The molarity of the solution containing 7.1g of `Na_(2)SO_(4)` in 100mL of aqueous solution is

Options

A. 2M
B. `0.5M`
C. `1M`
D. `0.05M`

Explanation

To find the molarity of a solution containing 7.1 g of Na₂SO₄ in 100 mL of aqueous solution, we can follow these steps:

### Step 1: Calculate the molar mass of Na₂SO₄
- The molar mass of Na₂SO₄ can be calculated by adding the atomic masses of its constituent elements:
- Sodium (Na): 23 g/mol (2 atoms)
- Sulfur (S): 32 g/mol (1 atom)
- Oxygen (O): 16 g/mol (4 atoms)

\[ \text{Molar mass of Na}_2\text{SO}_4 = (2 \times 23) + (1 \times 32) + (4 \times 16) = 46 + 32 + 64 = 142 \text{ g/mol} \]

### Step 2: Calculate the number of moles of Na₂SO₄
- The number of moles (n) can be calculated using the formula:

\[ n = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

Substituting the values:

\[ n = \frac{7.1 \text{ g}}{142 \text{ g/mol}} \approx 0.050 \text{ mol} \]

### Step 3: Convert the volume of the solution from mL to L
- The volume of the solution is given as 100 mL. We need to convert this to liters:

\[ \text{Volume in L} = \frac{100 \text{ mL}}{1000} = 0.1 \text{ L} \]

### Step 4: Calculate the molarity of the solution
- Molarity (M) is defined as the number of moles of solute per liter of solution:

\[ M = \frac{n}{\text{Volume in L}} = \frac{0.050 \text{ mol}}{0.1 \text{ L}} = 0.5 \text{ M} \]

### Final Answer:
The molarity of the solution is **0.5 M**.

---