Tutorial 8 Notes - Nov 13th 17th Tutorial Notes 1 Atkins...

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Nov. 13 th , 17 th Tutorial Notes 1) Atkins & dePaula, Exercise 5.7 a) We have been asked to calculate the freezing point of an aqueous solution, given that its osmotic pressure at 300 K is 120 kPa. Normally the freezing point of pure water is 273 K. However, the fact that this solution has osmotic pressure means that other colligative properties, including freezing point depression will also be present. To calculate the new freezing point of this solution we must use the equation for freezing point depression: solute solvent mass f f T K m n K ! = " = " The cryoscopic constant for water can be obtained from Table 7.2 in Atkins & dePaula and is 1.86 K kg mol -1 . However, we do not know the molality of this solution. Instead we have been given the magnitude of another colligative property which is also dependent on the amount of solute present, osmotic pressure: solute solvent cRT n RT V ! = = Solving for n solute : solvent solute V n RT ! = We do not know the volume of the solvent.
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Tutorial 8 Notes - Nov 13th 17th Tutorial Notes 1 Atkins...

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