500 ml of `2 M CH_(3)COOH` solution is mixid with `600 ml 12 % w//v CH_(3)COOH` solution then calculate the final molarity of solution.

Answer

To find the final molarity of the solution after mixing 500 ml of a 2 M CH₃COOH solution with 600 ml of a 12% w/v CH₃COOH solution, we can follow these steps:

### Step 1: Calculate the moles of CH₃COOH in the first solution
The first solution has a molarity (M₁) of 2 M and a volume (V₁) of 500 ml (or 0.5 L).

\[ \text{Moles of CH₃COOH in first solution} = M₁ \times V₁ = 2 \, \text{mol/L} \times 0.5 \, \text{L} = 1 \, \text{mol} \]

### Step 2: Calculate the moles of CH₃COOH in the second solution
The second solution is a 12% w/v solution. This means there are 12 grams of CH₃COOH in 100 ml of solution.

First, we need to find out how many grams are in 600 ml:

\[ \text{Grams of CH₃COOH in 600 ml} = \left( \frac{12 \, \text{g}}{100 \, \text{ml}} \right) \times 600 \, \text{ml} = 72 \, \text{g} \]

Next, we convert grams to moles. The molar mass of CH₃COOH (acetic acid) is approximately 60 g/mol.

\[ \text{Moles of CH₃COOH in second solution} = \frac{72 \, \text{g}}{60 \, \text{g/mol}} = 1.2 \, \text{mol} \]

### Step 3: Calculate the total moles of CH₃COOH
Now, we add the moles from both solutions:

\[ \text{Total moles of CH₃COOH} = 1 \, \text{mol} + 1.2 \, \text{mol} = 2.2 \, \text{mol} \]

### Step 4: Calculate the total volume of the mixed solution
The total volume (V₃) after mixing the two solutions is:

\[ V₃ = 500 \, \text{ml} + 600 \, \text{ml} = 1100 \, \text{ml} = 1.1 \, \text{L} \]

### Step 5: Calculate the final molarity of the solution
Finally, we can calculate the final molarity (M₃) of the mixed solution:

\[ M₃ = \frac{\text{Total moles of CH₃COOH}}{V₃} = \frac{2.2 \, \text{mol}}{1.1 \, \text{L}} = 2 \, \text{M} \]

### Final Answer
The final molarity of the solution is **2 M**.

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