250 ml of a solution contains 6.3 grams of oxalic acid (mol.wt. = 126). What is the volume (inlitres) owater to be added to this solution to make it a 0.1 N solution ?

Options

A. 750
B. `7.5`
C. `0.075`
D. `0.75`

Explanation

To solve the problem of determining the volume of water to be added to a solution containing oxalic acid to achieve a 0.1 N solution, we can follow these steps:

### Step 1: Calculate the number of grams equivalent of oxalic acid.
The formula for calculating the number of grams equivalent is:
\[ \text{Number of grams equivalent} = \frac{\text{Weight of solute}}{\text{Equivalent weight}} \]
The equivalent weight of oxalic acid can be calculated as:
\[ \text{Equivalent weight} = \frac{\text{Molar mass}}{n \text{-factor}} \]
Where:
- Molar mass of oxalic acid (C₂H₂O₄) = 126 g/mol
- n-factor for oxalic acid = 2 (since it can donate 2 protons)

Thus, the equivalent weight is:
\[ \text{Equivalent weight} = \frac{126}{2} = 63 \text{ g/equiv} \]

Now, substituting the values:
\[ \text{Number of grams equivalent} = \frac{6.3 \text{ g}}{63 \text{ g/equiv}} = 0.1 \text{ equiv} \]

### Step 2: Calculate the normality of the initial solution.
Normality (N) is given by:
\[ N = \frac{\text{Number of grams equivalent}}{\text{Volume of solution in liters}} \]
The volume of the solution is given as 250 mL, which is:
\[ \text{Volume in liters} = \frac{250}{1000} = 0.25 \text{ L} \]

Now substituting the values:
\[ N = \frac{0.1 \text{ equiv}}{0.25 \text{ L}} = 0.4 \text{ N} \]

### Step 3: Use the dilution formula to find the final volume needed.
The dilution formula is:
\[ N_1V_1 = N_2V_2 \]
Where:
- \(N_1 = 0.4 \text{ N}\) (initial normality)
- \(V_1 = 0.25 \text{ L}\) (initial volume)
- \(N_2 = 0.1 \text{ N}\) (final normality)
- \(V_2\) = final volume (unknown)

Substituting the known values:
\[ 0.4 \times 0.25 = 0.1 \times V_2 \]
\[ 0.1 = 0.1 \times V_2 \]
\[ V_2 = \frac{0.1}{0.1} = 1 \text{ L} \]

### Step 4: Calculate the volume of water to be added.
The final volume \(V_2\) is 1 L, and the initial volume \(V_1\) is 0.25 L. Therefore, the volume of water to be added is:
\[ \text{Volume of water} = V_2 - V_1 = 1 \text{ L} - 0.25 \text{ L} = 0.75 \text{ L} \]

### Final Answer:
The volume of water to be added is **0.75 liters**.

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