Pure chloroform, CHCl₃, has a boiling point of 61.23°C. Its boiling point elevation constant, K_b, is 3.63°C m^-1. A solution was made by taking 11.25 g of an unknown nonelectrolyte and dissolving it in 115.5 g of chloroform. The measured boiling point of the solution was 65.46°C. Calculate the molecular weight of the unknown substance.Hint: Find molality, then moles, then molar mass, being sure to keep track of units.

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Quizplus · Accepted Answer

The boiling point elevation is given by the equation:$$\Delta T_b = K_b imes m$$where:$$\Delta T_b$$ is the boiling point elevation$$K_b$$ is the boiling point elevation constant$$m$$ is the molality of the solutionWe can rearrange this equation to solve for the molality:$$m = \frac{\Delta T_b}{K_b}$$Substituting the given values into this equation, we get:$$m = \frac{65.46°C - 61.23°C}{3.63°C m^{-1}} = 1.15 m$$The molality is defined as the number of moles of solute per kilogram of solvent. We can use this to calculate the number of moles of solute in the solution:$$moles\ of\ solute = molality imes mass\ of\ solvent\ (in\ kg)$$Substituting the given values into this equation, we get:$$moles\ of\ solute = 1.15 m imes 115.5 g imes \frac{1 kg}{1000 g} = 0.133 mol$$Finally, we can calculate the molecular weight of the solute by dividing the mass of the solute by the number of moles of solute:$$molecular\ weight = \frac{mass\ of\ solute}{moles\ of\ solute}$$Substituting the given values into this equation, we get:$$molecular\ weight = \frac{11.25 g}{0.133 mol} = 84.6 g mol^{-1}$$Therefore, the molecular weight of the unknown substance is 84.6 g mol^-1, which is closest to choice A, 83.6 g mol^-1.

Pure Chloroform, CHCl3, Has a Boiling Point of 61

Pure Chloroform, CHCl₃, Has a Boiling Point of 61