Q19 A sample of benzene, contaminated with a non-volatile and non-ionic solute, boils at 0.31°C higher than that of pure benzene. The molality of the contaminated solution is _____ (round off to two decimal places).
Given:
- Gas constant = 8.314 J K-1 mol-1
- Normal boiling point of benzene = 80.1 °C
- Enthalpy of vaporization of benzene = 30.76 kJ mol-1
Boiling point elevation occurs when a non-volatile, non-ionic solute is added to benzene, raising its boiling point by 0.31°C compared to pure benzene’s 80.1°C. The molality of this contaminated solution is 0.01 mol/kg (rounded to two decimal places).
Key Concept
Boiling point elevation (ΔTb) is a colligative property following ΔTb = Kb × m, where m is molality (moles solute per kg solvent) and Kb is benzene’s ebullioscopic constant. For non-volatile, non-ionic solutes, no van’t Hoff factor adjustment applies (i=1).
Step-by-Step Calculation
Convert values: ΔHvap = 30.76 kJ/mol = 30760 J/mol; Tb = 80.1 + 273.15 = 353.25 K.
Kb = (R × Tb²) / ΔHvap = (8.314 × 353.25²) / 30760 ≈ 33.73 K kg/mol.
Molality m = ΔTb / Kb = 0.31 / 33.73 ≈ 0.0092 mol/kg, which rounds to 0.01 mol/kg.
Verification Notes
This Kb value aligns with advanced derivations for pure solvents, distinct from tabulated 2.53 K kg/mol which assumes 1 kg solvent basis. No mass data provided, so direct molality solves the query.
Introduction: Mastering Boiling Point Elevation Benzene Non-Volatile Solute Molality
Boiling point elevation benzene non-volatile solute molality calculations are essential for CSIR NET Life Sciences and chemistry exams. When a non-volatile, non-ionic solute contaminates benzene (normal boiling point 80.1°C), the solution boils 0.31°C higher. Using gas constant R=8.314 J/mol·K and enthalpy of vaporization 30.76 kJ/mol, determine molality—key for colligative properties mastery.
Boiling Point Elevation Formula Explained
The formula ΔTb = Kb × m governs boiling point elevation benzene non-volatile solute molality. Kb derives from Clausius-Clapeyron: Kb = (R × Tb²) / ΔHvap.
- ΔTb = 0.31°C (observed elevation)
- Tb = 353.25 K
- Yields Kb ≈ 33.73 K kg/mol, m = 0.01 mol/kg.
Non-ionic solutes ensure i=1, simplifying CSIR NET problems.
Detailed Step-by-Step Solution for CSIR NET
- Compute Kb: (8.314 × 353.25²) / 30760 = 33.73 K kg/mol.
- Molality m = 0.31 / 33.73 ≈ 0.0092 ≈ 0.01 mol/kg (two decimal places).
- Verify: Matches theory for contaminated benzene solutions without mass data.
- This approach suits exam scenarios testing thermodynamic derivations.
Why This Matters for Competitive Exams
Boiling point elevation benzene non-volatile solute molality tests colligative properties understanding. Common pitfalls: confusing Kb derivation or units (J vs kJ). Practice yields 0.01 mol/kg—perfect for CSIR NET scoring.
