Question

How many unit cells are present in a cube-shaped ideal crystal of NaCl of mass 1.00g? [Atomic masses : Na = 23, Cl = 35.5]

• A. 2.57 × 1021
• B. 5.14 × 1021
• C. 1.28 × 1021
• D. 1.71 × 1021

Answer 1

Answer: (A)

2.57 × 1021

Answer 2

mass (m) = density × volume = 1.00 g

Mol. wt. (M) of NaCl = 23 + 15.5 = 58.5

Number of unit cells present in a cube shaped crystal of NaCl of

mass 1.00 g =$\frac{\rho \times a^{3} \times Na}{M \times Z} = \frac{m \times N_{A}}{M \times Z} = \frac{1 \times 6.023 \times 10^{23}}{58.5 \times 4}$

(In NaCl each unit cells has 4 NaCl units. Hence Z = 4)

\ Number of unit cells = 0.02573 × 1023 = 2.57 × 1021 unit cells.