Q.18 g of glucose, $C_6H_{12}O_6$, is dissolved in 1 kg of water in a saucepan. At what temperature will water boil at 1.013 bar? $K_b$ for water is 0.52 K kg mol$^{-1}$.

Concept: Boiling Point Elevation — the increase in boiling point when a non-volatile solute is added to a solvent.

Step 1: Find moles of glucose

Molar mass of $C_6H_{12}O_6 = 180 \text{ g mol}^{-1}$

Moles $= \frac{18}{180} = 0.1 \text{ mol}$

Step 2: Find molality

Mass of solvent $= 1 \text{ kg}$

Molality $m = \frac{0.1}{1} = 0.1 \text{ mol kg}^{-1}$

Step 3: Apply boiling point elevation formula

$\Delta T_b = K_b \cdot m = 0.52 \times 0.1 = 0.052 \text{ K}$

Step 4: New boiling point

At 1.013 bar, pure water boils at $100^\circ\text{C}$.

Boiling point $= 100 + 0.052 = 100.052^\circ\text{C}$

Molality. Molar mass of glucose $\mathrm{C_6H_{12}O_6}=180\ \text{g mol}^{-1}$:

Elevation in boiling point.

Boiling point. Pure water boils at $373.15$ K at $1.013$ bar, so

Method: Boiling Point Elevation Formula

This is a direct application of the boiling point elevation formula for non-volatile solutes.

Concept (Why this works)

When a non-volatile solute like glucose is dissolved in a solvent (water), the vapour pressure of the solvent decreases. To make the solution boil (i.e., reach atmospheric pressure), we need to raise the temperature above the normal boiling point. The increase is called boiling point elevation, $\Delta T_b$.

Formula

Where:

• $\Delta T_b$ = elevation in boiling point (in K or °C)
• $K_b$ = ebullioscopic constant of solvent (given: 0.52 K kg mol⁻¹)
• $m$ = molality of solution (mol solute per kg solvent)

Steps

Step 1: Find moles of glucose

• Molar mass of glucose ($C_6H_{12}O_6$) = $6(12) + 12(1) + 6(16) = 180$ g/mol
• Moles = $\frac{18 \text{ g}}{180 \text{ g/mol}} = 0.1$ mol

Step 2: Find molality

• Mass of solvent (water) = 1 kg
• Molality, $m = \frac{0.1 \text{ mol}}{1 \text{ kg}} = 0.1$ mol/kg

Step 3: Calculate $\Delta T_b$

Step 4: Find boiling point of solution

• Normal boiling point of water at 1.013 bar = 100 °C (or 373.15 K)
• Boiling point of solution = $100 + 0.052 = \mathbf{100.052 \,^\circ C}$

Final Answer

The water will boil at 100.052 °C (or 373.202 K).

Here are the common mistakes students make with boiling point elevation problems — and how to avoid each one.

1. Forgetting that boiling point elevation is not the final boiling point

Mistake:

Students calculate $\Delta T_b$ and stop there, writing the answer as $0.052 \,^\circ\text{C}$.

Why it’s wrong:

The question asks: At what temperature will water boil?

You must add $\Delta T_b$ to the normal boiling point of water ($100 \,^\circ\text{C}$ at 1.013 bar).

How to avoid:

Always write the final step explicitly:

Here:

Key result: $\boxed{100.052 \,^\circ\text{C}}$

2. Using mass of solute instead of moles in the molality formula

Mistake:

Plugging 18 g directly into $\Delta T_b = K_b \cdot m$ without converting to moles.

Why it’s wrong:

Molality $m$ is moles of solute per kg of solvent, not grams per kg.

How to avoid:

Always compute moles first:

Then:

3. Using the wrong molar mass for glucose

Mistake:

Using $C_6H_{12}O_6$ molar mass as 160, 200, or forgetting to add oxygen.

Why it’s wrong:

Glucose = $6(12) + 12(1) + 6(16) = 72 + 12 + 96 = 180 \text{ g mol}^{-1}$.

How to avoid:

Write out the atomic masses clearly before calculating:

• Carbon: $6 \times 12 = 72$
• Hydrogen: $12 \times 1 = 12$
• Oxygen: $6 \times 16 = 96$
• Total: $180 \text{ g mol}^{-1}$

4. Confusing molality with molarity

Mistake:

Using volume of solution (which isn’t given) instead of mass of solvent.

Why it’s wrong:

Boiling point elevation uses molality ($m$), not molarity ($M$).

Here, solvent mass is given as 1 kg — that’s perfect for molality.

How to avoid:

Check the units: if the problem gives kg of solvent, use molality.

If it gives volume of solution, you’d need density to convert — but that’s rare for this concept.

5. Forgetting that $K_b$ units are K kg mol$^{-1}$ — not just K

Mistake:

Plugging numbers without checking unit cancellation.

Why it’s wrong:

You need $\Delta T_b$ in K (or $^\circ$C, same magnitude).

$K_b$ times molality gives:

How to avoid:

Write the units alongside each step:

6. Assuming glucose dissociates (like salt)

Mistake:

Using $i = 2$ or another van’t Hoff factor.

Why it’s wrong:

Glucose is a non-electrolyte — it does not dissociate in water.

$i = 1$ always for covalent molecular solutes like sugar.

How to avoid:

For molecular solutes (glucose, urea, sucrose), always use $i = 1$.

Only use $i > 1$ for ionic compounds (NaCl, CaCl$_2$, etc.).

Quick checklist to avoid all mistakes

Final answer: