10 Appendices

APPENDIX Appendix A List and Uses of Apparatus



A pipette is a narrow glass tube used for measuring liquid or for moving small amounts of 



liquid from one place to another. A pipetol is a rubbery-like material used to fill the pipette. An iron stand is a metal rod attached to a metal base. The heavy base keeps the stand



stable, and the vertical metal rod allows for easy height adjustment of the iron ring or 



clamp used to hold the substances to be heated above the Bunsen burner during heating. A thermometer is a device that measures temperature or a temperature gradient. ork is an impermeable, buoyant material, a prime-subset of bark tissue that is harvested



for commercial use, used as stopper in laboratory. A graduated cylinder, measuring cylinder or mi!ing cylinder is a common piece of 



laboratory equipment used to measure the volume of a liquid. "t has a narrow cylindrical shape. #ach marked line on the graduated cylinder represents the amount of liquid that



has been measured. A Bunse Bunsen n burne burnerr, name named d afte afterr $obe $obert rt Buns Bunsen en,, is a comm common on piece piece of labo laborat rator ory y equipment that produces a single open gas flame, which is used for heating, sterili%ation, and combustion. The gas can be natural gas &which is mainly methane' or a liquefied



 petroleum gas, such as propane, butane, or a mi!ture of both. (ire gau%e is placed on top of the iron ring to support beakers and flasks when heating them. A clay triangle is placed on top of the iron ring to support a crucible when it is



 being heated. )ropper is a laboratory tool commonly used in chemistry, biology and medicine to



transport a measured volume of liquid, often as a media dispenser. An analytical balance &often called a *lab balance*' is a class of balance designed to measure small mass in the sub-milligram range.



Test tubes are widely used by chemists to hold, mi!, or heat small quantities of solid or  liquid liquid chemicals, chemicals, especially especially for qualitative qualitative e!periments e!periments and assays. assays. Their round bottom and straight sides minimi%e mass loss when pouring, make them easier to clean, and allow convenient monitoring of the contents.



Test tubes are widely used by chemists to hold, mi!, or heat small quantities of solid or  liquid liquid chemicals, chemicals, especially especially for qualitative qualitative e!periments e!periments and assays. assays. Their round bottom and straight sides minimi%e mass loss when pouring, make them easier to clean, and allow convenient monitoring of the contents.

Appendix B Definition of Terms Terms



+olubility is the property of a solid, liquid, or gaseous chemical substance called solute to dissolve in a solid, liquid, or gaseous solvent to form a solution of the solute in the



solvent iscible is a word used by chemists to e!plain why some liquids mi! together well, while others do not. iscible liquids form homogeneous solutions, which is another way



of saying liquids that have a uniform nature. henol, also known as carbolic acid, is an aromatic organic compound with the molecular 



formula /01/. "t is a white crystalline solid that is volatile. /ydrochloric acid is a clear, colorless, highly pungent solution of hydrogen chloride &/l' in water. "t is a highly corrosive, strong mineral acid with many industrial uses. /ydrochloric acid is found naturally in gastric acid.

Appendix C Computations Phenol – Water

$un 2o. 34 (t. of henol in 3m5 678 phenol solution4

m = ρV =

(

1.05 g

mL

)(

)(

)

1 mL 0.80 = 0.84 g

(t. of (ater (ater in 3m5 678 phenol ph enol solution4

m = ρV =

(

1.05 g

mL

)(

)(

 Average  Average Temperature emperature =

wt phenol =

)

1 mL 0.20 = 0.21 g

24 + 23 2

=23.5 ℃

  0.84 wt phenol x 100 =  x 100 =7.6018 wt phenol + wt water water 10.21 + .84

$un 2o. 94 (t. of henol in 3m5 678 phenol solution4

m = ρV =

(

1.05 g

mL

)(

)(

)

2 mL 0.80 = 1.68 g

(t. of (ater (ater in 3m5 678 phenol ph enol solution4

m= ρV =

(

1.05 g

mL

)(

)(

2 mL 0.20

)=0.42 g

 Average  Average Temperature emperature =

wt phenol =

54 + 49 2

= 51.5 ℃

  1.68 wt phenol x 100 =  x 100=13.8843 wt phenol + wt water water 10.42 + 1.68

$un 2o. : (t. of henol in 3m5 678 phenol solution4

m = ρV =

(

1.05 g

mL

)(

)(

)

3 mL 0.80 = 2.52 g

(t. of (ater (ater in 3m5 678 phenol ph enol solution4

m = ρV =

(

1.05 g

mL

)(

)(

 Average  Average Temperature emperature =

wt phenol =

)

3 mL 0.20 = 0.63 g

62+ 58 2

=60 ℃

wt phenol   2.52 x 100 =  x 100=19.1635 10.63 + 2.52 wt phenol + wt water water

$un 2o. ;4 (t. of henol in 3m5 678 phenol solution4

m= ρV =

(

1.05 g

mL

)(

)(

4 mL 0.80

)=3.36 g

(t. of (ater (ater in 3m5 678 phenol ph enol solution4

m= ρV =

(

1.05 g

mL

)(

)(

4 mL 0.20

)=0.84 g

64 + 60.5

 Average Temperature =

wt phenol =

2

=62.25 ℃

  3.36 wt phenol x 100 =  x 100=23.6620 wt phenol + wt water 10.84 + 3.36

$un 2o. 04 (t. of henol in 3m5 678 phenol solution4

m = ρV =

(

1.05 g

mL

)(

)(

)

5 mL 0.80 = 4.2 g

(t. of (ater in 3m5 678 phenol solution4

m = ρV =

(

1.05 g

mL

)(

)(

 Average Temperatu ℜ=

wt phenol =

)

5 mL 0.20 = 1.05 g

66 + 62 2

=64 ℃

wt phenol 4.2 x 100 =  x 100=27.5410 11.05 + 4.2 wt phenol + wt water

$un 2o.  (t. of henol in 3m5 678 phenol solution4

m = ρV =

(

1.05 g

mL

)(

)(

)

6 mL 0.80 = 5.04 g

(t. of (ater in 3m5 678 phenol solution4

m = ρV =

(

1.05 g

mL

)(

)(

6 mL 0.20

)=1.26 g

 Average Temperature =

wt phenol =

67 + 64 2

= 65.5 ℃

  5.04 wt phenol x 100 =  x 100= 30.9202 wt phenol + wt water 11.26 + 5.04

$un 2o. < (t. of henol in 3m5 678 phenol solution4

m = ρV =

(

1.05 g

mL

)(

)(

)

7 mL 0.80 = 5.88 g

(t. of (ater in 3m5 678 phenol solution4

m = ρV =

(

1.05 g

mL

)(

)(

 Average Temperature =

wt phenol =

)

7 mL 0.20 = 1.47 g

68 + 65 2

=66.5 ℃

wt phenol 5.88  x 100 =  x 100=33.9905 11.47 + 5.88 wt phenol + wt wat er

$un 2o. 6 (t. of henol in 3m5 678 phenol solution4

m = ρV =

(

1.05 g

mL

)(

)(

8 mL 0.80

)=6.72 g

(t. of (ater in 3m5 678 phenol solution4

m = ρV =

(

1.05 g

mL

)(

)(

)

8 mL 0.20 = 1.68 g

 Average Temperature =

wt phenol =

69 + 66.5 2

=67.75 ℃

  6.72 wt phenol x 100 =  x 100 =36.5217 wt phenol + wt water 11.68 + 6.72

$un 2o. =4 (t. of henol in 3m5 678 phenol solution4

m = ρV =

(

1.05 g

mL

)(

)(

9 mL 0.80

)=7.56 g

(t. of (ater in 3m5 678 phenol solution4

m = ρV =

(

1.05 g

mL

)(

)(

 Average Temperature =

wt phenol =

)

9 mL 0.20 = 1.89 g

69.5 + 65 2

=67.25 ℃

wt phenol   7.56 x 100 =  x 100=38.8689 11.89 + 7.56 wt phenol + wt water

$un 2o. 37 (t. of henol in 3m5 678 phenol solution4

m = ρV =

(

1.05 g

mL

)(

)(

)

10 mL 0.80 =8.4 g

(t. of (ater in 3m5 678 phenol solution4

m = ρV =

(

1.05 g

mL

)(

)(

)

10 mL 0.20 =2.1 g

 Average Temperature =

wt phenol =

69.5 + 65 2

=67.25 ℃

8.4 wt phenol  x 100 =  x 100= 40.9796 wt phenol + wt water 12.1 + 8.4

$un 2o. 33 (t. of henol in 3m5 678 phenol solution4

m = ρV =

(

1.05 g

mL

)(

)(

)

11 mL 0.80 = 9.24 g

(t. of (ater in 3m5 678 phenol solution4

m = ρV =

(

1.05 g

mL

)(

)(

 Average Temperature =

wt phenol =

)

11 mL 0.20 =2.31 g

69.5 + 65 2

=67.25 ℃

wt phenol   9.24 x 100 =  x 100 =42.8770 12.31 + 9.24 wt phenol + wt water

$un 2o. 39 (t. of henol in 3m5 678 phenol solution4

m = ρV =

(

1.05 g

mL

)(

)(

)

12 mL 0.80 =10.08 g

(t. of (ater in 3m5 678 phenol solution4

m = ρV =

(

1.05 g

mL

)(

)(

)

12 mL 0.20 =2.52 g

 Average Temperature =

wt phenol =

69.5 + 65 2

=67.25 ℃

10.08 wt phenol x 100 =  x 100= 44.6018 wt phenol + wt water 12.52 + 10.08

Phenol – Water

$un 2o. 3

wt of phenol=

wt of water =

1.05 g

mL

105 g

mL

( 10 mL) ( 0.80 )= 8.4 g

( 10 mL) ( 0.2 )=2.1 g

wt of total water=3.1 g

wt phenol =

8.4 wt phenol x 100 =  x 100=73.0435 wt phenol + wt water 8.4 + 3.1

$un 2o. 9

wt of phenol=

wt of water =

1.05 g

mL

105 g

mL

( 10 mL) ( 0.80 )= 8.4 g

( 10 mL) ( 0.2 )=2.1 g

wt of total water = 4.1 g

wt phenol =

$un 2o. :

wt phenol 8.4 x 100 =  x 100= 67.2 8.4 + 4.1 wt phenol + wt water

wt of phenol=

wt of water =

1.05 g

mL

105 g

mL

( 10 mL) ( 0.80 )= 8.4 g

( 10 mL) ( 0.2 )=2.1 g

wt of total water=5.1 g

wt phenol =

8.4 wt phenol x 100 =  x 100 =62.2222 wt phenol + wt water 8.4 + 5.1

$un 2o. ;

wt of phenol=

wt of water =

1.05 g

mL

105 g

mL

( 10 mL) ( 0.80 )= 8.4 g

( 10 mL) ( 0.2 )=2.1 g

wt of total water = 6.1 g

wt phenol =

wt phenol 8.4 x 100 =  x 100=57.9310 8.4 + 6.1 wt phenol + wt water

$un 2o. 0

wt of phenol=

wt of water =

1.05 g

mL

105 g

mL

( 10 mL) ( 0.80 )= 8.4 g

( 10 mL) ( 0.2 )=2.1 g

wt of total water=7.1 g

wt phenol =

8.4 wt phenol x 100 =  x 100 =54.1935 8.4 + 7.1 wt phenol + wt water

$un 2o. 

wt of phenol=

wt of water =

1.05 g

mL

105 g

mL

( 10 mL) ( 0.80 )= 8.4 g

( 10 mL) ( 0.2 )=2.1 g

wt of total water= 8.1 g

wt phenol =

8.4 wt phenol x 100 =  x 100=50.9091 wt phenol + wt water 8.4 + 8.1

$un 2o. <

wt of phenol=

wt of water =

1.05 g

mL

105 g

mL

( 10 mL) ( 0.80 )= 8.4 g

( 10 mL) ( 0.2 )=2.1 g

wt of total water = 9.1 g

wt phenol =

wt phenol 8.4  x 100 =  x 100= 48 8.4 + 9.1 wt phenol + wt wate r

$un 2o. 6

wt of phenol=

wt of water =

1.05 g

mL

105 g

mL

( 10 mL) ( 0.80 )= 8.4 g

( 10 mL) ( 0.2 )=2.1 g

wt of total water=10.1 g

wt phenol =

wt phenol 8.4  x 100 =  x 100= 45.4054 8.4 + 10.1 wt phenol + wt wate r

$un 2o. =

wt of phenol=

wt of water =

1.05 g

mL

105 g

mL

( 10 mL) ( 0.80 )= 8.4 g

( 10 mL) ( 0.2 )=2.1 g

wt of total water =11.1 g

wt phenol =

8.4 wt phenol  x 100=  x 100 = 43.0769 wt phenol + wt w ater 8.4 + 11.1

$un 2o. 37

wt of phenol=

wt of water =

1.05 g

mL

105 g

mL

( 10 mL) ( 0.80 )= 8.4 g

( 10 mL) ( 0.2 )=2.1 g

wt of total water=12.1 g

wt phenol =

wt phenol 8.4 x 100 =  x 100 =40.9736 8.4 + 12.1 wt phenol + wt water

$un 2o. 33

wt of phenol=

wt of water =

1.05 g

mL

105 g

mL

( 10 mL) ( 0.80 )= 8.4 g

( 10 mL) ( 0.2 )=2.1 g

wt of total water=13.1 g

wt phenol =

wt phenol 8.4 x 100 =  x 100=39.0698 8.4 + 13.1 wt phenol + wt water

$un 2o. 39

wt of phenol=

wt of water =

1.05 g

mL

105 g

mL

( 10 mL) ( 0.80 )= 8.4 g

( 10 mL) ( 0.2 )=2.1 g

wt of total water =14.1 g

wt phenol =

8.4 wt phenol x 100 =  x 100 =37.3333 wt phenol + wt water 8.4 + 14.1

43.0769

Compositionat Critical Sol’ n Temperature =

94 43.0769 94

¿ 12.6569

+

100− 43.0769 18

 x 100

Appendix D Ans!er to Pro"lems

3. At :7> a mi!ture of /01/ and /91 is made up containing 78 by weight /91. The mi!ture splits into two layers, the /01/ layer containing .

?iven4 :7 8wt aniline +oln4 Basis 377 g solution From graph4 aniline layer @ =; 8wt aniline (ater layer @ = 8wt water  5et C @ g of aniline layer  D @g of water layer  :7 @ 7.=;C E 7.7;D