A hydrate of `Na_(2)SO_(3)` has 50% water by mass. It is

Options

A. `Na_(2)SO_(3).5H_(2)O`
B. `Na_(2)SO_(3).6H_(2)O`
C. `Na_(2)SO_(3).7H_(2)O`
D. `Na_(2)SO_(3).2H_(2)O`

Explanation

To solve the problem, we need to determine the formula of the hydrate of sodium sulfite (Na₂SO₃) that contains 50% water by mass. We can follow these steps:

### Step 1: Understand the Composition of the Hydrate
A hydrate of Na₂SO₃ can be represented as Na₂SO₃·xH₂O, where x is the number of water molecules associated with each formula unit of Na₂SO₃.

### Step 2: Set Up the Mass Relationship
Since the hydrate has 50% water by mass, this means that the mass of water is equal to the mass of Na₂SO₃ in the hydrate.

Let:
- Mass of Na₂SO₃ = m
- Mass of water = m (because it is 50% by mass)

Thus, the total mass of the hydrate is:
\[ \text{Total mass} = m + m = 2m \]

### Step 3: Calculate Molar Masses
Now, we need to find the molar masses:
- Molar mass of Na₂SO₃:
- Na: 23 g/mol × 2 = 46 g/mol
- S: 32 g/mol
- O: 16 g/mol × 3 = 48 g/mol
- Total = 46 + 32 + 48 = 126 g/mol

- Molar mass of H₂O:
- H: 1 g/mol × 2 = 2 g/mol
- O: 16 g/mol
- Total = 2 + 16 = 18 g/mol

### Step 4: Set Up the Equation
Since the mass of Na₂SO₃ is equal to the mass of water, we can express this in terms of moles:
- Moles of Na₂SO₃ = \(\frac{m}{126}\)
- Moles of H₂O = \(\frac{m}{18}\)

### Step 5: Establish the Ratio
The ratio of moles of Na₂SO₃ to moles of H₂O is:
\[ \text{Ratio} = \frac{\frac{m}{126}}{\frac{m}{18}} = \frac{18}{126} = \frac{1}{7} \]

### Step 6: Determine the Value of x
From the ratio, we find that for every 1 mole of Na₂SO₃, there are 7 moles of water. Thus, we can conclude:
\[ x = 7 \]

### Step 7: Write the Formula of the Hydrate
The formula of the hydrate is:
\[ \text{Na}_2\text{SO}_3 \cdot 7\text{H}_2\text{O} \]

### Final Answer
The hydrate of Na₂SO₃ that has 50% water by mass is Na₂SO₃·7H₂O.

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