What volume of `0.1 ` N `HNO_(3)` solution can be prepared from `6.3 " g of " HNO_(3)` ?

Options

A. 1 litre
B. 2 litres
C. `0.5` litres
D. 5 litres

Answer

To solve the problem of determining what volume of 0.1 N HNO₃ solution can be prepared from 6.3 g of HNO₃, we will follow these steps:

### Step 1: Determine the equivalent weight of HNO₃
The equivalent weight of a substance is calculated using the formula:
\[ \text{Equivalent weight} = \frac{\text{Molar mass}}{n} \]
For HNO₃, the molar mass is approximately 63 g/mol and the n-factor for HNO₃ (which is a strong acid) is 1.

So, the equivalent weight of HNO₃ is:
\[ \text{Equivalent weight} = \frac{63 \text{ g/mol}}{1} = 63 \text{ g/equiv} \]

### Step 2: Calculate the number of equivalents in 6.3 g of HNO₃
To find the number of equivalents, we use the formula:
\[ \text{Number of equivalents} = \frac{\text{Weight of solute}}{\text{Equivalent weight}} \]
Substituting the values:
\[ \text{Number of equivalents} = \frac{6.3 \text{ g}}{63 \text{ g/equiv}} = 0.1 \text{ equiv} \]

### Step 3: Use the normality formula to find the volume
Normality (N) is defined as:
\[ N = \frac{\text{Number of equivalents}}{\text{Volume in liters}} \]
We know the normality (0.1 N) and the number of equivalents (0.1 equiv). Rearranging the formula to find the volume:
\[ \text{Volume in liters} = \frac{\text{Number of equivalents}}{N} \]
Substituting the values:
\[ \text{Volume in liters} = \frac{0.1 \text{ equiv}}{0.1 \text{ N}} = 1 \text{ L} \]

### Conclusion
The volume of 0.1 N HNO₃ solution that can be prepared from 6.3 g of HNO₃ is **1 liter**.

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