The hydrated salt `Na_(2)CO_(3).xH_(2)O` undergoes ` 63%` loss in mass on heating and becomes anhydrous . The value x is :

Options

A. 10
B. 12
C. 8
D. 18

Answer

To solve the problem of finding the value of \( x \) in the hydrated salt \( \text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O} \) that undergoes a 63% loss in mass upon heating, we can follow these steps:

### Step 1: Understand the Mass Loss
When the hydrated salt is heated, it loses water and becomes anhydrous. A 63% loss in mass means that 37% of the original mass remains after heating.

### Step 2: Set Up the Equation
Let the molar mass of the hydrated salt \( \text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O} \) be \( Y \). The molar mass of \( \text{Na}_2\text{CO}_3 \) is \( 106 \, \text{g/mol} \) and the molar mass of water \( \text{H}_2\text{O} \) is \( 18 \, \text{g/mol} \). Therefore, we can express \( Y \) as:
\[ Y = 106 + x \cdot 18 \]

### Step 3: Calculate the Remaining Mass
Since 37% of the mass remains after heating, we can write:
\[ 0.37Y = 106 \]

### Step 4: Substitute for \( Y \)
Substituting the expression for \( Y \) into the equation gives:
\[ 0.37(106 + x \cdot 18) = 106 \]

### Step 5: Solve for \( x \)
Expanding and rearranging the equation:
\[ 39.62 + 0.37 \cdot 18x = 106 \]
\[ 0.37 \cdot 18x = 106 - 39.62 \]
\[ 0.37 \cdot 18x = 66.38 \]
\[ 18x = \frac{66.38}{0.37} \]
\[ 18x \approx 179.68 \]
\[ x \approx \frac{179.68}{18} \approx 9.98 \]

### Step 6: Round to the Nearest Whole Number
Since \( x \) must be a whole number, we round \( 9.98 \) to \( 10 \).

### Conclusion
Thus, the value of \( x \) is approximately \( 10 \).

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