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Number of atoms per unit cell, or the type of cubic cell, from density data

This type of question, asked 8 times: 10 May 2023 S1, 15 May 2023 S2, 10 May 2024 S2, 21 Apr 2025 S1, 21 Apr 2025 S2, 25 Apr 2025 S1, 13 Apr 2026 S1, 14 May 2026 S2. All 8 questions below, oldest first.

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  1. Question 1

    10 May 2023 S1

    92
    Calculate number of atoms per unit cell of an element having molar mass 92.0 g mol−192.0\,\text{g mol}^{-1} and density 8.6 g cm−38.6\,\text{g cm}^{-3} forming cubic unit cell structure. a3×NA=21.5 cm3mol−1a^3\times N_A=21.5\,\text{cm}^3\text{mol}^{-1}

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  2. Question 2

    15 May 2023 S2

    77
    Unit cell of an element has edge length of 55Å with density 4 g cm−34\text{ g cm}^{-3}, if its atomic mass is 149, identify the crystal structure.

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  3. Question 3

    10 May 2024 S2

    100
    Calculate the number of atoms present in unit cell of an element having molar mass 190 g mol−1190\,g\,mol^{-1} and density 20 g cm−320\,g\,cm^{-3}, a3NA=38 cm3 mol−1a^3 N_A = 38\,cm^3\,mol^{-1}

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  4. Question 4

    21 Apr 2025 S1

    87
    The mass of an atom present in unit cell is 4.4×10−23 g4.4 \times 10^{-23}\ \mathrm{g} and the product of density and volume of unit cell is 1.792×10−22 g1.792 \times 10^{-22}\ \mathrm{g}. What is the type of cubic unit cell?

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  5. Question 5

    21 Apr 2025 S2

    80
    Calculate the number of atoms present per unit cell if product of density and volume of unit cell is 1.8×10−22 g1.8 \times 10^{-22}\ \mathrm{g}. [Mass of an atom =4.5×10−23 g= 4.5 \times 10^{-23}\ \mathrm{g}]

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  6. Question 6

    25 Apr 2025 S1

    56
    Calculate the number of particles present per unit cell if mass of a particle is 8.0×10−23 g[ρ×a3=3.2×10−22 g]8.0 \times10^{- 23}\text{ }g\left\lbrack \rho\times a^{3}= 3.2 \times10^{- 22}\text{ }g \right\rbrack

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  7. Question 7

    13 Apr 2026 S1

    17
    An element with a molar mass 27g mol−1\mathrm{mol^{-1}} forms a cubic unit cell. Calculate the number of atoms present in a unit cell if the density of metal is 2.7g cm−3\mathrm{cm^{-3}}. [a3×NA=40 cm3 mol−1][a^3 \times N_A = 40\ \mathrm{cm^3\,mol^{-1}}]

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  8. Question 8

    14 May 2026 S2

    19
    Find the type of unit cell if the density is 6.25 g/cm36.25\ \mathrm{g/cm^3}, edge length 400400 pm and the atomic mass of an element is 120120 g/mol. (Consider NA=6.02×1023N_A = 6.02 \times 10^{23})

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