Question

How many unit cells are present in a cube-shaped ideal crystal of NaCl of mass 1.00 kg?

[Atomic masses of Na and Cl are 23 g/mol and 35.5 g/mol respectively.]

• A. $2.57\times {10}^{24}$
• B. $6.25\times {10}^{24}$
• C. $3.75\times {10}^{24}$
• D. $1.95\times {10}^{24}$

Answer 1

The correct option is A $2.57\times {10}^{24}$

Molar mass (M) of NaCl $=23+35.8=58.5$ g/mol.

An fcc unit cell has 4 atoms per unit cell. Z $=$4.

Mass (m) of one unit cell $=$ volume $\times$ density

$m=\frac{MZ}{N_0}$

$m=\frac{58.5\times 4}{6.02\times {10}^{23}}$

$m=3.88\times {10}^{-22}$ g

But 1 kg $=$ 1000 g.

So convert the unit of mass from g to kg.

$m=3.88\times {10}^{-22}g\times \frac{1kg}{1000g}=3.88\times {10}^{-25}$ kg

Number of unit cells present in 1 kg of NaCl $=\frac{1}{m}=\frac{1}{3.88\times {10}^{-25}}=2.57\times {10}^{24}$

Hence, the correct option is $\text{A}$