Molality And Mole Fraction

Question

A solution is prepared by dissolving 18 g of glucose ($\text{C}_6\text{H}_{12}\text{O}_6$, molar mass = 180 g/mol) in 500 g of water. What is the molality of this solution?

RankGuru · Accepted Answer

Molality is defined as the number of moles of solute per kilogram of solvent (not solution). It is temperature-independent and preferred for colligative property calculations. Moles of glucose $= 18/180 = 0.1$ mol. Mass of solvent (water) $= 500$ g $= 0.5$ kg. Molality $m = 0.1/0.5 = 0.2$ mol/kg $= 0.2$ m. Option 0.1 m would result if the solvent were taken as 1 kg (1000 g), but here only 500 g is used. Option 0.5 m would result from incorrectly using 0.2 mol of glucose instead of 0.1 mol. Option 1.0 m would require either 0.1 mol in 100 g of water, or 0.5 mol in 500 g — neither matches. This question applies the NCERT definition of molality, distinguishing it from molarity which uses solution volume. Plausibility check: molality is expressed in mol/kg; $0.1 \, \text{mol} \div 0.5 \, \text{kg} = 0.2 \, \text{m}$, with consistent units throughout.

Molality And Mole Fraction

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