When NaCl is dopped with `10^(-5) "mole % of" SrCl_(2)`, what is the no. of cationic vacanies?

Options

A. `10^(-5)xxN_(A)`
B. `10^(-7)xxN_(A)`
C. `2 xx 10^(-7)xxN_(A)`
D. none of these

Answer

To solve the problem of determining the number of cationic vacancies produced when NaCl is doped with \(10^{-5}\) mole percent of SrCl\(_2\), we can follow these steps:

### Step 1: Understand the Doping Process
When NaCl is doped with SrCl\(_2\), the Sr\(^{2+}\) ions replace Na\(^+\) ions in the crystal lattice. For every Sr\(^{2+}\) ion that replaces two Na\(^+\) ions, one cationic vacancy is created.

### Step 2: Calculate the Amount of SrCl\(_2\)
Given that the doping concentration is \(10^{-5}\) mole percent, we can express this in terms of moles.

1 mole percent means 1 mole of solute in 100 moles of solution. Therefore, \(10^{-5}\) mole percent means:
\[ \text{Moles of SrCl}_2 = \frac{10^{-5}}{100} = 10^{-7} \text{ moles} \]

### Step 3: Relate SrCl\(_2\) to Sr\(^{2+}\) Ions
Each mole of SrCl\(_2\) produces 1 mole of Sr\(^{2+}\) ions. Thus, \(10^{-7}\) moles of SrCl\(_2\) will produce:
\[ 10^{-7} \text{ moles of Sr}^{2+} \]

### Step 4: Determine the Number of Cationic Vacancies
Since each Sr\(^{2+}\) ion creates one cationic vacancy when it replaces two Na\(^+\) ions, the number of cationic vacancies created will be equal to the number of Sr\(^{2+}\) ions:
\[ \text{Cationic vacancies} = 10^{-7} \text{ moles of Sr}^{2+} \]

### Step 5: Convert Moles to Number of Vacancies
To find the total number of cationic vacancies, we multiply the number of moles by Avogadro's number (\(N_A\)):
\[ \text{Number of cationic vacancies} = 10^{-7} \times N_A \]

### Final Answer
Thus, the total number of cationic vacancies produced when NaCl is doped with \(10^{-5}\) mole percent of SrCl\(_2\) is:
\[ 10^{-7} N_A \]