AND CALCULATIONS EXPERIMENT 17: Viscosity of Liquids Question Assume that ∆P and L are constants. The viscosity is therefore a function of r only. Integrating equation 2 with respect to t we get.

V=

Rearranging to make the subject, we get

Absolute error =

Relative error = Absolute error/ actual =

Therefore, the relative is fourfold the radius of the viscometer.

Question 2

Values obtained from this formula

Density = mass/volume

Mass = density x volume

Viscosity = density x area/time

Since a similar viscometer was used, the cross-sectional area is constant and therefore the viscosity formula reduces to density/time.

Table 1: Methanol + Water

% methanol

Average time

(sec)

Density

(gcm-3

Viscosity

Gcm-1s-1

Volume of Methanol (ml)

Volume of water (ml)

0

53.0375

0.977

0.018

0

10

20

85.925

0.9392

0.011

2

8

40

109.3575

0.9014

0.008

4

6

60

102.15

0.8636

0.008

6

4

80

83.9275

0.8258

0.01

8

2

100

49.885

0.788

0.016

10

0

% Toluene

Average Time

(sec)

Density

Viscosity

Volume of Toluene

Volume of Xylene

0

84.3475

0.857

0.01

0

10

20

84.347

0.8578

0.01

2

8

40

81.6875

0.8586

0.01

4

6

60

80.605

0.8594

0.01

6

4

80

79.045

0.8602

0.01

8

2

100

78.53

0.861

0.01

10

0

Data Analysis

From the results above, viscosity decreases with increase in methanol content. For Toluene and Xylene, viscosity is constant. Viscosity reduces with increase in intermolecular forces and vice versa. If the resulting solution forms stronger intermolecular forces as in methanol and water the viscosity is lowered. For weak intermolecular forces, as in toluene and Xylene mixture remains constant, i.e. equal to the viscosity of the individual liquids.

Question 3

Viscosity is inversely proportional to the intermolecular forces in a solution. Viscosity of water is higher than that of methanol. Water molecules combine with methanol molecules to form stronger intermolecular forces which results to a decrease in viscosity.

EXPERIMENT 32: The Monomer Linkage Properties and Molecular Weight of Poly (vinyl alcohol)

Question 1

Mv/Mn = [(1+a)ᴦ(1+a)]1/a = S

From gamma function tables for a=0.5,

ᴦ(1+a) = ᴦ1.5 = 0.886227

For a = 0.6’

ᴦ(1+a) = ᴦ1.6 = 0.893515

S = [1+0.6(0.893515)]1/0.6 = 1.81

Question 2

Decrease in molecular weight =1/ Mn’ – 1/Mno

Where Mn’ is the number average molecular weight of polymer in the sample after cleavage. Mno is the number average molecular weight of the polymer in sample before cleavage.

The molecular weight of the monomer unit = Mo.

Change in the average molecular weight is expressed as a ratio ƒ.

Ƒ =

Simplifying

Ƒ = (Mo/ Mn’ ) – (Mo/ Mno)

Multiplying both sides by Mv

Mv Ƒ = [(Mo/ Mn’ ) – (Mo/ Mno)] Mv

But Mv/Mn = S From equation 17.

Therefore, Mv Ƒ = MoS’ – Moso

Ƒ = (MoS’ – Moso)/ Mv

Factorizing, Ƒ = MoS (1/ Mv’ – 1/Mvo)

Question 3

An experiment which measure the direct force generated during polymerization of the protein or biopolymer molecules using an optical trap can be conducted.

Question 4

For PVOH, whose structure is spherical the assumption is appropriate. The assumption yields poor results in non-spherical non-polar molecular structures with longer hydrocarbon chain.

Question 5

The water molecule in vapor form enters the PVOH molecule to minimize the solvolytic associations. The molecule consequently expands and swells out. In aqueous solutions, the polymer knots up and avoid interaction with water molecules. This is as a result of the temperature difference between the two media.

Question 6