Calculate the amound of benzoic acid `(C_(6)H_(5)COOH)` required for preparing `250mL` of `0.15 M` solution in methanol.

Explanation

To calculate the amount of benzoic acid (C₆H₅COOH) required to prepare 250 mL of a 0.15 M solution in methanol, we can follow these steps:

### Step 1: Convert Volume from mL to Liters
The volume of the solution is given as 250 mL. We need to convert this to liters because molarity is expressed in moles per liter.

\[ \text{Volume in liters} = \frac{250 \text{ mL}}{1000} = 0.250 \text{ L} \]

### Step 2: Use the Molarity Formula
Molarity (M) is defined as the number of moles of solute divided by the volume of the solution in liters. We can rearrange this formula to find the number of moles of benzoic acid required.

\[ \text{Molarity} = \frac{\text{Number of moles of solute}}{\text{Volume of solution in liters}} \]

Rearranging gives us:

\[ \text{Number of moles of solute} = \text{Molarity} \times \text{Volume of solution in liters} \]

Substituting the values:

\[ \text{Number of moles of benzoic acid} = 0.15 \, \text{M} \times 0.250 \, \text{L} = 0.0375 \, \text{moles} \]

### Step 3: Calculate the Molar Mass of Benzoic Acid
Next, we need to calculate the molar mass of benzoic acid (C₆H₅COOH). The molar mass can be calculated by adding the atomic masses of all the atoms in the formula.

- Carbon (C): 12 g/mol
- Hydrogen (H): 1 g/mol
- Oxygen (O): 16 g/mol

Calculating the molar mass:

\[ \text{Molar mass of C₆H₅COOH} = (6 \times 12) + (5 \times 1) + (2 \times 16) = 72 + 5 + 32 = 109 \, \text{g/mol} \]

### Step 4: Calculate the Mass of Benzoic Acid
Now that we have the number of moles and the molar mass, we can find the mass of benzoic acid required using the formula:

\[ \text{Mass} = \text{Number of moles} \times \text{Molar mass} \]

Substituting the values:

\[ \text{Mass} = 0.0375 \, \text{moles} \times 122 \, \text{g/mol} = 4.575 \, \text{g} \]

### Conclusion
The amount of benzoic acid required to prepare 250 mL of a 0.15 M solution in methanol is approximately **4.575 grams**.

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