JEE Main 2023 — Solid State Question with Solution

Given:

• Atomic substance A with molar mass $M=12{\text{g mol}}^{-1}$


• Cubic crystal structure with edge length $a=300\text{pm}=300\times 10^{-12}\text{m}$


• Density $\rho =3.0{\text{g mL}}^{-1}$


• Avogadro's number $N_A=6.02\times 10^{23}{\text{mol}}^{-1}$

First, calculate the volume of the unit cell $V=a^3=(300\times 10^{-12}\text{m}{)}^3=27\times 10^{-30}{\text{m}}^3$.

Next, calculate the mass of the unit cell using the given density. Density is mass over volume, so mass

$m=\rho \cdot V=3.0{\text{g mL}}^{-1}\cdot 27\times 10^{-30}{\text{m}}^3\cdot \frac{1\text{mL}}{10^{-6}{\text{m}}^3}=81\times 10^{-24}\text{g}$.

Then, calculate the number of moles in one unit cell. The molar mass is mass over number of moles, so

$n=\frac{m}{M}=\frac{81\times 10^{-24}\text{g}}{12{\text{g mol}}^{-1}}=6.75\times 10^{-26}\text{mol}$.

Finally, calculate the number of atoms in one unit cell. The Avogadro constant is the number of atoms per mole,

so number of atoms $=n\cdot N_A=6.75\times 10^{-26}\text{mol}\cdot 6.02\times 10^{23}{\text{mol}}^{-1}=4.06$.

Rounding to the nearest integer, we get 4 atoms. So, there are 4 atoms present in one unit cell of substance A.