What is the molality of a solution prepared by dissolving 0.114 mol of chloroform,CHCl₃,in 476 g of toluene,C₆H₅CH₃? A)0.0287 m B)0.0216 m C)0.239 m D)0.0220 m E)0.543 m

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Quizplus · Accepted Answer

Molality (m) is defined as the number of moles of solute per kilogram of solvent. Given 0.114 mol of chloroform and 476 g of toluene, first convert the mass of toluene to kilograms: 476 g = 0.476 kg. Then, calculate molality: $m = \frac{0.114 \, \text{mol}}{0.476 \, \text{kg}} = 0.239 \, \text{m}$.

What Is the Molality of a Solution Prepared by Dissolving