The specific gravity of `98%H_(2)SO_(4)` is 1.8 g/cc. 50 ml of this solution is mixed with 1750 ml of pure water. Molarity of resulting solution is

Options

A. 0.2 M
B. 0.5 M
C. 0.1 M
D. 1 M

Explanation

To find the molarity of the resulting solution after mixing 50 ml of 98% H₂SO₄ with 1750 ml of pure water, we can follow these steps:

### Step 1: Calculate the mass of the 50 ml of 98% H₂SO₄ solution.
Given:
- Specific gravity of 98% H₂SO₄ = 1.8 g/cc
- Volume of the solution = 50 ml

Using the formula for mass:
\[ \text{Mass} = \text{Volume} \times \text{Density} \]
\[ \text{Mass} = 50 \, \text{ml} \times 1.8 \, \text{g/ml} = 90 \, \text{g} \]

### Step 2: Calculate the mass of H₂SO₄ in the solution.
Since the solution is 98% H₂SO₄ by mass:
\[ \text{Mass of H₂SO₄} = 0.98 \times \text{Total mass of solution} \]
\[ \text{Mass of H₂SO₄} = 0.98 \times 90 \, \text{g} = 88.2 \, \text{g} \]

### Step 3: Calculate the number of moles of H₂SO₄.
The molar mass of H₂SO₄ (sulfuric acid) is approximately 98 g/mol.
\[ \text{Number of moles of H₂SO₄} = \frac{\text{Mass of H₂SO₄}}{\text{Molar mass of H₂SO₄}} = \frac{88.2 \, \text{g}}{98 \, \text{g/mol}} \approx 0.9 \, \text{mol} \]

### Step 4: Calculate the total volume of the resulting solution.
The total volume after mixing 50 ml of H₂SO₄ solution with 1750 ml of water:
\[ \text{Total volume} = 50 \, \text{ml} + 1750 \, \text{ml} = 1800 \, \text{ml} = 1.8 \, \text{L} \]

### Step 5: Calculate the molarity of the resulting solution.
Molarity (M) is defined as the number of moles of solute per liter of solution:
\[ \text{Molarity} = \frac{\text{Number of moles of H₂SO₄}}{\text{Total volume in liters}} = \frac{0.9 \, \text{mol}}{1.8 \, \text{L}} = 0.5 \, \text{M} \]

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

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