Q.Calculate molality of 2.5 g of ethanoic acid ($CH_3COOH$) in 75 g of benzene.

Concept: Molality Calculation

Molality ($m$) is defined as moles of solute per kilogram of solvent:

Step 1: Find moles of ethanoic acid.

Molar mass of $CH_3COOH = 12 + 3(1) + 12 + 16 + 16 + 1 = 60 \text{ g mol}^{-1}$

Step 2: Convert mass of benzene to kg.

$75 \text{ g} = 0.075 \text{ kg}$ …

Molality measures concentration in a way that doesn't change with temperature, unlike molarity. It asks: how many moles of solute are dissolved in exactly one kilogram of solvent? The definition is straightforward:

This problem hands you masses, so you need to convert the solute mass to moles using its molar mass, and the solvent mass to kilograms.

Step-by-step calculation

1. Find the molar mass of ethanoic acid, $CH_3COOH$.

   Count the atoms: 2 carbon, 4 hydrogen, 2 oxygen.

2. Convert the mass of ethanoic acid to moles.

   You have 2.5 g of $CH_3COOH$:

3. Convert the mass of benzene (solvent) to kilograms.

   The solvent is 75 g of benzene:

   $\text{mass of solvent} = 75 \text{ g} = 0.075 \text{ kg}$ …

Method: Molality Formula Method

Molality ($m$) is defined as the number of moles of solute per kilogram of solvent.

Steps

Step 1: Identify the solute and solvent

• Solute = ethanoic acid ($CH_3COOH$) — 2.5 g
• Solvent = benzene — 75 g

Step 2: Convert solvent mass to kilograms

Step 3: Calculate molar mass of ethanoic acid

• C: $2 \times 12 = 24$
• H: $4 \times 1 = 4$
• O: $2 \times 16 = 32$
• Molar mass = $24 + 4 + 32 = 60 \text{ g/mol}$

Step 4: Calculate moles of solute

Step 5: Apply the molality formula …

Here are the common mistakes students make in this exact molality problem, along with how to avoid each one.

1. Confusing Molality with Molarity

The Mistake:

Students often divide moles of solute by liters of solution (molarity) instead of kilograms of solvent (molality).

How to Avoid:

• Memorize the definition:

• The solvent here is benzene (75 g), not the total solution.
• Always check: Am I dividing by solvent mass or solution volume?

2. Forgetting to Convert Grams of Solvent to Kilograms

The Mistake:

Using 75 g directly in the formula without converting to kg.

How to Avoid:

• Molality requires kg of solvent.
• Convert:

• Write the conversion step explicitly — never skip it.

3. Incorrect Molar Mass of Ethanoic Acid ($CH_3COOH$)

The Mistake:

Counting atoms wrong — e.g., forgetting the two oxygen atoms or the extra hydrogen in the carboxyl group.

How to Avoid:

• Write the formula clearly: $C_2H_4O_2$
• Calculate step by step:
  • Carbon: $2 \times 12 = 24$
  • Hydrogen: $4 \times 1 = 4$
  • Oxygen: $2 \times 16 = 32$
  • Total = 60 g/mol
• Double-check with a quick mental sum: 24 + 4 + 32 = 60.

4. Rounding Too Early in the Calculation

The Mistake:

Rounding moles (e.g., 0.04167 → 0.04) before dividing by kg, leading to a wrong final answer.

How to Avoid:

• Keep at least 4 decimal places in intermediate steps.
• Moles of ethanoic acid:

• Then divide by 0.075 kg:

• Round only at the final answer: 0.556 mol/kg (three significant figures — the value NCERT prints; 0.56 is its two-significant-figure form).

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