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 mixing 100 mL of 3% (w/v) NaOH solution with 100 mL of 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**:
\[ \text{Weight} = \frac{3 \text{ g}}{100 \text{ mL}} \times 100 \text{ mL} = 3 \text{ g} \]
- **Molecular weight of NaOH**:
\[ \text{Molecular weight} = 23 \text{ (Na)} + 16 \text{ (O)} + 1 \text{ (H)} = 40 \text{ g/mol} \]
- **Number of moles in 3% solution**:
\[ \text{Moles} = \frac{\text{Weight}}{\text{Molecular weight}} = \frac{3 \text{ g}}{40 \text{ g/mol}} = 0.075 \text{ moles} \]

### Step 2: Calculate the molarity of the 3% NaOH solution
- **Volume in liters**:
\[ 100 \text{ mL} = 0.1 \text{ L} \]
- **Molarity (M1)**:
\[ M_1 = \frac{\text{Moles}}{\text{Volume}} = \frac{0.075 \text{ moles}}{0.1 \text{ L}} = 0.75 \text{ M} \]

### Step 3: Calculate the number of moles of NaOH in the 9% solution
- **Weight of NaOH in 100 mL of 9% solution**:
\[ \text{Weight} = \frac{9 \text{ g}}{100 \text{ mL}} \times 100 \text{ mL} = 9 \text{ g} \]
- **Number of moles in 9% solution**:
\[ \text{Moles} = \frac{9 \text{ g}}{40 \text{ g/mol}} = 0.225 \text{ moles} \]

### Step 4: Calculate the molarity of the 9% NaOH solution
- **Volume in liters**:
\[ 100 \text{ mL} = 0.1 \text{ L} \]
- **Molarity (M2)**:
\[ M_2 = \frac{0.225 \text{ moles}}{0.1 \text{ L}} = 2.25 \text{ M} \]

### Step 5: Calculate the total moles in the final solution
- **Total moles**:
\[ \text{Total moles} = \text{Moles from 3% solution} + \text{Moles from 9% solution} = 0.075 + 0.225 = 0.3 \text{ moles} \]

### Step 6: Calculate the total volume of the final solution
- **Total volume**:
\[ \text{Total volume} = 100 \text{ mL} + 100 \text{ mL} = 200 \text{ mL} = 0.2 \text{ L} \]

### Step 7: Calculate the molarity of the final solution
- **Final molarity (M)**:
\[ M = \frac{\text{Total moles}}{\text{Total volume}} = \frac{0.3 \text{ moles}}{0.2 \text{ L}} = 1.5 \text{ M} \]

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

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