Sp18 Sample Quiz 2 - Colligative Properties - key PDF

Title Sp18 Sample Quiz 2 - Colligative Properties - key
Author Michael Foster
Course General Chemistry Lab I
Institution Georgetown University
Pages 1
File Size 52.2 KB
File Type PDF
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General Chemistry 010 – Spring 2018 Sample Quiz 2: Colligative Properties Constants: 1.

Kf of benzene = 5.12oC/m

Name ________ Key_______________ Section __________ R= 0.0821 L atm K-1 mol-1

Conversion factor: K= oC + 273.15

Which of the following aqueous solutions will have the highest boiling point? a.

0.10 m Na2SO4

b. 0.20 m glucose

c. 0.25 m sucrose

d. 0.10 m NaCl

e. 0.10 m SrSO4

For #2-5. Calculate the following and give answers with correct significant figures. Show all your work. 2.

3.

A solution of 3.80 g of CH3COOH in 80.0 g C6H6 has a freezing point of 3.50 oC. Calculate the molar mass of the solute (CH3COOH). The normal freezing point of pure benzene (C6H6) is 5.50 oC. Tf = Tf (solution) - Tf (solvent) = 3.50oC – 5.50oC = - 2.00oC

(0.0800 kg C6H6) (0.390625 mol solute/kg C6H6) = 0.03125 mol solute

Tf = - iKf m m= -Tf = - (-2.00 oC) = 0.390625 m iKf (1) (5.12oC/m)

Molar mass =

3.80 g = 122 g/mol 0.03125 mol

For dilute aqueous solutions in which the density of the solution is roughly equal to that of the pure solvent, the molarity of the solution is equal to its molality. Show that this statement is correct for a 0.010M aqueous urea (NH2)2CO solution. dsolution = dwater = 1 g/mL = 1 kg/L (NH2)2CO is a molecular compound. M = mol urea/1 L solution = 0.010 M mol urea = (0.010 mol/L) x (1 L) = 0.010 mol molality = mol/1 kg solvent = 0.010 mol/1 kg solvent = 0.010 m

4.

Calculate the number of moles of solute present in a 124.0 g of a solution that is 6.45 % glucose (180.2 g/mol) by mass. 6.45 g glucose = x g glucose 100 g glucose 124 g solution x = 8.00 g glucose 8.00 g glucose x 1 mol glucose = 4.4 x 10-2 mol glucose 180.2 g

5.

Calculate the osmotic pressure (in atm) formed by dissolving 2.8 g of KCl (74.55 g/mol) in 1.0 L of water at 25.0 oC. Assume complete dissociation of the KCl.  = MRT = 2.8 g KCl x 1 mol KCl x 2 mol ions L 74.55 g 1 mol KCl

x 0.0821 L atm x 298.15 K = 1.8 atm K mol

This sample quiz includes topics from Chapters 13 sections 13.4-13.5. You should already have studied when you take the sample quiz (20 min). Do not refer to your notes or books while taking the sample quiz. The questions or difficulty of the actual quiz will vary from this sample quiz. Expt 19 is included in the quiz so you must demonstrate a thorough understanding of the concepts and calculations of the experiment you performed last week....


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