Liquid `NH_(3)` dissociation to a slight extent, At a certain temp. its self dissociation constant `K_(SDC(NH_(3))`=`10^(-30)`. The number of `NH_(4)^(+)` ions are present per `100 cm^(3)` of pure liquid are :

Options

A. `10^(-15)`
B. `6.022xx10^(8)`
C. `6.022xx10^(7)`
D. `6.022xx10^(6)`

Answer

To solve the problem of finding the number of \( \text{NH}_4^+ \) ions present per 100 cm³ of pure liquid ammonia (NH₃), we will follow these steps:

### Step 1: Write the dissociation reaction
The dissociation of liquid ammonia can be represented as:
\[ 2 \text{NH}_3 \rightleftharpoons \text{NH}_4^+ + \text{NH}_2^- \]

### Step 2: Define the self-dissociation constant
The self-dissociation constant (\( K_{SDC} \)) for ammonia is given by:
\[ K_{SDC} = [\text{NH}_4^+][\text{NH}_2^-] \]
Since both ions are produced in equal amounts, we can denote their concentrations as \( [\text{NH}_4^+] = [\text{NH}_2^-] = x \). Thus, we can rewrite the equation as:
\[ K_{SDC} = x^2 \]

### Step 3: Substitute the given value of \( K_{SDC} \)
We are given that \( K_{SDC} = 10^{-30} \). Therefore, we can set up the equation:
\[ x^2 = 10^{-30} \]

### Step 4: Solve for \( x \)
To find \( x \), we take the square root of both sides:
\[ x = \sqrt{10^{-30}} = 10^{-15} \text{ M} \]

### Step 5: Calculate the number of ions in 1000 cm³
To find the number of \( \text{NH}_4^+ \) ions in 1000 cm³ (1 L), we use the concentration and Avogadro's number:
\[ \text{Number of } \text{NH}_4^+ \text{ ions} = x \times \text{Avogadro's number} \times \text{Volume in L} \]
\[ = 10^{-15} \text{ mol/L} \times 6.022 \times 10^{23} \text{ ions/mol} \times 1 \text{ L} \]
\[ = 6.022 \times 10^{8} \text{ ions} \]

### Step 6: Calculate the number of ions in 100 cm³
Since we need the number of ions in 100 cm³, we will scale down our previous result:
\[ \text{Number of } \text{NH}_4^+ \text{ ions in 100 cm}^3 = \frac{6.022 \times 10^{8} \text{ ions}}{10} = 6.022 \times 10^{7} \text{ ions} \]

### Final Answer
The number of \( \text{NH}_4^+ \) ions present per 100 cm³ of pure liquid ammonia is:
\[ \boxed{6.022 \times 10^{7}} \]