Molecular weight of oxalic acid is 126. the weight of oxalic acid required to neutralise 100cc of normal solution of NaOH is

Options

A. 6.3 gm
B. 126 gm
C. 530 gm
D. 63 gm

Answer

To find the weight of oxalic acid required to neutralize 100 cc of a normal solution of NaOH, we can follow these steps:

### Step 1: Understand the Reaction
Oxalic acid (H₂C₂O₄) reacts with sodium hydroxide (NaOH) in a neutralization reaction. The balanced reaction is:
\[ \text{H}_2\text{C}_2\text{O}_4 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{C}_2\text{O}_4 + 2 \text{H}_2\text{O} \]
This shows that one molecule of oxalic acid reacts with two molecules of NaOH.

### Step 2: Determine the Normality and Volume of NaOH
Given that the NaOH solution is 1 normal (1 N) and the volume is 100 cc (which is equivalent to 100 mL).

### Step 3: Calculate the Equivalent of NaOH
The equivalent of NaOH can be calculated using the formula:
\[ \text{Equivalent of NaOH} = \text{Normality} \times \text{Volume (in L)} \]
Converting 100 mL to liters:
\[ 100 \text{ mL} = 0.1 \text{ L} \]
Now substituting the values:
\[ \text{Equivalent of NaOH} = 1 \, \text{N} \times 0.1 \, \text{L} = 0.1 \, \text{equivalents} \]

### Step 4: Determine the n-factor of Oxalic Acid
Oxalic acid (H₂C₂O₄) can donate 2 protons (H⁺ ions), so its n-factor is 2.

### Step 5: Calculate the Equivalent Weight of Oxalic Acid
The equivalent weight of oxalic acid can be calculated using the formula:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{n-factor}} \]
Given that the molecular weight of oxalic acid is 126 g/mol:
\[ \text{Equivalent weight} = \frac{126 \, \text{g/mol}}{2} = 63 \, \text{g/equiv} \]

### Step 6: Calculate the Equivalent of Oxalic Acid
Using the equivalence principle (equivalents of acid = equivalents of base), we can set up the equation:
\[ \text{Equivalent of oxalic acid} = \text{Equivalent of NaOH} \]
Let \( W \) be the weight of oxalic acid required:
\[ \frac{W}{\text{Equivalent weight of oxalic acid}} = 0.1 \]
Substituting the equivalent weight:
\[ \frac{W}{63} = 0.1 \]

### Step 7: Solve for W
Now, we can solve for \( W \):
\[ W = 0.1 \times 63 = 6.3 \, \text{grams} \]

### Final Answer
The weight of oxalic acid required to neutralize 100 cc of a normal solution of NaOH is **6.3 grams**.

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