50 mL of 0.5 M oxalic acid is needed to neutralize 25 mL of sodium hydroxide solution. The amount of NaOH in 50 mL of the given sodium hydroxide solution is _________.

Options

A. 80g
B. 20 g
C. 10 g
D. 4 gm

Explanation

To solve the problem, we need to determine the amount of sodium hydroxide (NaOH) in 50 mL of the given sodium hydroxide solution that is neutralized by 50 mL of 0.5 M oxalic acid (H2C2O4).

### Step-by-Step Solution:

1. **Write the balanced chemical equation**:
The reaction between oxalic acid and sodium hydroxide can be represented as:
\[ H_2C_2O_4 + 2 NaOH \rightarrow Na_2C_2O_4 + 2 H_2O \]
From the equation, we see that 1 mole of oxalic acid reacts with 2 moles of sodium hydroxide.

2. **Calculate the number of moles of oxalic acid**:
We know the molarity (M) and volume (V) of oxalic acid:
\[ \text{Moles of } H_2C_2O_4 = M \times V = 0.5 \, \text{mol/L} \times 0.050 \, \text{L} = 0.025 \, \text{moles} \]

3. **Determine the moles of sodium hydroxide needed**:
From the balanced equation, 1 mole of oxalic acid reacts with 2 moles of NaOH. Therefore, the moles of NaOH required are:
\[ \text{Moles of } NaOH = 2 \times \text{Moles of } H_2C_2O_4 = 2 \times 0.025 = 0.050 \, \text{moles} \]

4. **Calculate the molarity of sodium hydroxide solution**:
We know that the volume of the sodium hydroxide solution used is 25 mL (0.025 L). The molarity (M) of NaOH can be calculated as:
\[ M_{NaOH} = \frac{\text{Moles of } NaOH}{\text{Volume of solution in L}} = \frac{0.050 \, \text{moles}}{0.025 \, \text{L}} = 2 \, \text{M} \]

5. **Calculate the amount (mass) of sodium hydroxide in 50 mL of the solution**:
The molar mass of NaOH is approximately 40 g/mol. Therefore, the mass of NaOH in 50 mL of the sodium hydroxide solution can be calculated as follows:
\[ \text{Mass of } NaOH = \text{Moles} \times \text{Molar mass} = 0.050 \, \text{moles} \times 40 \, \text{g/mol} = 2 \, \text{g} \]

6. **Final Calculation**:
Since the question asks for the amount of NaOH in 50 mL of the given sodium hydroxide solution, we need to consider that the molarity we calculated (2 M) corresponds to the 25 mL used in the reaction. Therefore, we scale it up to 50 mL:
\[ \text{Mass of NaOH in 50 mL} = 2 \, \text{g} \times 2 = 4 \, \text{g} \]

### Conclusion:
The amount of NaOH in 50 mL of the given sodium hydroxide solution is **4 grams**.