The compound `Na_(2)CO_(3).xH_(2)O` has 50% `H_(2)O` by mass. The value of 'x' is

Options

A. 4
B. 5
C. 6
D. 7

Explanation

To find the value of 'x' in the compound Na₂CO₃·xH₂O, which has 50% H₂O by mass, we can follow these steps:

### Step 1: Calculate the molar mass of Na₂CO₃
- The molar mass of sodium (Na) = 23 g/mol
- The molar mass of carbon (C) = 12 g/mol
- The molar mass of oxygen (O) = 16 g/mol

The formula for Na₂CO₃ contains:
- 2 sodium atoms: 2 × 23 g/mol = 46 g/mol
- 1 carbon atom: 1 × 12 g/mol = 12 g/mol
- 3 oxygen atoms: 3 × 16 g/mol = 48 g/mol

Now, adding these together:
\[ \text{Molar mass of Na₂CO₃} = 46 + 12 + 48 = 106 \text{ g/mol} \]

### Step 2: Set up the equation based on the mass percentage
Given that the compound has 50% H₂O by mass, we can express this as:
\[ \frac{\text{mass of H₂O}}{\text{mass of Na₂CO₃} + \text{mass of H₂O}} = 0.50 \]

Let the mass of H₂O be \( x \times 18 \) g (since the molar mass of H₂O is 18 g/mol). The total mass of the compound is:
\[ 106 + 18x \]

Thus, we can write the equation:
\[ \frac{18x}{106 + 18x} = 0.50 \]

### Step 3: Cross-multiply to solve for x
Cross-multiplying gives:
\[ 18x = 0.50(106 + 18x) \]

Expanding the right side:
\[ 18x = 53 + 9x \]

### Step 4: Rearranging the equation
Now, rearranging the equation to isolate x:
\[ 18x - 9x = 53 \]
\[ 9x = 53 \]

### Step 5: Solve for x
Dividing both sides by 9:
\[ x = \frac{53}{9} \approx 5.89 \]

Since 'x' must be a whole number, we round this to the nearest whole number:
\[ x \approx 6 \]

### Conclusion
Thus, the value of 'x' is 6.

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