What is the molality of a glucose solution prepared by dissolving 27.0 g of glucose,C₆H₁₂O₆,in 125.9 g of water? A)1.19 × 10^-3 m B)0.214 m C)0.981 m D)1.19 m

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Quizplus · Accepted Answer

Molality (m) is defined as the number of moles of solute per kilogram of solvent. The molar mass of glucose (C6H12O6) is calculated as follows: (12.01*6) + (1.008*12) + (16.00*6) = 180.16 g/mol. The number of moles of glucose is 27.0 g / 180.16 g/mol = 0.1499 moles. The mass of water as the solvent in kilograms is 125.9 g = 0.1259 kg. Therefore, the molality is 0.1499 moles / 0.1259 kg = 1.19 m.

What Is the Molality of a Glucose Solution Prepared by Dissolving