Q.Boiling point of water at 750 mm Hg is 99.63$^\circ$C. How much sucrose is to be added to 500 g of water such that it boils at 100$^\circ$C.

This problem involves Boiling Point Elevation, a colligative property where the boiling point of a solvent increases upon the addition of a non-volatile solute.

1. First, calculate the elevation in boiling point ($\Delta T_b$):

   $\Delta T_b = T_b - T_b^0 = 100^\circ C - 99.63^\circ C = 0.37^\circ C$.

2. Next, use the boiling point elevation formula to find the molality ($m$) of the sucrose solution. The molal elevation constant ($K_b$) for water is $0.52 \text{ K kg mol}^{-1}$ (or $0.52 ^\circ C \text{ kg mol}^{-1}$).

   $\Delta T_b = K_b \cdot m$

   $0.37^\circ C = 0.52^\circ C \text{ kg mol}^{-1} \cdot m$

   $m = \frac{0.37}{0.52} \approx 0.7115 \text{ mol kg}^{-1}$

3. Now, calculate the moles of sucrose needed. The mass of water (solvent) is $500 \text{ g} = 0.5 \text{ kg}$.

   Moles of sucrose = $m \times \text{mass of water (kg)}$

   Moles of sucrose = $0.7115 \text{ mol kg}^{-1} \times 0.5 \text{ kg} = 0.35575 \text{ mol}$

4. Finally, convert moles of sucrose to mass using its molar mass ($C_{12}H_{22}O_{11}$), which is $342 \text{ g mol}^{-1}$.

   Mass of sucrose = Moles of sucrose $\times$ Molar mass of sucrose

   Mass of sucrose = $0.35575 \text{ mol} \times 342 \text{ g mol}^{-1} \approx 121.66 \text{ g}$

Boiling-point elevation is a colligative property — it depends only on the number of solute particles dissolved in a fixed mass of solvent, not on their identity. For a non-volatile, non-electrolyte solute such as sucrose, the relationship is

where $\Delta T_b$ is the elevation in boiling point, $K_b$ is the molal elevation (ebullioscopic) constant of the solvent, and $m$ is the molality of the solution.

Step 1: Find the required boiling-point elevation

The water must boil at 100°C instead of its actual boiling point of 99.63°C at 750 mm Hg:

Step 2: Find the required molality

Using $K_b = 0.52\ \text{K kg mol}^{-1}$ (as given for water):

Step 3: Find the moles of sucrose needed

Molality is defined per kilogram of solvent. Here the solvent is 500 g = 0.500 kg of water:

Step 4: Convert moles to mass

The molar mass of sucrose ($\text{C}_{12}\text{H}_{22}\text{O}_{11}$) is $342\ \text{g mol}^{-1}$:

So dissolving about 121.7 g of sucrose in 500 g of water raises its boiling point by 0.37 K, bringing it exactly to 100°C at 750 mm Hg.

Method: Boiling Point Elevation Formula

This is a colligative property problem — the boiling point rises because solute particles (sucrose) lower the vapour pressure of the solvent (water).

Step 1 — Identify the given data

• Normal boiling point of water (at 750 mm Hg) = $99.63^\circ\text{C}$
• Desired boiling point = $100^\circ\text{C}$
• Mass of solvent (water) = $500\ \text{g}$
• Solute = sucrose ($\text{C}_{12}\text{H}_{22}\text{O}_{11}$), molar mass = $342\ \text{g/mol}$
• $K_b$ for water = $0.52\ \text{K kg mol}^{-1}$ (standard value, must be known)

Step 2 — Calculate the elevation in boiling point

Step 3 — Apply the boiling point elevation formula

where $m$ = molality of the solution (mol solute per kg solvent).

Step 4 — Find moles of sucrose needed

Molality $m$ = $\dfrac{\text{moles of solute}}{\text{mass of solvent in kg}}$

Mass of water = $500\ \text{g} = 0.5\ \text{kg}$

Step 5 — Convert moles to mass

Final Answer:

Key concept check: Sucrose is non-volatile and does not dissociate — so $i = 1$. If the solute were ionic (like NaCl), you would multiply by van’t Hoff factor $i$.

🧠 The Core Concept First

Boiling point elevation is a colligative property — it depends only on the number of solute particles, not their identity.

The formula is:

Where:

• $\Delta T_b$ = elevation in boiling point ($T_b - T_b^0$)
• $K_b$ = ebullioscopic constant of the solvent (for water, $K_b = 0.52 \, \text{K kg mol}^{-1}$)
• $m$ = molality of the solution (moles of solute per kg of solvent)

🔍 Step-by-Step Solution (for reference)

Given:

• Initial boiling point of water = $99.63^\circ\text{C}$ at 750 mm Hg
• Final boiling point = $100^\circ\text{C}$
• Mass of water = $500 \, \text{g} = 0.5 \, \text{kg}$
• Solute = sucrose ($\text{C}_{12}\text{H}_{22}\text{O}_{11}$, molar mass = $342 \, \text{g/mol}$)

Step 1: Find $\Delta T_b$

Step 2: Use $\Delta T_b = K_b \cdot m$

Step 3: Find moles of sucrose

Step 4: Find mass of sucrose

✓ Final answer: Approximately 121.7 g of sucrose.

✗ Common Mistakes & How to Avoid Them

1. Using the wrong $K_b$ value

• Mistake: Using $K_b = 0.512$ or $0.52$ without checking units or if it's for water.
• How to avoid: Memorise: $K_b$ for water = 0.52 K kg mol⁻¹. Always confirm in the problem if given. If not, use the standard value.

2. Confusing molality with molarity

• Mistake: Using volume of solution instead of mass of solvent.
• How to avoid: Remember: molality = moles of solute / kg of solvent. Here, solvent is water — use its mass in kg, not volume.

3. Forgetting to convert grams to kg

• Mistake: Using 500 g directly as 500 kg in the molality formula.
• How to avoid: Always convert: $500 \, \text{g} = 0.5 \, \text{kg}$. Write it explicitly.

4. Using the wrong molar mass of sucrose

• Mistake: Taking 342 as 342 g (correct) but then using it in molality without converting.
• How to avoid: Double-check: $\text{C}_{12}\text{H}_{22}\text{O}_{11}$ = $12(12) + 22(1) + 11(16) = 144 + 22 + 176 = 342 \, \text{g/mol}$. Write it down.

5. Rounding off too early

• Mistake: Rounding $\frac{0.37}{0.52}$ to 0.71 instead of 0.7115, then multiplying by 0.5 to get 0.355 instead of 0.3558.
• How to avoid: Keep at least 4 decimal places during intermediate steps. Round only at the final answer.

6. Assuming $\Delta T_b$ is the final temperature

• Mistake: Writing $\Delta T_b = 100^\circ\text{C}$.
• How to avoid: $\Delta T_b$ is the change, not the final temperature. Always subtract: $\Delta T_b = T_b - T_b^0$.

7. Ignoring the given pressure (750 mm Hg)

• Mistake: Thinking the pressure is irrelevant and using $100^\circ\text{C}$ as the normal boiling point.
• How to avoid: The problem gives the boiling point at 750 mm Hg as $99.63^\circ\text{C}$. That is your $T_b^0$. The normal boiling point ($100^\circ\text{C}$ at 760 mm Hg) is not used here. The elevation is from $99.63^\circ\text{C}$ to $100^\circ\text{C}$.

✓ Quick Checklist to Avoid Errors

Final tip: Always write the formula first, then plug in numbers with units. This catches most errors before they happen. Good luck!