The solution of sulphuric acid contains 80% by weight `H_(2)SO_(4)`. Specific gravity of this solution is 1.71. Its normality is about

Options

A. (a)18.0
B. (b)27.9
C. (c )1.0
D. (d)10.0

Explanation

To find the normality of the sulfuric acid solution, we can follow these steps:

### Step 1: Understand the Given Information
- The solution contains 80% by weight of H₂SO₄.
- The specific gravity of the solution is 1.71.

### Step 2: Calculate the Mass of the Solution
Assuming we have 100 g of the solution:
- Mass of H₂SO₄ = 80 g (since it is 80% by weight)
- Mass of the solution = 100 g

### Step 3: Calculate the Volume of the Solution
Using the specific gravity (SG), we can find the volume of the solution:
- Specific gravity = mass/volume
- Therefore, volume = mass/specific gravity = 100 g / 1.71 g/mL = 58.48 mL

### Step 4: Calculate the Equivalent Weight of H₂SO₄
The molecular weight of H₂SO₄ is calculated as follows:
- H: 1 g/mol × 2 = 2 g/mol
- S: 32 g/mol
- O: 16 g/mol × 4 = 64 g/mol
- Total = 2 + 32 + 64 = 98 g/mol

Now, we need to find the equivalent weight:
- H₂SO₄ can donate 2 protons (H⁺ ions), so the n-factor = 2.
- Equivalent weight = Molecular weight / n-factor = 98 g/mol / 2 = 49 g/equiv.

### Step 5: Calculate Normality
Using the formula for normality:
\[ \text{Normality} = \frac{10 \times \text{Specific Gravity} \times \text{Percentage Purity}}{\text{Equivalent Weight}} \]
Substituting the values:
\[ \text{Normality} = \frac{10 \times 1.71 \times 80}{49} \]

Calculating this:
\[ \text{Normality} = \frac{1368}{49} \approx 27.91 \, \text{N} \]

### Final Answer
The normality of the sulfuric acid solution is approximately **27.91 N**.

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