Question

58.5 g of NaCl and 180 g of glucose were separately dissolved in 1000 mL of water. Identify the correct statement regarding the elevation of boiling point of the resulting solution.

• A. NaCl solution will show higher elevation of boiling point.
• B. Glucose solution will show higher elevation of boiling point.
• C. Both solutions will show equal elevation of boiling point.
• D. None will show boiling point elevation.

Hint:

Ionic compounds like NaCl dissociate in water, increasing the effect on colligative properties.

Answer 1

The Correct Option is A

Step 1: Calculating Moles of Solute
\[ {Moles of NaCl} = \frac{58.5}{58.5} = 1 { mol} \] \[ {Moles of Glucose} = \frac{180}{180} = 1 { mol} \] Step 2: Dissociation of NaCl
NaCl dissociates into two ions (\( Na^+ \) and \( Cl^- \)), increasing the effective particle concentration. Glucose does not dissociate, so the number of particles remains the same. Thus, NaCl solution will have a higher elevation in boiling point. Thus, the correct answer is (A).

Answer 2

Step 1: Understand boiling point elevation
Boiling point elevation is a colligative property that depends on the number of solute particles in solution.
It is given by ΔT_b = i × K_b × m,
where i = van’t Hoff factor, K_b = ebullioscopic constant, and m = molality.

Step 2: Calculate moles of solute for each solution
For NaCl:
Molar mass of NaCl = 58.5 g/mol
Moles of NaCl = 58.5 g / 58.5 g/mol = 1 mole

For glucose:
Molar mass of glucose (C₆H₁₂O₆) = 180 g/mol
Moles of glucose = 180 g / 180 g/mol = 1 mole

Step 3: Consider dissociation and van’t Hoff factor
NaCl dissociates into Na⁺ and Cl⁻ ions (2 particles), so i = 2.
Glucose does not dissociate (i = 1).

Step 4: Compare the total number of particles
Both solutions have 1 mole of solute,
NaCl solution effectively has 2 moles of particles,
while glucose has 1 mole of particles.

Step 5: Conclusion
Since boiling point elevation depends on the number of particles, NaCl solution will have a higher elevation of boiling point than glucose solution.