 \times N_A)}{Z}\)
\(M = \frac{9.3 \times 22.6}{1} = \mathbf{210.18\text{ g mol}^{-1} \approx 210.2\text{ g mol}^{-1}}\)
MHT-CET 2023
Q59. Find the radius of an atom in fcc unit cell having edge length 405 pm.
Correct Answer
(C) 143.21 pm
Calculation:
For FCC, \(4r = \sqrt{2}a \implies r = \frac{a}{2\sqrt{2}}\)
\(r = \frac{405}{2 \times 1.414} = \frac{405}{2.828} \approx \mathbf{143.21\text{ pm}}\)
For FCC, \(4r = \sqrt{2}a \implies r = \frac{a}{2\sqrt{2}}\)
\(r = \frac{405}{2 \times 1.414} = \frac{405}{2.828} \approx \mathbf{143.21\text{ pm}}\)
MHT-CET 2023
Q60. Which from following metal has hcp crystal structure?
Correct Answer
(B) Zn
Reasoning: Copper (Cu) and Silver (Ag) have FCC (CCP) structures. Polonium (Po) is Simple Cubic. Zinc (Zn) and Magnesium (Mg) are standard examples of metals that crystallize in the Hexagonal Close Packed (HCP) system.
MHT-CET 2023
Q61. Calculate the radius of metal atom in bcc unit cell having edge length 287 pm.
Correct Answer
(A) 124.27 pm
Calculation:
For BCC, \(4r = \sqrt{3}a \implies r = \frac{\sqrt{3}}{4} a\)
\(r = 0.433 \times 287 \approx \mathbf{124.27\text{ pm}}\)
For BCC, \(4r = \sqrt{3}a \implies r = \frac{\sqrt{3}}{4} a\)
\(r = 0.433 \times 287 \approx \mathbf{124.27\text{ pm}}\)
MHT-CET 2023
Q62. Calculate the number of atoms present in unit cell if an element having molar mass 23 g mol⁻¹ and density 0.96 g cm⁻³. (a³ × N_A = 48 cm³ mol⁻¹)
Correct Answer
(B) 2
Calculation:
\(Z = \frac{\rho \times (a^3 \times N_A)}{M}\)
\(Z = \frac{0.96 \times 48}{23} = \frac{46.08}{23} \approx \mathbf{2}\)
\(Z = \frac{\rho \times (a^3 \times N_A)}{M}\)
\(Z = \frac{0.96 \times 48}{23} = \frac{46.08}{23} \approx \mathbf{2}\)
MHT-CET 2023
Q63. Which of the following statements is NOT true about polymorphism?
Correct Answer
(D) The crystal shape of polymorphic substances is identical to each other
Reasoning: Statement D is false. By definition, polymorphism is the ability of a substance to exist in different crystalline forms (shapes). For example, Carbon can exist as graphite (hexagonal) or diamond (cubic).
MHT-CET 2023
Q64. Find the radius of metal atom in bcc unit cell having edge length 450 pm.
Correct Answer
(B) 194.85 pm
Calculation:
\(r = \frac{\sqrt{3}}{4} a = \frac{1.732}{4} \times 450\)
\(r = 0.433 \times 450 = \mathbf{194.85\text{ pm}}\)
\(r = \frac{\sqrt{3}}{4} a = \frac{1.732}{4} \times 450\)
\(r = 0.433 \times 450 = \mathbf{194.85\text{ pm}}\)
MHT-CET 2023
Q65. Calculate the molar mass of an element having density 2.8 g cm⁻³ and forms fcc unit cell. (a³ × N_A = 38.5 cm³ mol⁻¹)
Correct Answer
(A) 26.95 g mol⁻¹
Calculation:
For FCC, Z = 4
\(M = \frac{\rho \times (a^3 \times N_A)}{Z}\)
\(M = \frac{2.8 \times 38.5}{4} = 0.7 \times 38.5 = \mathbf{26.95\text{ g mol}^{-1}}\)
For FCC, Z = 4
\(M = \frac{\rho \times (a^3 \times N_A)}{Z}\)
\(M = \frac{2.8 \times 38.5}{4} = 0.7 \times 38.5 = \mathbf{26.95\text{ g mol}^{-1}}\)
MHT-CET 2023
Q66. What is the number of different types of unit cells present in tetragonal crystal system?
Correct Answer
(B) 2
Reasoning: The tetragonal crystal system has two Bravais lattices: Primitive (Simple) and Body-centered.
MHT-CET 2023
Q67. Find the edge length (in cm) of bcc unit cell if radius of metal atom is 126 pm.
Correct Answer
(A) 2.91 × 10⁻⁸ cm
Calculation:
For BCC, \(a = \frac{4r}{\sqrt{3}}\)
\(a = \frac{4 \times 126}{1.732} = \frac{504}{1.732} \approx 291\text{ pm}\)
Convert to cm: \(291 \times 10^{-10}\text{ cm} = \mathbf{2.91 \times 10^{-8}\text{ cm}}\)
For BCC, \(a = \frac{4r}{\sqrt{3}}\)
\(a = \frac{4 \times 126}{1.732} = \frac{504}{1.732} \approx 291\text{ pm}\)
Convert to cm: \(291 \times 10^{-10}\text{ cm} = \mathbf{2.91 \times 10^{-8}\text{ cm}}\)
MHT-CET
Q68. A metal crystallizes in BCC structure with molar mass 56 g mol⁻¹ and density 7.8 g cm⁻³. Find the volume of unit cell.
Correct Answer
(A) 2.38 × 10⁻²³ cm³
Formula: \(V = a^3 = \frac{Z \times M}{\rho \times N_A}\)
Given: BCC (Z = 2), M = 56 g mol⁻¹, ρ = 7.8 g cm⁻³, N_A = 6.022 × 10²³
Calculation: \[V = \frac{2 \times 56}{7.8 \times 6.022 \times 10^{23}} = \frac{112}{46.97 \times 10^{23}} \approx \mathbf{2.38 \times 10^{-23}\text{ cm}^3}\]
Given: BCC (Z = 2), M = 56 g mol⁻¹, ρ = 7.8 g cm⁻³, N_A = 6.022 × 10²³
Calculation: \[V = \frac{2 \times 56}{7.8 \times 6.022 \times 10^{23}} = \frac{112}{46.97 \times 10^{23}} \approx \mathbf{2.38 \times 10^{-23}\text{ cm}^3}\]
MHT-CET
Q69. Which of the following compounds exhibits Frenkel defect?
Correct Answer
(A) Calcium fluoride
Reasoning: Frenkel defects occur when an ion (usually the smaller cation) is displaced from its lattice site to an interstitial site. While common in AgCl and ZnS, Calcium fluoride (CaF₂) exhibits an "Anion Frenkel" defect where the F⁻ ions occupy interstitial positions. The other options (NaCl, KCl, LiCl) are alkali halides that predominantly show Schottky defects.
MHT-CET
Q70. Which of the following is NOT a covalent network solid?
Correct Answer
(D) Calcium fluoride (CaF₂)
Reasoning: Silica (SiO₂), Boron Nitride (BN), and Silicon Carbide (SiC) all form giant, three-dimensional covalent networks. Calcium fluoride (CaF₂) is an ionic solid held together by electrostatic forces of attraction between Ca²⁺ and F⁻ ions.
MHT-CET
Q71. An element forms FCC unit cell with density 21 g cm⁻³. If a³ × N_A = 36 cm³ mol⁻¹, find its molar mass.
Correct Answer
(B) 189.00 g mol⁻¹
Formula: \(M = \frac{\rho \times (a^3 \times N_A)}{Z}\)
Given: FCC (Z = 4), ρ = 21 g cm⁻³, a³ × N_A = 36 cm³ mol⁻¹
Calculation: \[M = \frac{21 \times 36}{4} = 21 \times 9 = \mathbf{189\text{ g mol}^{-1}}\]
Given: FCC (Z = 4), ρ = 21 g cm⁻³, a³ × N_A = 36 cm³ mol⁻¹
Calculation: \[M = \frac{21 \times 36}{4} = 21 \times 9 = \mathbf{189\text{ g mol}^{-1}}\]
MHT-CET
Q72. Which of the following metals crystallizes in CCP (Cubic Close Packed) structure?
Correct Answer
(A) Cu
Reasoning: Copper (Cu) and Silver (Ag) are the most common examples of metals that crystallize in the Cubic Close Packed (CCP) or Face-Centred Cubic (FCC) system. Zn and Mg are HCP, and Po is simple cubic.
MHT-CET
Q73. A simple cubic crystal has edge length 334.7 pm. Find the radius of the atom.
Correct Answer
(A) 167.35 pm
Formula: In a Simple Cubic cell, a = 2r ⇒ r = a/2
Given: a = 334.7 pm
Calculation: \[r = \frac{334.7}{2} = \mathbf{167.35\text{ pm}}\]
Given: a = 334.7 pm
Calculation: \[r = \frac{334.7}{2} = \mathbf{167.35\text{ pm}}\]
MHT-CET
Q74. An element crystallizes in BCC structure with molar mass 93 g mol⁻¹ and density 6.20 g cm⁻³. Find the edge length. (Take N_A = 6 × 10²³)
Correct Answer
(D) \(\sqrt[3]{50} \times 10^{-8}\text{ cm}\)
Formula: \(a^3 = \frac{Z \times M}{\rho \times N_A}\)
Given: BCC (Z = 2), M = 93, ρ = 6.20 g cm⁻³, N_A = 6 × 10²³
Calculation: \[a^3 = \frac{2 \times 93}{6.2 \times 6 \times 10^{23}} = \frac{186}{37.2 \times 10^{23}} = 5 \times 10^{-23} = 50 \times 10^{-24}\text{ cm}^3\] Taking cube root: \(a = \sqrt[3]{50} \times 10^{-8}\text{ cm}\)
Given: BCC (Z = 2), M = 93, ρ = 6.20 g cm⁻³, N_A = 6 × 10²³
Calculation: \[a^3 = \frac{2 \times 93}{6.2 \times 6 \times 10^{23}} = \frac{186}{37.2 \times 10^{23}} = 5 \times 10^{-23} = 50 \times 10^{-24}\text{ cm}^3\] Taking cube root: \(a = \sqrt[3]{50} \times 10^{-8}\text{ cm}\)
MHT-CET
Q75. A BCC crystal has edge length 2.93 Å. Find the radius of the atom.
Correct Answer
(B) 1.268 Å
Formula: For BCC, \(4r = \sqrt{3}a \implies r = \frac{\sqrt{3}}{4} a\)
Given: a = 2.93 Å
Calculation: \[r = \frac{1.732 \times 2.93}{4} = 0.433 \times 2.93 \approx \mathbf{1.268\text{ Å}}\]
Given: a = 2.93 Å
Calculation: \[r = \frac{1.732 \times 2.93}{4} = 0.433 \times 2.93 \approx \mathbf{1.268\text{ Å}}\]
MHT-CET
Q76. What is the percentage of unoccupied volume in a simple cubic unit cell?
Correct Answer
(B) 47.64%
Reasoning:
- Packing efficiency of Simple Cubic = 52.4%
- Unoccupied (Void) space = 100% - 52.4% = 47.6%
MHT-CET
Q77. An element crystallizes in BCC structure with molar mass 23 g mol⁻¹. If a³ × N_A = 45.62 cm³ mol⁻¹, find its density.
Correct Answer
(B) 1.0 g cm⁻³
Formula: \(\rho = \frac{Z \times M}{a^3 \times N_A}\)
Given: BCC (Z = 2), M = 23 g mol⁻¹, a³ × N_A = 45.62 cm³ mol⁻¹
Calculation: \[\rho = \frac{2 \times 23}{45.62} = \frac{46}{45.62} \approx \mathbf{1.0\text{ g cm}^{-3}}\]
Given: BCC (Z = 2), M = 23 g mol⁻¹, a³ × N_A = 45.62 cm³ mol⁻¹
Calculation: \[\rho = \frac{2 \times 23}{45.62} = \frac{46}{45.62} \approx \mathbf{1.0\text{ g cm}^{-3}}\]
MHT-CET
Q78. Calculate the total volume occupied by atoms in a BCC unit cell having edge length 4 × 10⁻⁸ cm.
Correct Answer
(A) 4.352 × 10⁻²³ cm³
Method: Volume occupied = Packing Efficiency × Volume of unit cell
Given: a = 4 × 10⁻⁸ cm. Packing efficiency for BCC = 68% (0.68)
Calculation: V_cell = a³ = (4 × 10⁻⁸)³ = 64 × 10⁻²⁴ cm³
V_occupied = 0.68 × 64 × 10⁻²⁴ = 43.52 × 10⁻²⁴ = 4.352 × 10⁻²³ cm³
Given: a = 4 × 10⁻⁸ cm. Packing efficiency for BCC = 68% (0.68)
Calculation: V_cell = a³ = (4 × 10⁻⁸)³ = 64 × 10⁻²⁴ cm³
V_occupied = 0.68 × 64 × 10⁻²⁴ = 43.52 × 10⁻²⁴ = 4.352 × 10⁻²³ cm³
MHT-CET
Q79. Which of the following exhibits substitutional impurity defect?
Correct Answer
(C) Brass
Reasoning: In Brass (an alloy of Copper and Zinc), Zinc atoms substitute for Copper atoms in the crystal lattice. NaCl, ZnS, and AgBr primarily show stoichiometric point defects (Schottky/Frenkel).
MHT-CET
Q80. An element has edge length 288 pm. Find the volume of its unit cell.
Correct Answer
(A) 2.39 × 10⁻²³ cm³
Calculation:
a = 288 pm = 2.88 × 10⁻⁸ cm
V = (2.88 × 10⁻⁸)³ ≈ 23.887 × 10⁻²⁴ cm³ = 2.39 × 10⁻²³ cm³
V = (2.88 × 10⁻⁸)³ ≈ 23.887 × 10⁻²⁴ cm³ = 2.39 × 10⁻²³ cm³
MHT-CET
Q81. How many unit cells are present in 1 mole of a BCC metal?
Correct Answer
(C) 3.011 × 10²³
Given: 1 mole = N_A atoms (6.022 × 10²³). For BCC, Z = 2 atoms/cell.
Calculation: Unit cells = Total atoms / Z = (6.022 × 10²³) / 2 = 3.011 × 10²³
Calculation: Unit cells = Total atoms / Z = (6.022 × 10²³) / 2 = 3.011 × 10²³
MHT-CET
Q82. Ice is an example of which type of solid?
Correct Answer
(C) Molecular solid
Reasoning: Ice (H₂O) consists of molecules held together by strong intermolecular hydrogen bonds. It is a classic example of a hydrogen-bonded molecular solid.
MHT-CET
Q83. How many tetrahedral voids are present in 0.6 mole of a CCP structure?
Correct Answer
(C) 7.23 × 10²³
Formula: Number of tetrahedral voids = 2 × Number of atoms
Calculation: Atoms = 0.6 × 6.022 × 10²³ = 3.6132 × 10²³
Voids = 2 × 3.6132 × 10²³ = 7.2264 × 10²³ ≈ 7.23 × 10²³
Calculation: Atoms = 0.6 × 6.022 × 10²³ = 3.6132 × 10²³
Voids = 2 × 3.6132 × 10²³ = 7.2264 × 10²³ ≈ 7.23 × 10²³
MHT-CET
Q84. Which type of unit cell is present in hexagonal crystal system?
Correct Answer
(D) Simple
Reasoning: The hexagonal system only has one Bravais lattice: Simple (Primitive).
MHT-CET
Q85. Which combination produces a p-type semiconductor?
Correct Answer
(C) Ge + B
Reasoning: A p-type semiconductor is formed when a Group 14 element (like Germanium) is doped with a Group 13 element (like Boron) which has 3 valence electrons, creating "positive" holes.
MHT-CET
Q86. In a compound, y atoms form HCP lattice. x atoms occupy 1/3 of tetrahedral voids. Find the formula.
Correct Answer
(B) x₂y₃
Reasoning:
Let number of y atoms = N
Total tetrahedral voids = 2N
x occupies 1/3 of voids: x = (1/3) × 2N = (2/3)N
Ratio x : y = (2/3)N : N = 2 : 3 ⇒ x₂y₃
Total tetrahedral voids = 2N
x occupies 1/3 of voids: x = (1/3) × 2N = (2/3)N
Ratio x : y = (2/3)N : N = 2 : 3 ⇒ x₂y₃
MHT-CET
Q87. An element forms FCC unit cell with density 2.7 g cm⁻³. If a³ × N_A = 40 cm³ mol⁻¹, find its molar mass.
Correct Answer
(D) 27 g mol⁻¹
Formula: M = [ρ × (a³ × N_A)] / Z
Given: ρ = 2.7 g cm⁻³, FCC (Z = 4), a³ × N_A = 40 cm³ mol⁻¹
Calculation: M = (2.7 × 40) / 4 = 2.7 × 10 = 27 g mol⁻¹
Given: ρ = 2.7 g cm⁻³, FCC (Z = 4), a³ × N_A = 40 cm³ mol⁻¹
Calculation: M = (2.7 × 40) / 4 = 2.7 × 10 = 27 g mol⁻¹
MHT-CET
Q88. What is the minimum number of spheres required to form an octahedral void?
Correct Answer
(C) 6
Reasoning: An octahedral void is a vacancy formed at the center of 6 spheres—4 in one plane, 1 above, and 1 below.
MHT-CET
Q89. Find the volume of BCC unit cell if the radius of atom is 216.5 pm.
Correct Answer
(B) 1.25 × 10⁻²² cm³
Formula: For BCC, a = (4r)/√3 and V = a³
Given: r = 216.5 pm
Calculation: a = (4 × 216.5)/1.732 = 866/1.732 ≈ 500 pm = 5 × 10⁻⁸ cm
V = (5 × 10⁻⁸)³ = 125 × 10⁻²⁴ = 1.25 × 10⁻²² cm³
Given: r = 216.5 pm
Calculation: a = (4 × 216.5)/1.732 = 866/1.732 ≈ 500 pm = 5 × 10⁻⁸ cm
V = (5 × 10⁻⁸)³ = 125 × 10⁻²⁴ = 1.25 × 10⁻²² cm³
MHT-CET
Q90. An FCC unit cell has volume 1.25 × 10⁻²² cm³. Find the volume occupied by atoms.
Correct Answer
(D) 9.25 × 10⁻²³ cm³
Method: V_occupied = Packing Efficiency × V_cell
Given: V_cell = 1.25 × 10⁻²² cm³. For FCC, efficiency = 74% (0.74)
Calculation: V_occupied = 0.74 × (1.25 × 10⁻²²) = 0.925 × 10⁻²² = 9.25 × 10⁻²³ cm³
Given: V_cell = 1.25 × 10⁻²² cm³. For FCC, efficiency = 74% (0.74)
Calculation: V_occupied = 0.74 × (1.25 × 10⁻²²) = 0.925 × 10⁻²² = 9.25 × 10⁻²³ cm³
MHT-CET
Q91. What is the packing fraction of a simple cubic lattice?
Correct Answer
(B) 0.524
Fact: In a Simple Cubic lattice, the packing efficiency is 52.4%.
The fraction is 0.524 (represented as 0.523 or 0.524 in options).
The fraction is 0.524 (represented as 0.523 or 0.524 in options).
MHT-CET
Q92. Find the number of atoms per unit cell if density = 9.8 g cm⁻³, a³ = 3.6 × 10⁻²³ cm³, M = 210 g mol⁻¹.
Correct Answer
(C) 1
Formula: Z = (ρ × a³ × N_A) / M
Given: ρ = 9.8 g cm⁻³, a³ = 3.6 × 10⁻²³ cm³, M = 210 g mol⁻¹
Calculation: Z = (9.8 × 3.6 × 10⁻²³ × 6.022 × 10²³) / 210 = (9.8 × 3.6 × 6.022) / 210 ≈ 212.4/210 ≈ 1
Given: ρ = 9.8 g cm⁻³, a³ = 3.6 × 10⁻²³ cm³, M = 210 g mol⁻¹
Calculation: Z = (9.8 × 3.6 × 10⁻²³ × 6.022 × 10²³) / 210 = (9.8 × 3.6 × 6.022) / 210 ≈ 212.4/210 ≈ 1
MHT-CET
Q93. How many tetrahedral voids are present in 0.2 mole of a close-packed structure?
Correct Answer
(B) 2.4088 × 10²³
Formula: Total tetrahedral voids = 2 × Number of atoms
Calculation: Atoms = 0.2 × 6.022 × 10²³ = 1.2044 × 10²³
Voids = 2 × 1.2044 × 10²³ = 2.4088 × 10²³
Calculation: Atoms = 0.2 × 6.022 × 10²³ = 1.2044 × 10²³
Voids = 2 × 1.2044 × 10²³ = 2.4088 × 10²³
MHT-CET
Q94. Which of the following is NOT a polymorph of carbon?
Correct Answer
(B) Aragonite
Reasoning: Diamond, Graphite, and Fullerenes are all allotropic/polymorphic forms of pure carbon. Aragonite is a mineral form of Calcium Carbonate (CaCO₃).
MHT-CET
Q95. How many unit cells are present in 1 cm³ of a metal having edge length 200 pm?
Correct Answer
(C) 1.25 × 10²³
Given: a = 200 pm = 2 × 10⁻⁸ cm
Calculation: V_cell = a³ = (2 × 10⁻⁸)³ = 8 × 10⁻²⁴ cm³
Unit cells = Total Volume / V_cell = 1 / (8 × 10⁻²⁴) = 0.125 × 10²⁴ = 1.25 × 10²³
Calculation: V_cell = a³ = (2 × 10⁻⁸)³ = 8 × 10⁻²⁴ cm³
Unit cells = Total Volume / V_cell = 1 / (8 × 10⁻²⁴) = 0.125 × 10²⁴ = 1.25 × 10²³
MHT-CET
Q96. An FCC unit cell has volume 1.2 × 10⁻²² cm³. Find the volume occupied by one particle.
Correct Answer
(A) 2.22 × 10⁻²³ cm³
Method: Volume of one particle = (Packing Efficiency × V_cell) / Z
Given: V_cell = 1.2 × 10⁻²² cm³; for FCC, Z = 4 and efficiency = 0.74
Calculation: V_one particle = (0.74 × 1.2 × 10⁻²²) / 4 = (0.888 × 10⁻²²) / 4 = 0.222 × 10⁻²² = 2.22 × 10⁻²³ cm³
Given: V_cell = 1.2 × 10⁻²² cm³; for FCC, Z = 4 and efficiency = 0.74
Calculation: V_one particle = (0.74 × 1.2 × 10⁻²²) / 4 = (0.888 × 10⁻²²) / 4 = 0.222 × 10⁻²² = 2.22 × 10⁻²³ cm³
MHT-CET
Q97. An element forms FCC unit cell with molar mass 63 g mol⁻¹. If a³ × N_A = 28 cm³ mol⁻¹, find its density.
Correct Answer
(B) 9.0 g cm⁻³
Formula: ρ = (Z × M) / (a³ × N_A)
Given: FCC (Z = 4), M = 63 g mol⁻¹, a³ × N_A = 28 cm³ mol⁻¹
Calculation: ρ = (4 × 63) / 28 = 252 / 28 = 9.0 g cm⁻³
Given: FCC (Z = 4), M = 63 g mol⁻¹, a³ × N_A = 28 cm³ mol⁻¹
Calculation: ρ = (4 × 63) / 28 = 252 / 28 = 9.0 g cm⁻³
MHT-CET
Q98. A BCC unit cell has volume 1.5 × 10⁻²² cm³. Find the void volume.
Correct Answer
(A) 4.8 × 10⁻²³ cm³
Method: V_void = (1 - Packing Efficiency) × V_cell
Given: V_cell = 1.5 × 10⁻²² cm³; BCC efficiency = 68% (0.68)
Calculation: V_void = (1 - 0.68) × (1.5 × 10⁻²²) = 0.32 × 1.5 × 10⁻²² = 0.48 × 10⁻²² = 4.8 × 10⁻²³ cm³
Given: V_cell = 1.5 × 10⁻²² cm³; BCC efficiency = 68% (0.68)
Calculation: V_void = (1 - 0.68) × (1.5 × 10⁻²²) = 0.32 × 1.5 × 10⁻²² = 0.48 × 10⁻²² = 4.8 × 10⁻²³ cm³
MHT-CET
Q99. A simple cubic unit cell has volume 5.5 × 10⁻²² cm³. Find the void volume.
Correct Answer
(C) 2.619 × 10⁻²² cm³
Method: V_void = (1 - 0.524) × V_cell
Given: V_cell = 5.5 × 10⁻²² cm³; SC efficiency = 52.4% (0.524)
Calculation: V_void = 0.476 × (5.5 × 10⁻²²) ≈ 2.618 × 10⁻²² cm³ ≈ 2.619 × 10⁻²² cm³
Given: V_cell = 5.5 × 10⁻²² cm³; SC efficiency = 52.4% (0.524)
Calculation: V_void = 0.476 × (5.5 × 10⁻²²) ≈ 2.618 × 10⁻²² cm³ ≈ 2.619 × 10⁻²² cm³
MHT-CET 2024
Q100. What is the relation between edge length (a) and total volume occupied by atoms (V) in bcc unit cell?
Correct Answer
(B) V = (√3 π a³)/8
Reasoning:
In BCC, there are 2 atoms. Total occupied volume V = 2 × (4/3)πr³ = (8/3)πr³
Relation between r and a: 4r = √3a ⇒ r = (√3 a)/4
Substituting: V = (8/3)π × (√3 a/4)³ = (8/3)π × (3√3 a³)/64 = (√3 π a³)/8
Relation between r and a: 4r = √3a ⇒ r = (√3 a)/4
Substituting: V = (8/3)π × (√3 a/4)³ = (8/3)π × (3√3 a³)/64 = (√3 π a³)/8
MHT-CET 2024
Q101. Calculate the volume of fcc unit cell if radius of a particle is 106.05 pm.
Correct Answer
(C) 2.7 × 10⁻²³ cm³
Calculation:
For FCC, a = 2√2 r = 2 × 1.414 × 106.05 ≈ 300 pm = 3 × 10⁻⁸ cm
V = a³ = (3 × 10⁻⁸)³ = 2.7 × 10⁻²³ cm³
V = a³ = (3 × 10⁻⁸)³ = 2.7 × 10⁻²³ cm³
MHT-CET 2024
Q102. In an ionic solid, equal number of cations and anions are missing from their regular positions. What defect is this?
Correct Answer
(C) Schottky defect
Reasoning: When equal numbers of cations and anions are missing, electrical neutrality is maintained but density decreases. This is a Schottky defect.
MHT-CET 2024
Q103. Metallic silver has fcc structure. If radius of Ag atom is 144 pm, what is the edge length in cm?
Correct Answer
(A) 4.07 × 10⁻⁸ cm
Calculation:
a = 2√2 r = 2.828 × 144 pm ≈ 407.2 pm
407.2 pm = 4.07 × 10⁻⁸ cm
407.2 pm = 4.07 × 10⁻⁸ cm
MHT-CET 2024
Q104. What type of unit cell is common to all seven types of crystal systems?
Correct Answer
(A) Simple
Reasoning: Every crystal system (Cubic, Tetragonal, Orthorhombic, etc.) has a Simple (Primitive) unit cell as its basic arrangement.
MHT-CET 2024
Q105. B forms ccp structure. Atoms of A occupy 1/3 of tetrahedral voids. Find the formula.
Correct Answer
(A) A₂B₃
Calculation:
Number of B atoms = 4 (for CCP)
Tetrahedral voids = 2 × 4 = 8
Number of A atoms = (1/3) × 8 = 8/3
Ratio A : B = (8/3) : 4 = 8 : 12 = 2 : 3 ⇒ A₂B₃
Tetrahedral voids = 2 × 4 = 8
Number of A atoms = (1/3) × 8 = 8/3
Ratio A : B = (8/3) : 4 = 8 : 12 = 2 : 3 ⇒ A₂B₃
MHT-CET 2024
Q106. An element has a = 5 Å, density = 4 g cm⁻³, atomic mass = 149. Identify the crystal structure.
Correct Answer
(B) Body centred cubic
Calculation:
Z = (ρ × a³ × N_A) / M
a = 5 Å = 5 × 10⁻⁸ cm, a³ = 125 × 10⁻²⁴ cm³
Z = (4 × 125 × 10⁻²⁴ × 6.022 × 10²³) / 149 = (301.1) / 149 ≈ 2
Z = 2 corresponds to Body centred cubic (BCC)
a = 5 Å = 5 × 10⁻⁸ cm, a³ = 125 × 10⁻²⁴ cm³
Z = (4 × 125 × 10⁻²⁴ × 6.022 × 10²³) / 149 = (301.1) / 149 ≈ 2
Z = 2 corresponds to Body centred cubic (BCC)
MHT-CET 2024
Q107. Calculate the number of unit cells in 0.9 g metal if it forms bcc structure. (ρ × a³ = 3 × 10⁻²² g)
Correct Answer
(C) 3.0 × 10²¹
Reasoning: The term ρ × a³ represents the Mass of a single unit cell.
Calculation: Number of unit cells = Total mass / Mass of one unit cell
Number of unit cells = (0.9 g) / (3 × 10⁻²² g) = 0.3 × 10²² = 3.0 × 10²¹
Calculation: Number of unit cells = Total mass / Mass of one unit cell
Number of unit cells = (0.9 g) / (3 × 10⁻²² g) = 0.3 × 10²² = 3.0 × 10²¹
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