Q.Calculate the mass of ascorbic acid (Vitamin C, $C_6H_8O_6$) to be dissolved in 75 g of acetic acid to lower its melting point by 1.5$^\circ$C. $K_f = 3.9$ K kg mol$^{-1}$.

The problem involves the depression of the freezing point, a colligative property that depends on the number of solute particles. Ascorbic acid ($C_6H_8O_6$) is a non-electrolyte, meaning its van't Hoff factor ($i$) is 1 as it does not dissociate or associate in solution.

1. The given depression in melting point is $\Delta T_f = 1.5^\circ C$, which is $1.5 K$. The cryoscopic constant $K_f = 3.9$ K kg mol$^{-1}$. We calculate the molality ($m$) of the solution:

   $m = \frac{\Delta T_f}{i K_f} = \frac{1.5 \text{ K}}{1 \times 3.9 \text{ K kg mol}^{-1}} = 0.3846 \text{ mol kg}^{-1}$.

2. The mass of the solvent (acetic acid) is $W_1 = 75 \text{ g} = 0.075 \text{ kg}$. Using the definition of molality ($m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}$), the moles of ascorbic acid ($n_2$) required are: …

1. Molality from $\Delta T_f = K_f\, m$.

2. Moles of ascorbic acid in $75\ \text{g} = 0.075\ \text{kg}$ of acetic acid:

$n = m \times 0.075 = 0.3846 \times 0.075 = 0.02885\ \text{mol}$ …

Method: Freezing Point Depression (Cryoscopy)

This is a direct application of Raoult’s Law for colligative properties — specifically, the depression of freezing point caused by dissolving a non-volatile solute.

Step 1 — Write the formula

The freezing point depression is given by:

Where:

• $\Delta T_f$ = depression in freezing point (in K or °C, same magnitude)
• $K_f$ = cryoscopic constant (in K kg mol⁻¹)
• $m$ = molality of the solution (mol solute per kg solvent)

Step 2 — Identify given values

• $\Delta T_f = 1.5 \, ^\circ\text{C}$ (same as 1.5 K)
• $K_f = 3.9 \, \text{K kg mol}^{-1}$
• Mass of solvent (acetic acid) = $75 \, \text{g} = 0.075 \, \text{kg}$
• Solute = ascorbic acid, $C_6H_8O_6$

Step 3 — Calculate molality required

From $\Delta T_f = K_f \cdot m$:

Step 4 — Find moles of solute needed

Molality = moles of solute per kg of solvent:

Step 5 — Convert moles to mass

Molar mass of $C_6H_8O_6$: …

🧠 The Core Concept First

The problem uses freezing point depression:

where

• $\Delta T_f$ = depression in freezing point (here, 1.5 °C = 1.5 K, since the size of 1 °C = 1 K)
• $K_f$ = cryoscopic constant (given: 3.9 K kg mol⁻¹)
• $m$ = molality = moles of solute per kg of solvent

We need mass of solute (ascorbic acid).

✗ Common Mistake #1: Forgetting to convert solvent mass to kg

The error:

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

Why it’s wrong:

Molality is moles per kg of solvent, not per gram.

✓ How to avoid:

Always write the unit conversion explicitly:

✗ Common Mistake #2: Using $\Delta T_f$ in °C without realising it equals K

The error:

Thinking 1.5 °C needs conversion to Kelvin (e.g., adding 273).

Why it’s wrong:

A difference of 1.5 °C is exactly equal to a difference of 1.5 K. The Kelvin scale has the same “step size” as Celsius.

✓ How to avoid:

Remember: $\Delta T$ in °C = $\Delta T$ in K. Only absolute temperatures need the +273 conversion.

✗ Common Mistake #3: Incorrect molar mass of ascorbic acid ($C_6H_8O_6$)

The error:

Miscounting atoms or using wrong atomic masses.

Why it’s wrong:

A wrong molar mass gives a wrong final mass.

✓ How to avoid:

Calculate step by step:

• C: $6 \times 12 = 72$
• H: $8 \times 1 = 8$
• O: $6 \times 16 = 96$

Double-check — this is a standard value.

✗ Common Mistake #4: Mixing up the formula — using $\Delta T_f = K_f \cdot n$ instead of $K_f \cdot m$

The error:

Plugging moles directly without dividing by kg of solvent.

Why it’s wrong:

Molality $m = \frac{\text{moles}}{\text{kg solvent}}$, not just moles.

✓ How to avoid:

Write the full chain:

Then rearrange stepwise.

✓ Correct Solution (Quick Walkthrough)

Step 1: Find molality …