KBr is doped with $$10^{-5}$$ mole percent of SrBr$$_2$$. The number of cationic vacancies in 1 g of KBr crystal is $$10^{14}$$ ________. (Round off to the Nearest Integer). [Atomic Mass: K: 39.1u, Br: 79.9u, $$N_A = 6.023 \times 10^{23}$$]

To solve this problem, we use the principle of electrical neutrality in ionic crystals.

When $$\mathrm{SrBr_2}$$ is added to $$\mathrm{KBr}$$, the $$\mathrm{Sr^{2+}}$$ ion replaces $$\mathrm{K^+}$$ ions in the crystal lattice.

Since strontium carries a $$+2$$ charge while potassium carries a $$+1$$ charge, one $$\mathrm{Sr^{2+}}$$ ion replaces two $$\mathrm{K^+}$$ ions.

One lattice site is occupied by $$\mathrm{Sr^{2+}}$$ and the second site remains vacant.

Therefore:

$$\mathrm{1\ Sr^{2+}\ ion \longrightarrow 1\ cationic\ vacancy}$$

Step 1: Calculate molar mass of $$\mathrm{KBr}$$

Atomic mass of potassium:

$$\mathrm{K = 39.1\ u}$$

Atomic mass of bromine:

$$\mathrm{Br = 79.9\ u}$$

Therefore,

$$\mathrm{Molar\ mass\ of\ KBr = 39.1 + 79.9 = 119.0\ g\ mol^{-1}}$$

Step 2: Calculate moles of $$\mathrm{KBr}$$ in $$1\ g$$ sample

$$\mathrm{Moles\ of\ KBr = \frac{1}{119}}$$

$$\mathrm{Moles\ of\ KBr = 8.403\times10^{-3}\ mol}$$

Step 3: Apply doping concentration

Given concentration:

$$\mathrm{10^{-5}\ mole\ percent\ of\ SrBr_2}$$

This means:

$$\mathrm{\frac{10^{-5}}{100}=10^{-7}}$$

moles of $$\mathrm{SrBr_2}$$ are present per mole of $$\mathrm{KBr}$$.

Hence,

$$\mathrm{Moles\ of\ SrBr_2 = \frac{1}{119}\times10^{-7}}$$

$$\mathrm{Moles\ of\ SrBr_2 = 8.403\times10^{-10}\ mol}$$

Step 4: Calculate number of cationic vacancies

Since each $$\mathrm{Sr^{2+}}$$ ion creates one cationic vacancy:

$$\mathrm{Moles\ of\ vacancies = 8.403\times10^{-10}\ mol}$$

Using Avogadro number:

$$\mathrm{N_A = 6.023\times10^{23}}$$

Number of vacancies:

$$\mathrm{Vacancies = 8.403\times10^{-10}\times6.023\times10^{23}}$$

$$\mathrm{Vacancies = 5.061\times10^{14}}$$

Thus, the number of cationic vacancies is:

$$\boxed{\mathrm{5.061\times10^{14}}}$$

Rounded value:

$$\boxed{5}$$