An atomic substance A of molar mass $12 \mathrm{~g} \mathrm{~mol}^{-1}$ has a cubic crystal structure with edge length of $300 ~\mathrm{pm}$. The no. of atoms present in one unit cell of $\mathrm{A}$ is ____________. (Nearest integer)

Given the density of $\mathrm{A}$ is $3.0 \mathrm{~g} \mathrm{~mL}^{-1}$ and $\mathrm{N}_{\mathrm{A}}=6.02 \times 10^{23} \mathrm{~mol}^{-1}$

<p>Given:</p> <ul> <li>Atomic substance A with molar mass $M = 12 \, \text{g mol}^{-1}$</li><br/> <li>Cubic crystal structure with edge length $a = 300 \, \text{pm} = 300 \times 10^{-12} \, \text{m}$</li><br/> <li>Density $\rho = 3.0 \, \text{g mL}^{-1}$</li><br/> <li>Avogadro&#39;s number $N_A = 6.02 \times 10^{23} \, \text{mol}^{-1}$</li> </ul> <p>First, calculate the volume of the unit cell $V = a^3 = (300 \times 10^{-12} \, \text{m})^3 = 27 \times 10^{-30} \, \text{m}^3$.</p> <p>Next, calculate the mass of the unit cell using the given density. Density is mass over volume, so mass <br/><br/>$m = \rho \cdot V = 3.0 \, \text{g mL}^{-1} \cdot 27 \times 10^{-30} \, \text{m}^3 \cdot \frac{1 \, \text{mL}}{10^{-6} \, \text{m}^3} = 81 \times 10^{-24} \, \text{g}$.</p> <p>Then, calculate the number of moles in one unit cell. The molar mass is mass over number of moles, so <br/><br/>$n = \frac{m}{M} = \frac{81 \times 10^{-24} \, \text{g}}{12 \, \text{g mol}^{-1}} = 6.75 \times 10^{-26} \, \text{mol}$.</p> <p>Finally, calculate the number of atoms in one unit cell. The Avogadro constant is the number of atoms per mole,<br/><br/> so number of atoms $ = n \cdot N_A = 6.75 \times 10^{-26} \, \text{mol} \cdot 6.02 \times 10^{23} \, \text{mol}^{-1} = 4.06$.</p> <p>Rounding to the nearest integer, we get 4 atoms. So, there are 4 atoms present in one unit cell of substance A.</p>