2.0 g of an elements is reacted with aqueous solution containing KOH and `KNO_(3)` to yield `K_(2)XO_(2)` and `NH_(3).NH_(3)` thus liberated is absorbed in 200 " mL of " 0.05 M `H_(2)SO_(4)`. The excess acid required 10 " mL of " 1.5 M `NaOH` for complete neutralisation Which of the following statements is/are correct?

Options

A. The atomic weight of X is 100 g
B. The equivalent weight of X is 50 g
C. The equivalent weight of X is 25 g
D. The atomic weight of X is 200 g

Explanation

To solve the problem step by step, we will analyze the reactions involved and calculate the necessary values to determine the correct statements.

### Step 1: Identify the reactions
The element \( X \) reacts with KOH and KNO3 to produce \( K_2XO_2 \) and \( NH_3 \). The reaction can be summarized as follows:

\[ 4X + 7KOH + KNO_3 \rightarrow 4K_2XO_2 + NH_3 + 2H_2O \]

### Step 2: Calculate the moles of \( H_2SO_4 \)
Given:
- Volume of \( H_2SO_4 = 200 \, \text{mL} = 0.2 \, \text{L} \)
- Molarity of \( H_2SO_4 = 0.05 \, \text{M} \)

To find the number of equivalents of \( H_2SO_4 \):

\[ \text{Normality} = \text{Molarity} \times n \quad (n = 2 \text{ for } H_2SO_4) \]
\[ \text{Normality} = 0.05 \times 2 = 0.1 \, \text{N} \]
\[ \text{Milliequivalents of } H_2SO_4 = 0.1 \, \text{N} \times 0.2 \, \text{L} = 0.02 \, \text{equivalents} = 20 \, \text{milliequivalents} \]

### Step 3: Calculate the moles of \( NaOH \)
Given:
- Volume of \( NaOH = 10 \, \text{mL} = 0.01 \, \text{L} \)
- Molarity of \( NaOH = 1.5 \, \text{M} \)

To find the number of equivalents of \( NaOH \):

\[ \text{Milliequivalents of } NaOH = 1.5 \, \text{M} \times 0.01 \, \text{L} = 0.015 \, \text{equivalents} = 15 \, \text{milliequivalents} \]

### Step 4: Calculate the excess \( H_2SO_4 \)
The amount of \( H_2SO_4 \) that reacted with \( NH_3 \) can be found by subtracting the amount of \( NaOH \) used from the total amount of \( H_2SO_4 \):

\[ \text{Excess } H_2SO_4 = 20 \, \text{milliequivalents} - 15 \, \text{milliequivalents} = 5 \, \text{milliequivalents} \]

### Step 5: Relate \( NH_3 \) to \( H_2SO_4 \)
Since \( NH_3 \) reacts with \( H_2SO_4 \) in a 1:1 ratio, the milliequivalents of \( NH_3 \) produced will be equal to the milliequivalents of \( H_2SO_4 \) that reacted:

\[ \text{Milliequivalents of } NH_3 = 15 \, \text{milliequivalents} \]

### Step 6: Relate \( NH_3 \) to \( X \)
From the balanced equation, 1 mole of \( NH_3 \) corresponds to 4 moles of \( X \):

\[ \text{Milliequivalents of } X = 4 \times \text{milliequivalents of } NH_3 = 4 \times 15 = 60 \, \text{milliequivalents} \]

### Step 7: Calculate the atomic weight of \( X \)
Given that 2 grams of \( X \) corresponds to 60 milliequivalents:

\[ \text{Moles of } X = \frac{60 \, \text{milliequivalents}}{1000} = 0.06 \, \text{moles} \]
\[ \text{Atomic weight of } X = \frac{2 \, \text{g}}{0.06 \, \text{moles}} = \frac{2}{0.06} \approx 33.33 \, \text{g/mol} \]

### Step 8: Determine the equivalent weight of \( X \)
The equivalent weight of \( X \) is calculated as follows:

\[ \text{Equivalent weight} = \frac{\text{Atomic weight}}{n} = \frac{33.33}{2} \approx 16.67 \, \text{g/equiv} \]

### Conclusion
Based on the calculations, we can conclude that the atomic weight of \( X \) is approximately 33.33 g/mol, and the equivalent weight is approximately 16.67 g/equiv.