Q.An antifreeze solution is prepared from 222.6 g of ethylene glycol ($C_2H_6O_2$) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL$^{-1}$, then what shall be the molarity of the solution?

Concept: Molality and molarity calculations from mass composition and density.

First, find the moles of ethylene glycol. The molar mass of $\mathrm{C_2H_6O_2}$ is $2(12) + 6(1) + 2(16) = 62 \text{ g mol}^{-1}$.

Molality is moles of solute per kg of solvent:

For molarity, we need the volume of solution. The total mass is $222.6 + 200 = 422.6 \text{ g}$. …

The distinction between molality and molarity is central here. Molality ($m$) measures concentration as moles of solute per kilogram of solvent, making it temperature-independent because mass doesn't change with temperature. Molarity ($M$), on the other hand, is moles of solute per liter of solution, which depends on volume and thus temperature. Antifreeze problems often ask for both because molality governs colligative properties (freezing-point depression) while molarity is used in reaction stoichiometry.

The strategy is straightforward: calculate molality directly from the given masses, then use the solution density to convert total mass into volume for molarity.

Calculating Molality

1. Find moles of ethylene glycol

The molar mass of $C_2H_6O_2$ is:

Moles of ethylene glycol:

2. Convert mass of solvent to kilograms

Mass of water (solvent) = $200 \text{ g} = 0.200 \text{ kg}$

3. Calculate molality

Calculating Molarity

4. Find total mass of solution

5. Convert mass to volume using density

Given density $\rho = 1.072 \text{ g mL}^{-1}$:

6. Calculate molarity …

Method: Stepwise Formula-Based Approach for Molality & Molarity

We will solve this using direct formula substitution — first for molality, then for molarity using density.

Step 1 — Calculate Moles of Solute (Ethylene Glycol)

• Formula:

  $\text{Moles} = \frac{\text{mass}}{\text{molar mass}}$

• Molar mass of $C_2H_6O_2$:

  $2(12) + 6(1) + 2(16) = 24 + 6 + 32 = 62 \text{ g mol}^{-1}$

• Moles of ethylene glycol:

  $\frac{222.6}{62} = 3.59 \text{ mol}$

Step 2 — Calculate Molality

• Formula:

  $\text{Molality (m)} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}$

• Mass of water (solvent) = $200 \text{ g} = 0.200 \text{ kg}$

• Molality:

  $m = \frac{3.59}{0.200} = \mathbf{17.95 \text{ mol kg}^{-1}}$

Step 3 — Calculate Total Mass & Volume of Solution (for Molarity)

• Total mass of solution:

  $222.6 \text{ g (solute)} + 200 \text{ g (solvent)} = 422.6 \text{ g}$

• Density given: $1.072 \text{ g mL}^{-1}$

• Volume of solution:

  $\text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{422.6}{1.072} = 394.2 \text{ mL} = 0.3942 \text{ L}$

Step 4 — Calculate Molarity

• Formula: …

🧠 Concept First — What is Molality?

Molality ($m$) is moles of solute per kilogram of solvent (not solution).

Molarity ($M$) is moles of solute per litre of solution.

✗ Mistake 1: Using mass of solution instead of mass of solvent for molality

What students do wrong:

They take total mass of solution (solute + solvent) as the denominator.

Example of error:

$m = \frac{222.6 / 62}{222.6 + 200} \times 1000$ — wrong!

✓ How to avoid:

Always remember:

Molality denominator = solvent mass only (here, 200 g water = 0.200 kg).

Correct step:

Molar mass of ethylene glycol ($C_2H_6O_2$) =

$2(12) + 6(1) + 2(16) = 24 + 6 + 32 = 62$ g/mol

Moles of solute = $\frac{222.6}{62} = 3.59$ mol

Molality = $\frac{3.59}{0.200} = 17.95$ mol/kg

✓ Answer: $m = 17.95$ m

✗ Mistake 2: Forgetting to convert solvent mass to kg

What students do wrong:

They plug 200 g directly without dividing by 1000.

Example of error:

$m = \frac{3.59}{200} = 0.01795$ — off by factor of 1000!

✓ How to avoid:

Always write the unit:

$m = \frac{\text{moles}}{\text{mass in kg}}$

Convert grams to kg: $200 \text{ g} = 0.200 \text{ kg}$

✗ Mistake 3: Using density of solvent instead of density of solution for molarity

What students do wrong:

They take density of water (1 g/mL) to find volume of solution.

Example of error:

Volume = $\frac{222.6 + 200}{1} = 422.6$ mL — wrong!

This assumes solution density = 1 g/mL, which is not given.

✓ How to avoid:

Use the given density of solution (1.072 g/mL) to find volume:

Total mass of solution = $222.6 + 200 = 422.6$ g

Volume = $\frac{422.6}{1.072} = 394.2$ mL = $0.3942$ L

Molarity = $\frac{3.59}{0.3942} = 9.11$ M

✓ Answer: $M = 9.11$ M

✗ Mistake 4: Mixing up molality and molarity formulas

What students do wrong:

They use the same denominator for both.

✓ How to avoid:

Make a quick comparison table:

| Quantity | Denominator | Unit |

|----------|-------------|------| …