Animated Solution for Chemistry - States of Matter: KBr is doped with $10^{-5}$ mole per cent of ${\text{SrBr}}_2$. The number of cationic vacancies in $1\text{ g}$ of KBr crystal is ......... $10^{14}$ (Round off to the nearest integer). [Atomic mass : $\text{K}=39.1\text{ u}$, $\text{Br}=79.9\text{ u}$, $N_A=6.023\times 10^{23}$]

Imagine a pristine, perfect crystal of potassium bromide ($\text{KBr}$). It is a beautiful, repeating geometric grid where positively charged potassium ions (${\text{K}}^{+}$) and negatively charged bromide ions (${\text{Br}}^{-}$) sit in perfect harmony. Every charge is balanced, and every lattice site is occupied.

But what happens when we introduce a tiny amount of a foreign substance, like strontium bromide (${\text{SrBr}}_2$)? This process is known as doping. When a strontium ion (${\text{Sr}}^{2+}$) enters this perfect lattice, it disrupts the delicate balance.

Here is the fundamental catch of solid-state chemistry: electrical neutrality must be maintained at all costs.

A strontium ion carries a $+2$ charge, while a potassium ion only carries a $+1$ charge. To keep the overall charge of the crystal neutral, a single ${\text{Sr}}^{2+}$ ion must force two ${\text{K}}^{+}$ ions to leave the lattice.

However, the ${\text{Sr}}^{2+}$ ion can only physically occupy one of those newly emptied spots. The second spot is left completely empty. This empty spot is what we call a cationic vacancy. Therefore, the number of cationic vacancies created is exactly equal to the number of ${\text{Sr}}^{2+}$ ions added.

Moles of KBr

Before we can count the vacancies, we need to know how much of the host crystal we have. We are given $1\text{ g}$ of $\text{KBr}$.

First, let's calculate the molar mass of $\text{KBr}$: $M_{\text{KBr}}=39.1+79.9=119\text{ g/mol}$

Now, we can find the total moles of $\text{KBr}$ in our sample: $n_{\text{KBr}}=\frac{1}{119}\text{ mol}$

The problem states that the crystal is doped with $10^{-5}\text{ mole \%}$ of ${\text{SrBr}}_2$. This is where many students make a critical error. Mole percent is not the same as a simple molar ratio. Just like $5\%$ means $\frac{5}{100}$, $10^{-5}\%$ means we must divide by $100$.

Let's calculate the exact moles of the dopant: $n_{{\text{SrBr}}_2}=\frac{10^{-5}}{100}\times n_{\text{KBr}}$

Avogadro's Call

Since one ${\text{Sr}}^{2+}$ ion creates exactly one vacancy, the total number of vacancies is simply the total number of ${\text{Sr}}^{2+}$ ions. To convert moles into an exact particle count, we multiply by Avogadro's number ($N_A=6.022\times 10^{23}$).

Let's simplify the powers of ten: $\text{Vacancies}=\frac{6.022}{119}\times 10^{16}$

Now, performing the division: $\frac{6.022}{119}\approx 0.0506$

Substituting this back: $\text{Vacancies}\approx 0.0506\times 10^{16}=5.06\times 10^{14}$

The question asks us to round off to the nearest integer for the coefficient of $10^{14}$. Since $5.06$ is closest to $5$, our final answer is $5$.