`100 mL` of mixture of `NaOH` and `Na_(2)SO_(4)` is neutralised by `10 mL` of `0.5 M H_(2) SO_(4)`. Hence,the amount of NaOH in `100 mL` solution is

Options

A. `0.2 g`
B. `0.4 g`
C. `0.6 g`
D. None

Explanation

To find the amount of NaOH in the 100 mL solution that is neutralized by 10 mL of 0.5 M H₂SO₄, we can follow these steps:

### Step 1: Determine the number of equivalents of H₂SO₄ used.
The normality (N) of H₂SO₄ is given as 0.5 M. Since H₂SO₄ is a diprotic acid (it can donate two protons), its normality is twice its molarity.

\[ \text{Normality of H₂SO₄} = 0.5 \, \text{M} \times 2 = 1 \, \text{N} \]

Now, we calculate the number of equivalents of H₂SO₄ used:

\[ \text{Equivalents of H₂SO₄} = \text{Normality} \times \text{Volume (L)} = 1 \, \text{N} \times 0.010 \, \text{L} = 0.01 \, \text{equivalents} \]

### Step 2: Set up the reaction with NaOH.
The reaction between NaOH and H₂SO₄ can be represented as:

\[ \text{NaOH} + \text{H₂SO₄} \rightarrow \text{Na₂SO₄} + \text{H₂O} \]

From the balanced equation, we see that 1 equivalent of H₂SO₄ reacts with 2 equivalents of NaOH. Therefore, the equivalents of NaOH that reacted can be calculated as follows:

\[ \text{Equivalents of NaOH} = 2 \times \text{Equivalents of H₂SO₄} = 2 \times 0.01 = 0.02 \, \text{equivalents} \]

### Step 3: Calculate the normality of NaOH solution.
Now, we know that the total volume of the NaOH solution is 100 mL (or 0.1 L). We can find the normality of the NaOH solution using the formula:

\[ \text{Normality of NaOH (N₁)} = \frac{\text{Equivalents of NaOH}}{\text{Volume of NaOH solution (L)}} \]

Substituting the values:

\[ N₁ = \frac{0.02}{0.1} = 0.2 \, \text{N} \]

### Step 4: Calculate the mass of NaOH.
To find the mass of NaOH, we can use the formula:

\[ \text{Mass} = \text{Normality} \times \text{Volume (L)} \times \text{Equivalent weight} \]

The equivalent weight of NaOH is its molar mass (40 g/mol) since it provides one equivalent of OH⁻.

\[ \text{Mass of NaOH} = 0.2 \, \text{N} \times 0.1 \, \text{L} \times 40 \, \text{g/mol} \]

Calculating this gives:

\[ \text{Mass of NaOH} = 0.2 \times 0.1 \times 40 = 0.8 \, \text{g} \]

### Final Answer:
The amount of NaOH in the 100 mL solution is **0.8 grams**.

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