Exam_2_LB_172_Practice_Questions-JT

Exam_2_LB_172_Practice_Questions-JT - EXAM 2 LB 172...

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EXAM 2: LB 172 PRACTICE PROBLEMS JULIAN THWAINEY 1. The boiling point elevation of a solution of 0.177 mol of a sugar in 53 g of water is 1.713 °C. Calculate kb? Δ T b = miK b K b = molal boiling point elevation constant different for every every liquid different IMFs in each case! K b > 0 m = mol of solute kg of solvent Calculate K b Δ T b = 1.713 ° C m = .177mol .053kg = 3.34 m i = 1 (sugar is molecular) K b = ? Δ T b mi = K b 1.713 1(3.34) = K b = .513 °C m 2.Cyclohexane has a freezing point of 6.50 °C and a Kf of 20.0 °C/m . What is the freezing point of a solution made by dissolving 0.694g of biphenyl (C12H10) in 25.0g of cyclohexane? Δ T f = T f T f o Δ T f freezing point depression, must be negative,depends on # of dissolved particles T f =freezing point of solution T f o =freezing point of pure solvent

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= mol of solute kg of solvent Δ T f = miK f K f = molal freezing point elevation constant different for every every liquid different IMFs in each case! K f < 0 Calculate T f T f o = 6.50 ° C K f = 20.0 ° C/m mol biphenyl (C 12 H 10 )=0.694g × 1mol 154.2g = .00450mol biphenyl m = .00450mol .025kg = .180 m T f =? i = 1 (biphenyl is molecular, contains only H and C) Δ T f = 1(.180)( 20) = 3.60 3.60 = T f 6.50 T f = 2.90°C 3. An osmotic pressure device was used to determine the molar mass of a protein. A mass of 0.150 g of protein dissolved in 100 mL of H2O at 300 K supports a column of water 5.00 cm high by osmotic pressure. It takes 10.0 m of this water to equal 1.00 atm. What is the
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Exam_2_LB_172_Practice_Questions-JT - EXAM 2 LB 172...

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