`2.56 xx 10^(-3)` equivalent of KOH is required to neutralise `0.125 44g H_(2)XO_(4)`.The atomic mass of X ( in `g//` mol ) is `:`
[ Given `: H_(2) XO_(4)` is a dibasic acid]

Options

A. 16
B. 8
C. 7
D. 32

Answer

To find the atomic mass of X in the compound \( H_2XO_4 \), we can follow these steps:

### Step 1: Understand the reaction
The neutralization reaction between KOH and \( H_2XO_4 \) can be represented as:
\[ KOH + H_2XO_4 \rightarrow K_2XO_4 + 2H_2O \]
Since \( H_2XO_4 \) is a dibasic acid, it can donate 2 protons (H⁺ ions).

### Step 2: Use the equivalent law
According to the equivalent law:
\[ \text{Equivalent of KOH} = \text{Equivalent of } H_2XO_4 \]
Given that the equivalent of KOH is \( 2.56 \times 10^{-3} \).

### Step 3: Calculate the equivalent of \( H_2XO_4 \)
The equivalent of an acid is given by:
\[ \text{Equivalent} = \frac{\text{Given weight}}{\text{Equivalent weight}} \]
For \( H_2XO_4 \):
\[ \text{Equivalent of } H_2XO_4 = \frac{0.12544 \, \text{g}}{\text{Equivalent weight of } H_2XO_4} \]

### Step 4: Determine the equivalent weight of \( H_2XO_4 \)
The equivalent weight of \( H_2XO_4 \) can be calculated as:
\[ \text{Equivalent weight} = \frac{\text{Molar mass}}{n} \]
where \( n \) is the number of protons the acid can donate. Since \( H_2XO_4 \) is dibasic, \( n = 2 \).

### Step 5: Set up the equation
Let the molar mass of \( H_2XO_4 \) be \( M \):
\[ \text{Equivalent weight} = \frac{M}{2} \]
Substituting this into the equivalent equation gives:
\[ 2.56 \times 10^{-3} = \frac{0.12544}{\frac{M}{2}} \]

### Step 6: Solve for \( M \)
Rearranging the equation:
\[ 2.56 \times 10^{-3} = \frac{0.12544 \times 2}{M} \]
\[ M = \frac{0.12544 \times 2}{2.56 \times 10^{-3}} \]
Calculating the right side:
\[ M = \frac{0.25088}{2.56 \times 10^{-3}} = 98 \, \text{g/mol} \]

### Step 7: Find the atomic mass of X
Since the molar mass of \( H_2XO_4 \) is \( 98 \, \text{g/mol} \), we can express it as:
\[ M = 2 \times 1 + \text{Atomic mass of } X + 4 \times 16 \]
\[ 98 = 2 + x + 64 \]
\[ x = 98 - 66 = 32 \, \text{g/mol} \]

### Final Answer
The atomic mass of X is \( 32 \, \text{g/mol} \).

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