8/18/2019 Chem 16 Problem Set 10

1/3

Problem Set 10 Solutions

(30 Points Total – 3 points for each part)

1. An aqueous solution of ethylene glycol (antifreeze; C 2H6O2) is 16.0% ethylene glycol bymass. The density of the solution is 1.019 g/mL. Calculate:

a) molarity

Assuming 1.0 L solution:

Moles C 2H6O2 = 1000 mL ! 1.019 g1 mL

!16 g C 2H 6O2

100 g!

1 mol62.069 g

= 2.63 mol

Molarity = 2.63 M

b) molality

Assuming 1.0 L solution:Mass solution = 1000 mL !

1.019 g1 mL

= 1019 g

Mass C 2H6O2 = 1019 g ! 16 g C 2H 6O 2

100 g = 163 g

Mass of water = 1019 – 163 = 865 g

Molality =2.63 mol C 2H 6O2

0.856 kg H 2O = 3.07 m

c) mole fraction of water

! H 2 O =

856 g18.02 g/mol

2.63 +856 g

18.02 g/mol

= 0.943

d) freezing point

! T = m K f = 3.07 m ! 1.86°C/ m = 5.71°C

F.P. = 0°C – 5.71°C = – 5.71°C

8/18/2019 Chem 16 Problem Set 10

2/3

2. A 1.24– M solution of KI has a density of 1.15 g/cm 3.a) Calculate the mole fraction of water in this solution, assuming that the KI is completely

dissociated.

Assuming 1.0 L solution:

Mass solution = 1000 mL ! 1.15 g1 mL = 1150 g

Mass KI = 1.24 mol ! 166.0 g1 mol

= 206 g

Mass water = 944 g

! H 2 O =

944 g18.02 g/mol

2 1.24( ) + 944 g18.02 g/mol

= 0.955

b) Predict the vapor pressure of water over this solution at 25°C, assuming completedissociation of the KI.P

H 2 O = 0.955 ! 23.8 torr = 22.73 torr

c) Predict the freezing point of this solution, assuming complete dissociation of the KI.

! T = m K f

= 2 ! 1.24 mol0.944 kg

! 1.86°C/ m = 4.89°C

F.P. = 0°C – 4.89°C = –4.89°C

d) The freezing point of this solution is measured to be –4.46°C. What percent of the KI isactually dissociated in this solution?

Molality KI (nonelectrolyte) =1.24 mol KI

0.944 kg H 2O = 1.31 m

Calculate i: i =4.46 o C

1.31 m ! 1.86 o C/ m = 1.83

Let x be the fraction dissociated:

KI"

K+ + I

– 1 0 0

–x +x +x1–x x x

1.83 = 1–x + x + x = 1 + xx = 0.83

Therefore, 83% of the KI is dissociated.

8/18/2019 Chem 16 Problem Set 10

3/3

3. Lysozyme is an enzyme that breaks down bacterial cell walls. It is found in tears andegg whites and elsewhere. A solution containing 0.150 g of lysozyme in 210. mL of water solution has an osmotic pressure of 0.953 torr at 25°C. What is the molarmass of lysozyme?

Osmotic pressure ! V = nRT so:

n =! VRT =

"#$

%&'0.953760 atm (0.210 L)

(0.0821 L·atm/mol·K)(298 K) = 1.076 ( 10 –5 moles

so the molar mass is:

0.150 g1.076 ( 10 –5 mol

= 13900 g/mol or 1.39 ( 10 4 g/mol

4. Table sugar (sucrose, C 12H22O11 ) and table salt (NaCl) are both crystalline,colorless solids which are similar in outward appearance. You are given 100. gramsof a mixture of these two solids. You dissolve the mixture in 500. grams of waterand find that the resulting solution freezes at –2.25°C. Calculate the mass of sugarand salt in the original mixture.

2.25°C ="#$

%&'moles of dissolved solutes

0.500 kg water ( ( 1.86°C/m)

moles of dissolved solutes = 0.605 moles

For each mole of salt we get 2 moles of solutes . Let's do 2 equations 2 unknowns:

0.605 moles = n(sucrose) + 2n(salt) = m(sucrose)

342 g/mol + 2 m(salt)

58.5 g/mol100.g = m(sucrose) + m(salt)

Solving these two equations gives: m(sucrose) = 90g and m(salt) = 10g