`40.25g` of Glauber's salt `(Na_(2)SO_(4).10H_(2)O)` is dissolved in water to obtain `500mL` of solution of density `1077.5gdm^(-3)`. The molality of `Na_(2)SO_(4)` in solution is about:

Options

A. `0.48molkg^(-1)`
B. `0.24molkg^(-1)`
C. `0.12molkg^(-1)`
D. `0.84molkg^(-1)`

Explanation

To find the molality of Glauber's salt (Na₂SO₄·10H₂O) in the solution, follow these steps:

### Step 1: Calculate the number of moles of Na₂SO₄ in the given mass of Glauber's salt.

1. **Molar mass of Glauber's salt (Na₂SO₄·10H₂O)**:
- Na: 2 × 23 g/mol = 46 g/mol
- S: 1 × 32 g/mol = 32 g/mol
- O: 4 × 16 g/mol = 64 g/mol
- H₂O: 10 × (2 + 16) g/mol = 180 g/mol
- Total = 46 + 32 + 64 + 180 = 322 g/mol

2. **Mass of Glauber's salt**: 40.25 g

3. **Calculate moles of Na₂SO₄**:
- Since 1 mole of Glauber's salt contains 1 mole of Na₂SO₄, we first need to find the mass of Na₂SO₄ in 40.25 g of Glauber's salt.
- Mass of Na₂SO₄ = (mass of Glauber's salt) × (molar mass of Na₂SO₄ / molar mass of Glauber's salt)
- Molar mass of Na₂SO₄ = 142 g/mol
- Moles of Na₂SO₄ = (40.25 g) × (142 g/mol / 322 g/mol)

\[ \text{Moles of Na₂SO₄} = 40.25 \times \frac{142}{322} = 17.75 \text{ g} \]

### Step 2: Calculate the mass of the solution.

1. **Density of the solution**: 1.0775 g/dm³
2. **Volume of the solution**: 500 mL = 0.5 dm³

3. **Calculate mass of the solution**:
\[ \text{Mass of solution} = \text{Density} \times \text{Volume} = 1.0775 \, \text{g/dm}^3 \times 0.5 \, \text{dm}^3 = 538.75 \, \text{g} \]

### Step 3: Calculate the mass of the solvent (water).

1. **Mass of solvent (water)**:
\[ \text{Mass of solvent} = \text{Mass of solution} - \text{Mass of solute} \]
\[ \text{Mass of solvent} = 538.75 \, \text{g} - 17.75 \, \text{g} = 521 \, \text{g} \]

### Step 4: Convert the mass of the solvent to kg.

1. **Convert grams to kg**:
\[ \text{Mass of solvent in kg} = \frac{521 \, \text{g}}{1000} = 0.521 \, \text{kg} \]

### Step 5: Calculate the molality of Na₂SO₄.

1. **Molality (m)** is defined as the number of moles of solute per kg of solvent:
\[ \text{Molality} = \frac{\text{Number of moles of Na₂SO₄}}{\text{Mass of solvent in kg}} = \frac{0.125 \, \text{moles}}{0.521 \, \text{kg}} \approx 0.24 \, \text{mol/kg} \]

### Final Answer:
The molality of Na₂SO₄ in the solution is approximately **0.24 mol/kg**.

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