Q.Calculate the amount of benzoic acid ($C_6H_5COOH$) required for preparing 250 mL of 0.15 M solution in methanol.

The key idea is Molality Calculation — but here the question asks for molarity, so we use the molarity formula directly.

Step 1: Molar mass of benzoic acid

$C_6H_5COOH = 7 \times 12 + 6 \times 1 + 2 \times 16 = 84 + 6 + 32 = 122\ \text{g/mol}$.

Step 2: Moles required

Molarity $M = \frac{\text{moles}}{\text{volume in L}}$, so …

Why This Approach Works

Molarity tells us the number of moles of solute dissolved in one litre of solution. When you know the volume you want to prepare, you can scale the moles proportionally. Once you have the moles, multiplying by the molar mass of benzoic acid gives the mass you need to weigh out. The solvent (methanol here) doesn't affect the calculation — it just carries the solute.

The formula is straightforward:

But watch out: volume must be in litres, not millilitres. That's the most common slip.

Step-by-Step Calculation

1. Write down what's given.

• Molarity, $M = 0.15\ \text{mol L}^{-1}$
• Volume, $V = 250\ \text{mL} = 0.250\ \text{L}$ (convert by dividing by 1000)
• Solute: benzoic acid, $C_6H_5COOH$

2. Find the molar mass of benzoic acid.

Benzoic acid has the formula $C_7H_6O_2$ (since $C_6H_5COOH$ has 7 carbons, 6 hydrogens, 2 oxygens).

Atomic masses (approx.):

• Carbon: $12.0\ \text{g mol}^{-1}$
• Hydrogen: $1.0\ \text{g mol}^{-1}$
• Oxygen: $16.0\ \text{g mol}^{-1}$

So:

3. Calculate the moles of benzoic acid needed. …

Method: Molarity-to-Mass Conversion (via Moles)

This is a molarity → moles → mass problem. Molarity ($M$) gives moles per litre; we scale to the given volume, then convert moles to grams using molar mass.

Steps

1. Write the molarity formula

where $n$ = moles of solute, $V$ = volume in litres.

2. Convert volume to litres

3. Solve for moles of benzoic acid

4. Calculate molar mass of $C_6H_5COOH$
   • Carbon: $7 \times 12.01 = 84.07$
   • Hydrogen: $6 \times 1.008 = 6.048$
   • Oxygen: $2 \times 16.00 = 32.00$

5. Convert moles to mass …

🧪 Common Mistakes in Molality Calculation (Benzoic Acid in Methanol)

Let’s first understand the concept — then the mistakes.

✓ The Core Idea

The question asks for 0.15 M solution — that’s molarity, not molality.

• Molarity (M) = moles of solute per litre of solution
• Molality (m) = moles of solute per kg of solvent

Here, we need mass of benzoic acid for a given volume of solution at a given molarity.

✗ Mistake #1: Confusing Molarity with Molality

What students do:

They try to use the mass of solvent (methanol) instead of the volume of solution.

Why it’s wrong:

Molarity uses volume of solution, not solvent. Methanol is the solvent here, but the 250 mL is the final solution volume.

✓ How to avoid:

Always check the unit:

• If it says M → use volume of solution in litres
• If it says m → use mass of solvent in kg

Formula to use:

✗ Mistake #2: Forgetting to Convert mL to L

What students do:

Plug in 250 directly as litres.

Why it’s wrong:

Molarity is moles per litre, so volume must be in litres.

✓ How to avoid:

Always convert:

✗ Mistake #3: Using Wrong Molar Mass

What students do:

Use atomic masses incorrectly — e.g., forget that benzoic acid is $C_6H_5COOH$ (7 carbons, 6 hydrogens, 2 oxygens).

Why it’s wrong:

Wrong molar mass → wrong mass of solute.

✓ How to avoid:

Count atoms carefully:

• C: 7 atoms → $7 \times 12 = 84$
• H: 6 atoms → $6 \times 1 = 6$
• O: 2 atoms → $2 \times 16 = 32$

Molar mass = $84 + 6 + 32 = 122 \text{ g/mol}$

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