Properties of Liquids Pre-Lab

1. Surface Tension = resistance of a liquid to an increase in its surface area

a. Surface molecules are not involved in all possible intermolecular bonding

b. Requires energy to go the surface, so liquid resists increases in surface area

c. The higher the intermolecular forces, the higher the surface tension

2. Capillary Action = spontaneous rising of a liquid up a narrow tube

a. Adhesive Forces = polar liquid has intermolecular forces with polar surface

b. Cohesive Forces = intermolecular forces of the liquid for itself

c. Water: Adhesive (H-Bonding) > Cohesive, so concave meniscus

d. Mercury: Cohesive (London) > Adhesive, so convex meniscus

3. We will use the distance up a capillary the liquid climbs to find Surface Tension

a) We will use the known surface tension of water to find the tube radius

= surface tension [kg/s2] h = height [m]

d = density [kg/m3] g = accel. gravity = [9.8 m/s2]

b) We will then find the surface tension of an uknown (ethanol)

m 4.5x10)m/s )(9.8kg/m m)(1000(3.29x10

)kg/s 2(0.07259

hdg

2γr

2

rhdgγ

4232

2

22324

kg/s 0.02262

)m/s )(9.8kg/m m)(790m)(1.3x10 (4.5x10

2

rhdgγ

4. Viscosity = a liquid’s resistance to flow

a. Large intermolecular forces would cause high viscosity (glycerol)

b. Large, complex molecules can become physically entangled (grease)

= coefficient of viscosity[g/cm·s] r = radius of sphere = 0.80cm

do = density of fluid = 1.26 g/cm3 g = accel. gravity = 980 cm/s2

d = density of glass sphere = 2.23 g/cm3

v = velocity of sphere = length of fall [cm]/time of fall [s] = [cm/s]

5. Polarity and Solubility: “Like Dissolves Like”

a. The polarity of three solvents will be investigated

i. H2O

ii. CH3OH

iii. CHCl3

b. Two solids and one liquid will be used to determine polarity

i. Na2Cr2O7 = ionic = extremely polar

ii. I2 = nonpolar (only dispersion forces)

iii. Vegetable Oil = nonpolar

H2C

HC

H2C

OH

OH

OH

9v

)gd-(d2rη O

2

falling glass bead

glycerol

6. Calculation of Molecular Size

a. We will trap a single layer of oleic acid on a water surface

b. We will use the area of oleic acid to find its length and area/molecule

c. PreLab (page 5): 1 drop of 0.0060 M produces 190 cm2 film

d = 0.895 g/cm3 MW = 282.47 g/mol 40 drops = 1 ml

1a:

1b:

1c:

2a:

g4.24x10mol

g 282.47mol1.5x10

mol1.5x10drop 1drops 40

cm 1

cm 1000

L 1

L

mol 0.0060

57

73

3

355-3

cmx1073.4g 4.24x10g 0.895

cm 1

cmx1049.2cm 901

cmx1073.4

Area

Volumelength moleculethickness 7

2

35

/moleculenm 0.21cm 1x10

nm 1/moleculecm2.10x10

/moleculecm2.10x10

molmolecule 6.022x10

mol 1.5x10

cm 190

molecule

Area

22

7-215

21523

7

2

CO

OH