The density of `NH_(4)OH` solution is `0.6g//mL`. It contains `34%` by weight of `NH_(4)OH`. Calculate the normality of the solution:

Options

A. `4.8N`
B. `10N`
C. `0.5N`
D. `5.8N`

Explanation

To calculate the normality of the `NH₄OH` solution, we will follow these steps:

### Step 1: Calculate the mass of the solution
Given the density of the `NH₄OH` solution is `0.6 g/mL`, we can find the mass of `1 L` (or `1000 mL`) of the solution.

\[ \text{Mass of the solution} = \text{Density} \times \text{Volume} = 0.6 \, \text{g/mL} \times 1000 \, \text{mL} = 600 \, \text{g} \]

### Step 2: Calculate the mass of `NH₄OH` in the solution
The solution contains `34%` by weight of `NH₄OH`. Therefore, the mass of `NH₄OH` can be calculated as follows:

\[ \text{Mass of } NH₄OH = \frac{34}{100} \times 600 \, \text{g} = 204 \, \text{g} \]

### Step 3: Calculate the equivalent mass of `NH₄OH`
To find the normality, we need the equivalent mass of `NH₄OH`. The molar mass of `NH₄OH` is approximately `35 + 1 + 16 = 52 \, \text{g/mol}`. Since `NH₄OH` can donate one hydroxide ion (`OH⁻`), its acidity is `1`. Thus, the equivalent mass is calculated as:

\[ \text{Equivalent mass of } NH₄OH = \frac{\text{Molar mass}}{\text{Acidity}} = \frac{52 \, \text{g/mol}}{1} = 52 \, \text{g/equiv} \]

### Step 4: Calculate the normality of the solution
Normality (N) is calculated using the formula:

\[ N = \frac{\text{Mass of solute (g)}}{\text{Equivalent mass (g/equiv)} \times \text{Volume (L)}} \]

Substituting the values we have:

\[ N = \frac{204 \, \text{g}}{52 \, \text{g/equiv} \times 1 \, \text{L}} = \frac{204}{52} \approx 3.92 \, N \]

### Final Step: Conclusion
The normality of the `NH₄OH` solution is approximately `3.92 N`.

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