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
g = surface tension [kg/s2]
h = height [m]
d = density [kg/m3]
g = accel. gravity = [9.8 m/s2]
rhdg
γ
2
2γ
2(0.07259 kg/s 2 )
4
r


4.5x10
m
2
3
2
hdg (3.29x10 m)(1000 kg/m )(9.8 m/s )
b) We will then find the surface tension of an uknown (ethanol)
rhdg (4.5x10 4 m)(1.3x10 2 m)(790 kg/m 3 )(9.8 m/s 2 )
γ

 0.0226 kg/s 2
2
2
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)
h = 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]
2r 2 (d - d O )g
η
9v
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
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:
3
 0.0060 mol  1 L  1 cm 

1 drop   1.5x10 7 mol


3 
L

 1000 cm  40 drops 
1.5x10
7

 282.47 g 
5
mol 
  4.24x10 g
 mol 
 1 cm3 

 4.24x10 -5 g  4.73x10 5 cm3
 0.895 g 

1b:

5
3
 Volume   4.73x10 cm 
  2.49x10 7 cm
thickness  molecule length  
  
2
 Area   190 cm

1c:
Area
2a: molecule 

1.5x10
7
190 cm 2
 2.10x10 15 cm 2 /molecule
23
6.022x10 molecule
mol
mol



2
 1 nm 
2
2.10x10 15 cm 2 /molecule 
  0.21 nm /molecule
-7
 1x10 cm 