100mL, 3% (w/v) NaOH solution is mixed with 100 ml, 9% (w/v)
NaOH solution. The molarity of final solution is

Answer

To find the molarity of the final solution when 100 mL of a 3% (w/v) NaOH solution is mixed with 100 mL of a 9% (w/v) NaOH solution, we can follow these steps:

### Step 1: Calculate the number of moles of NaOH in the 3% solution.
- **Weight of NaOH in 100 mL of 3% solution**:
\[ \text{Weight} = \frac{3}{100} \times 100 \text{ mL} = 3 \text{ g} \]
- **Molar mass of NaOH**:
\[ \text{Molar mass} = 23 \text{ (Na)} + 16 \text{ (O)} + 1 \text{ (H)} = 40 \text{ g/mol} \]
- **Number of moles (N1)**:
\[ N1 = \frac{\text{Weight}}{\text{Molar mass}} = \frac{3 \text{ g}}{40 \text{ g/mol}} = 0.075 \text{ moles} \]

### Step 2: Calculate the number of moles of NaOH in the 9% solution.
- **Weight of NaOH in 100 mL of 9% solution**:
\[ \text{Weight} = \frac{9}{100} \times 100 \text{ mL} = 9 \text{ g} \]
- **Number of moles (N2)**:
\[ N2 = \frac{\text{Weight}}{\text{Molar mass}} = \frac{9 \text{ g}}{40 \text{ g/mol}} = 0.225 \text{ moles} \]

### Step 3: Calculate the total number of moles of NaOH in the final solution.
- **Total moles (N_total)**:
\[ N_{\text{total}} = N1 + N2 = 0.075 \text{ moles} + 0.225 \text{ moles} = 0.300 \text{ moles} \]

### Step 4: Calculate the total volume of the final solution.
- **Total volume**:
\[ V_{\text{total}} = 100 \text{ mL} + 100 \text{ mL} = 200 \text{ mL} = 0.200 \text{ L} \]

### Step 5: Calculate the molarity of the final solution.
- **Molarity (M)**:
\[ M = \frac{N_{\text{total}}}{V_{\text{total}}} = \frac{0.300 \text{ moles}}{0.200 \text{ L}} = 1.5 \text{ M} \]

### Final Answer:
The molarity of the final solution is **1.5 M**.

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