Ncert Exercises & Solutions

1.18 An element with molar mass 2.7 × 10^-2 kg mol^-1 forms a cubic unit cell with edge length 405 pm. If its density is 2.7 × 10³ kg m⁻³, what is the nature of the cubic unit cell?

It is given that density of the element, d = 2.7 × 10³ kg m⁻³

Molar mass, M = 2.7 × 10⁻² kg mol⁻¹
Edge length, a = 405 pm = 405 × 10⁻¹² m = 4.05 × 10⁻¹⁰ m
It is known that, Avogadro’s number, NA = 6.022 × 10²³ mol⁻¹

Applying the relation,

d = \frac{zM}{a^{3} \cdot N_{A}}

z = \frac{d \cdot a^{3} N_{A}}{M}

= \frac{2.7 \times 10^{3} kg m^{- 1} \times {(4.05 \times 10^{- 10} m)}^{3} \times 6.022 \times 10^{23} {mol}^{- 1}}{2.7 \times 10^{- 2} kg {mol}^{- 1}}

= 4.004

= 4

This implies that four atoms of the element are present per unit cell. Hence, the unit cell is face-centred cubic (fcc) or cubic close-packed (ccp).

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