The normality of 26% (wt/vol) solution of ammonia (density = 0.855 ) is approximately :

Options

A. 1.5
B. 0.4
C. 15.3
D. 4

Answer

To find the normality of a 26% (wt/vol) solution of ammonia, we can follow these steps:

### Step 1: Understand the given information
We are given a 26% (wt/vol) solution of ammonia, which means there are 26 grams of ammonia (NH₃) in 100 mL of solution.

### Step 2: Calculate the equivalent mass of ammonia
The molecular mass of ammonia (NH₃) is calculated as follows:
- Nitrogen (N) = 14 g/mol
- Hydrogen (H) = 1 g/mol (and there are 3 hydrogen atoms in ammonia)

Thus, the molecular mass of NH₃ = 14 + (1 × 3) = 17 g/mol.

Next, we need to find the equivalent mass. The n-factor for ammonia, which is the number of moles of H⁺ it can accept, is 1. Therefore, the equivalent mass of ammonia is the same as its molecular mass:
- Equivalent mass = Molecular mass / n-factor = 17 g/mol / 1 = 17 g/equiv.

### Step 3: Calculate the volume of the solution in liters
Since we have 100 mL of solution, we convert this to liters:
- Volume in liters = 100 mL × (1 L / 1000 mL) = 0.1 L.

### Step 4: Use the normality formula
The formula for normality (N) is given by:
\[ N = \frac{\text{mass of solute (g)}}{\text{equivalent mass (g/equiv)} \times \text{volume of solution (L)}} \]

Substituting the values we have:
- Mass of solute = 26 g
- Equivalent mass = 17 g/equiv
- Volume of solution = 0.1 L

So,
\[ N = \frac{26 \, \text{g}}{17 \, \text{g/equiv} \times 0.1 \, \text{L}} \]

### Step 5: Calculate normality
Now, we can calculate the normality:
\[ N = \frac{26}{17 \times 0.1} = \frac{26}{1.7} \approx 15.29 \]

Thus, the normality of the solution is approximately 15.3 N.

### Final Answer
The normality of the 26% (wt/vol) solution of ammonia is approximately **15.3 N**.

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