A 0.442g sample of a nonvolatile solid solute dissolvesin 15.0g of t-butanol. Th
ID: 675968 • Letter: A
Question
A 0.442g sample of a nonvolatile solid solute dissolvesin 15.0g of t-butanol. The freezing point if tohe solution is23.9degrees celcius. a. What is the molality of the solute in the solution b. Calculate the molar mass of the solute c. The same mass of solute is dissolved in 15.0g ofcyclohexane instead of t-butanol. what is the expected freezingpoint change of this solution? A 0.442g sample of a nonvolatile solid solute dissolvesin 15.0g of t-butanol. The freezing point if tohe solution is23.9degrees celcius. a. What is the molality of the solute in the solution b. Calculate the molar mass of the solute c. The same mass of solute is dissolved in 15.0g ofcyclohexane instead of t-butanol. what is the expected freezingpoint change of this solution?Explanation / Answer
Formula : Tf = Kf *m Where Tf is the depression in freezingpoint Kf is the molal depression in freezing constant mis the molality Kf for t-butanol of 8.37°C/m. and D T = Tpure -Tsolution T pure for t-butanol = 25.69 °C D T = 25.69 °C -23.9°C =1.79 °C a ) m = DT / Kf = 1.79 °C / 8.37°C/m. =0.2138 m b) molality = no.of moles of solute / 1 kg ofsolvent number of moles = wt ofthe solute / molar mass of the solute We have wt ofthe solute = 0.442 g wt of the solvent = 15.0 g = 0.015 kg 0.2138 m = 0.442 g / molar mass * 0.015 kg molar mass = 0.442 g / 0.2138 m * 0.015 kg =137.82 g /mol c) Tf = Kf *m Kf ofcyclohexane = 20 °C/m. Tf = 20 °C/m. * 0.442 g / molarmass of solute * 1 kg of solvent By substituting the values in the above equation we can getthe expected freezing point change of thissolution.Related Questions
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