`HNO_(3)` used as a reagent has specific gravity of `1.42 g mL^(-1)` and contains 70% by strength `HNO_(3)`. Calcualte:
Normality of acid

Explanation

To calculate the normality of HNO₃, we will follow these steps:

### Step 1: Determine the molecular weight of HNO₃
The molecular weight (molar mass) of HNO₃ can be calculated as follows:
- Hydrogen (H) = 1 g/mol
- Nitrogen (N) = 14 g/mol
- Oxygen (O) = 16 g/mol (and there are 3 oxygen atoms)

So, the molecular weight of HNO₃ is:
\[ \text{Molecular weight of HNO}_3 = 1 + 14 + (3 \times 16) = 1 + 14 + 48 = 63 \text{ g/mol} \]

### Step 2: Use the formula for normality
Normality (N) can be calculated using the formula:
\[ N = \frac{\text{Percentage by weight} \times 10 \times \text{Density}}{\text{Equivalent weight}} \]

### Step 3: Identify the values to substitute
- Percentage by weight of HNO₃ = 70%
- Density of the solution = 1.42 g/mL
- Equivalent weight of HNO₃ = Molecular weight (since the valency factor is 1) = 63 g/equiv

### Step 4: Substitute the values into the formula
Now substituting the values into the formula:
\[ N = \frac{70 \times 10 \times 1.42}{63} \]

### Step 5: Calculate the normality
Calculating the numerator:
\[ 70 \times 10 \times 1.42 = 994 \]
Now divide by the equivalent weight:
\[ N = \frac{994}{63} \approx 15.78 \]

### Final Answer
The normality of the HNO₃ solution is approximately **15.78 N**.

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