Q. 26 A non–volatile solute has a molecular weight of 180 g mol–1. Assume that the solute
does not associate or dissociate in water, and the boiling–point constant
(ebullioscopic constant) of water is 0.51 K kg mol–1.
The amount (in g) of solute added to 500 g of water to elevate the boiling point by
0.153 K is ______. (answer in integer)
Boiling Point Elevation Calculation: 180 g/mol Solute in 500g Water by 0.153 K
Final Answer: 108 g
🔬 Problem Solution
Boiling point elevation follows the formula ΔT_b = K_b × m, where ΔT_b = 0.153 K is the elevation, K_b = 0.51 K kg mol⁻¹ is the ebullioscopic constant for water, and m is molality (moles of solute per kg of solvent).
Molality m = ΔT_b / K_b = 0.153 / 0.51 = 0.3 mol kg⁻¹.
For 500 g (0.5 kg) water, moles of solute = 0.3 × 0.5 = 0.15 mol. With molecular weight 180 g mol⁻¹, mass = 0.15 × 180 = 108 g (integer).
📋 Step-by-Step Derivation
- Rearrange for molality: m = ΔT_b / K_b
- Calculate moles: n = m × W_solvent (kg)
- Mass of solute: w = n × M = (ΔT_b × W_solvent × M) / K_b
- No dissociation: van’t Hoff factor i = 1
🎯 CSIR NET Preparation Guide
Boiling point elevation calculation is key for CSIR NET life sciences, applying colligative properties to non-volatile solutes like this 180 g/mol compound in 500g water.
Using ebullioscopic constant 0.51 K kg mol⁻¹, elevation of 0.153 K requires precise molality determination for exam success.
Key Calculations:
- ΔT_b = K_b m confirms non-associating solute behavior
- Molality 0.3 mol kg⁻¹ leads to 0.15 mol in 0.5 kg solvent
- Final mass 108 g matches integer answer format
This boiling point elevation example reinforces biochemistry concepts for competitive exams.
