Boiling Point Elevation

Question

What is the boiling point of a solution containing 6 g of a non-volatile, non-electrolyte solute dissolved in 100 g of water, given the molar mass of the solute is 40 g/mol and K_b for water is 0.512 K·kg/mol?

RankGuru · Accepted Answer

The boiling point elevation is a colligative property described by the equation ΔT_b = K_b × m, where K_b is the ebullioscopic constant and m is the molality of the solution. The moles of solute = 6/40 = 0.15 mol. The mass of solvent = 100 g = 0.100 kg. Therefore, molality (m) = 0.15 / 0.100 = 1.5 mol/kg. ΔT_b = 0.512 × 1.5 = 0.768 K. Since water boils at 100°C under standard conditions, the boiling point of the solution = 100 + 0.768 = 100.768°C. Option 100.256°C results from using m = 0.5 mol/kg (wrong moles), perhaps dividing mass of solute by 12. Option 101.024°C uses m = 2.0, doubling the actual molality. Option 100.512°C comes from using m = 1.0 mol/kg by mistakenly using 150 g of water. This is a standard NCERT-style application of the boiling point elevation formula from Solutions chapter. Plausibility check: the boiling point increase is less than 1°C, which is physically reasonable for a dilute solution.

Boiling Point Elevation

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