Prelab for Colligative Properties

Prelab for Colligative Properties - Chemistry 112 Lab...

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Chemistry 112 Lab Section 005 Prelab for Colligative Properties 1. Addition of solute to a pure solvent will decrease its freezing point and increase its boiling point. 2. Estimate the van’t Hoff factor (i) for the following species. (Hint: Two species are ionic.) Barium chloride n-propanol ammonium sulfate hexane BaCl 2 CH 3 CH 2 CH 2 OH (NH 4 ) 2 SO 4 CH 3 (CH 2 ) 4 CH 3 3 1 3 1
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3. Given the following information: Solvent Formula Boiling point (°C) Freezing point (°C) K b (°C/m) K f (°C/m) Cyclohexane C 6 H 12 80.74 6.55 2.79 20.0 a. Calculate the boiling point and freezing point of a solution containing 100. mL of cyclohexane (density = 0.779 g/mL) and 5.1 g of ethanol (CH 3 CH 2 OH). Boiling point = 80.74 ° C + ΔT = 80.74 ° C + 3.96 ° C = 84.70 °C ΔT= i K b m = 1 2.79 ° C / m m = 1 2.79 ° C / m 1.42 mol / kg = 3.96 °C m = molsolute kg solvent = 0.110705913 mol ( 100 mL 0.779 g mL ) / 1000 = 0.110705913 mol 0.0779 kg = 1.421128537 mol / kg mol solute = g molarmass = 5.1 g 46.068
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Unformatted text preview: mol = 0.110705913 mol molar mass = g mol = 46.068 g 1 molC H 3 C H 2 OH = 46.068 g / mol Freezing point = 6.55 °C − ΔT = 6.55 ° − 28.42 °C = − 21. 87 °C Δ±= i K b m = 1 ∗ 20.0 ° C / m ∗ m = 1 ∗ 20.00 ° C / m ∗ 1.42 mol / kg = 28.42 ° C b. A soluTon containing 0.200 g of an unknown substance in 2.50 g of cyclohexane is found to freeze at 5.1 °C. What is the molar mass (MM) of the unknown substance if it is a non-electrolyte? m = ΔT i K f = −( 5.1 ° C − 6.55 ° C ) 1 ∗ 20.0 °C / m = 1.45 °C 20.0 ° C / m = 0.0725 m 0.0725 m = molsolute kg solvent = molsolute 0.00250 kg mol solute = 0.0725 m ∗ 0.00250 kg = 0.00018125 molof unknown molar mass = g mol = 0.200 g 0.00018125 mol = 1103.448276 g / mol = 1.10 ∗ 10 3 g / mol...
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  • Spring '08
  • LEMASTER
  • Chemistry, Mole, Colligative properties, Solvent, Hoff, Van 't Hoff factor, mol solute

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