CHM 118_Sec_A_B_Spring_2015

General Chemistry Laboratory CHM 116Dr. Mohammad Harun-Ur-Rashid

Room# 110, IUBAT

Experiment # 02.

Name of Experiment: Flame test.

Objective of the test:

1. To explore the line spectra of a variety of elements byheating them in a Bunsen burner flame.

2. The data gathered from the flame colors created by theelements will be used to identify some elements present inunknown solutions.

3. To get the idea about qualitative analysis.

4. Identification of Metal ion in compound.

Theory:

The flame test is used to visually determine the identity of anunknown metal or metalloid ion based on the characteristic colorthe salt turns the flame of a Bunsen burner. The heat of theflame excites the metals ions, causing them to emit visiblelight. The characteristic emission spectra can be used todifferentiate between some elements.

Apparatus:

1. Bunsen burner 2. Watch glass dish

3. Platinum wire

4. Beaker

5. Salts/compound

6. Hydrochloric acid solution (dilute) 1


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7. Distilled water

Procedure:

1. Dip the wire in a acid solution2. Hold the wire in Bunsen flame (to check if wire is clean)

3. Place some of the salt on the wire

4. Hold the salt in the flame

5. Note the colour imparted to the flame.

6. Repeat for other salts.

7. Record your results in flame test chart (first table) , using different colours

8. Use the first table to fill in the blanks of the second table.

How to Interpret the Results

The sample is identified by comparing the observed flame coloragainst known values from a table or chart.

Flame Test Colors

Symbol Element Color

As Arsenic BlueB Boron Bright greenBa Barium Pale/Yellowish GreenCa Calcium Orange to red

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Cs Cesium BlueCu(I Copper(I) BlueCu(II)

Copper(II) non-halide Green

Cu(II)

Copper(II) halide Blue-green

Fe Iron GoldIn Indium BlueK Potassium VioletLi Lithium Magenta to carmineMg Magnesium Bright whiteMn(II) Manganese(II) Yellowish green

Mo Molybdenum Yellowish green

Na Sodium Intense yellow/golden yellow

P Phosphorus Pale bluish greenPb Lead BlueRb Rubidium Red to purple-redSb Antimony Pale greenSe Selenium Azure blueSr Strontium CrimsonTl Thallium Pure green

Zn Zinc Bluish green to whitish green

Result:

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Sl#

Name ofCompound

Molecular

formula

Metalion

Observed color

1 Sodium carbonate

Na2CO3 Na+ Yellow

2 Potassium iodide

KI K+ Violet

3 Copper (II) nitrate

Cu(NO3)2 Cu++ Green

Limitations of the Flame Test

The test cannot detect low concentrations of most ions. The brightness of the signal varies from one sample to

another. For example, the yellow emission from sodium ismuch brighter than the red emission from the same amount oflithium.

Impurities or contaminants affect the test results. Sodium,in particular, is present in most compounds and will colorthe flame. Sometimes a blue glass is used to filter out theyellow of sodium.

The test cannot differentiate between all elements. Severalmetals produce the same flame color. Some compounds do notchange the color of the flame at all.

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Experiment # 03.

Name of Experiment: Preparation of Stock solutions

having various strengths and the calculation of

Molecular Weight of Following Compounds

SL# Name of

Compound

Type

(solid/

liquid/

powder/

crystal)

Color Molecular

Formula

Molecular

Weight of

Individual

Element

Calcu

latio

n of

Molec

ular

Weigh

t

Molecular

Weight of

Compound

(g/mol)

1 Ammonium

Chloride

Solid

Crystal

White NH4Cl N=14, H=1,

Cl=35.5

14+(1

x4)+3

5.5

53.5

2 Aluminium

Sulphate

Solid

Crystal

White Al2(SO4)3 342.15

3 Calcium

Hydroxide

Powder White Ca(OH)2 74.10

4 Cadmium

Chloride

Solid

Crystal

White CdCl2 183.32

5 Copper(II)Nitr

ate

Solid

Crystal

Blue Cu(NO3)2 187.55

6 Copper Sulfate Solid

Crystal

Blue CuSO4 159.62

5

http://en.wikipedia.org/wiki/Sulfate

http://en.wikipedia.org/wiki/Aluminium


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7 Calcium

Chloride

Powder White CaCl2 110.98

8 Ferrous

Sulphate

Solid

Crystal

Green FeSO4 151.908

9 Ferric

Chloride

Solid

Crystal

Brown FeCl3 162.2

SL

#

Name of

Compound

Type

(solid/

liquid/

powder/

crystal)

Color Molecul

ar

Formula

Molecul

ar

Weight

of

Individ

ual

Element

Calculation

of

Molecular

Weight

Molecular

Weight of

Compound

(g/mol)

10 Magnesium

sulphate

Solid

Crystal

White MgSO4 120.366

11 Potassium

dichromate

Solid

Crystal

red-

orange

K2Cr2O7 294.185

12 Potassium

Chloride

Solid

Crystal

White KCl 74.5

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13 Potassium

Nitrate

Solid

Crystal

White KNO3 101.10

14 Sodium Sulphate Solid

Crystal

White Na2SO4 142.04

15 Sodium

Hydroxide

Solid

Crystal

White NaOH 40

16 Sodium Nitrate Solid

Crystal

Colorles

s

NaNO3 85

17 Sodium Nitrite Solid

Crystal

White NaNO2 67

18 Sodium

Bicarbonate

Solid

Crystal

White NaHCO3 84

19 Sodium

Carbonate

Powder White Na2CO3 106

20 Zinc Sulphate Solid

Crystal

White ZnSO4 161.47

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Experiment # 04.

Name of Experiment: Preparation of Stock solutions

having various strengths.

Required Equation Amount of Reagent, A = V x C x Mw ,

where V is the volume of stock solution in liter (L), C is the concentration or the strength of solution in molar (M) and Mw is the molecular weight of required reagent or compound in gram (g).

Example: 1: How much NaCl is required to prepare 400 ml of 0.24M NaCl solution (the MW of NaCl = 58.44 g/mole)?

Solution: We know that,

Amount of Reagent = V x C x Mw

= 0.4 L x 0.24 moles/L x 58.44 g/mole

= 5.61 g

Answer: 5.61 g of NaCl is required to prepare 400 ml of 0.24 M NaCl solution.

Example: 2: How much NaNO3 is required to prepare 2.0 liters ofa 1.5 M sodium nitrate (NaNO3) Solution (the MW of NaNO3 = 85g/mole)?


Amount of Reagent = V x C x Mw8


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= 2.0 L x 1.5 moles/L x 85 g/mole

= 255 g

Answer: 255 g of NaNO3 is required to prepare 2.0 liters of a 1.5M sodium nitrate (NaNO3) Solution.

Example: 3: How much Ca(OH)2 is required to prepare 5.0 litersof a 0.1 M Ca(OH)2 solution (the MW of Ca(OH)2 = 74 g/mole)?


Amount of Reagent = V x C x Mw

= 5.0 L x 0.1 moles/L x 74 g/mole

= 37 g

Answer: 37 g of Ca(OH)2 is required to prepare 5.0 liters of a 0.1 M Ca(OH)2 solution.

Molarity of Solution, C = A / (V x Mw) ,

where V is the volume of stock solution in liter (L), A is the Amount of Reagentin gram (g) and Mw is the molecular weight of required reagent or compound in gram (g).

Example: 4: What is the molarity of a solution in which 5.0g of sodium carbonate (Na2CO3) are dissolved in 200 mL of solution (the MW of Na2CO3 = 106 g/mole)?

Solution: We know that, 9


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Molarity of Solution, C = A / (V x Mw)

= 5.0 g / (0.2 L x 106 g/mole)

= 0.2358 ≈ 0.24 M

Answer: The molarity of the solution is 0.24 M.

Example: 5: What is the molarity of a solution containing 150 gof zinc sulfate (ZnSO4) per liter? (the MW of ZnSO4 = 161.5g/mole)?


Molarity of Solution, C = A / (V x Mw)

= 150.0 g / (1.0 L x 161.5 g/mole)

= 0.9287 ≈ 0.93 M

Answer: The molarity of the solution is 0.93 M.

Volume of Solution, V = A / (C x Mw) ,

where C is the concentration or the strength of solution in molar (M), A is the Amount of Reagent in gram (g) and Mwis the molecular weight of required reagent or compound in gram (g).

Example: 6: A 1.0 M solution containing 80 g of Copper Sulfate (CuSO4). Calculate the volume of the solution in liters (the MW of CuSO4 = 160 g/mole).


Volume of Solution, V = A / (C x Mw)

= 80.0 g / (1.0 M x 160.0 g/mole)

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= 0.5 L

Answer: The volume of the solution is 0.5 L.

Dilution of Stock solution

If you're working in a chemistry lab, it's essential to know how to calculate a dilution. Here's a review of how to prepare a dilution from a stock solution.

The calculation for the dilution:

MdilutionVdilution = MstockVstock

where Mdilution, Mstock the molarity of diluted and stock solution and Vdilution, Vstock the volume of diluted and stock solution respectively.

Example: 7: What volume of stock solution of 0.24 M NaCl is required to prepare 75 mL of 0.1 M NaCl?


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In the case of dilution, MdilutionVdilution = MstockVstock

0.1 x 0.075 = 0.24 x Vstock

Vstock = (0.1 x 0.075)/0.24

= 0.03125 L or 31.25 mL

Answer: 31.25 mL stock solution is required.

Example: 8: What volume of stock solution of 2.0 M NaOH is required to prepare 50 mL of 1.0 M NaOH?


In the case of dilution, MdilutionVdilution = MstockVstock

1.0 x 0.050 = 2.0 x Vstock

Vstock = (1.0 x 0.050)/ 2.0

= 0.025 L or 25 mL

Answer: 25 mL stock solution is required.

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Experiment # 05.

Name of Experiment: Acid-Base Titration.

Objective of the test:

To determine the concentration of an unknown acid or

base

To determine the pH of a solution

To demonstrate the basic laboratory technique of

titration

To learn to calculate molarity based on titrations

Theory:

In chemistry, acid-base titration is used to determine the

concentration of a solution. The process of titrating utilizes a

solution of unknown concentration, which is either an acid or a

base. A solution of known concentration is then added to the

solution being testing. The added solution is the opposite of

what the tested solution is: If one is a base, the other is an

acid. The type of chemical reaction occurring between the two

solutions is called a "neutralization reaction." The amount of

one required to exactly neutralize the other tells you when the

amount of acid and base are equal. The known concentration can

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therefore be used to calculate the unknown concentration. Acid-

base titrations are the most common type of titration.

Equivalence Point

The point, at which the titrant fully neutralizes all the

titrand, leaving neutral water, is called the "equivalence

point." This is when the titrant has "used up" all of the

titrand. The acid and base have fully canceled each other out. An

example of this sort of mutual cancellation is illustrated in

this chemical formula:

HCl + NaOH-------> NaCl + H2O

At equilibrium, the solution's pH is 7.0.

Procedure:

The solution of unknown concentration is called the "titrand."

The added solution is called the "titrant." In acid-base

titration, enough titrant is added to the titrand to neutralize

it. So if the titrand is a base, an acid is added as the titrant.

A color indicator is added to the titrand before starting to

indicate to the lab technician when the neutralization point is

approaching. This is important because if he adds the titrant too

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fast, the technician can go right by the neutralization point and

not know exactly how much titrant was needed to reach it.

Titration : Name of the procedure

Titrant : The standard or known solution (we know the

concentration of solution). In the experiment HCl acid

was used as titrant. Concentration was 0.1 N. Titrant

was taken in burette.

Titrand : The analyte or unknown solution (we do not know the

concentration of solution). In the experiment NaOH base

was used as titrand. Titrand was taken in conical flask

by pipette.

Table:

SL# Initial

burette

reading

(IBR)

Final

burette

reading

(FBR)

FBR-IBR Vacid

(Volume

of acid)

Mean

123

Calculation:

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Vacid X Sacid = Vbase X Sbase , where Vacid ---volume of

acid=from burette reading

Sacid---Strength of acid= known

Vbase ---volume of base=from pipette

Sbase---Strength of base=unknown

Sbase = (Vacid X Sacid) / Vbase

Result and Discussions:

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Experiment # 06.

Name of Experiment: To Determine the Heat of Solution.

Purpose:

To apply the concepts of specific heat and temperature change in

the experimental determination of the heat of solution of a

soluble salt.

Theory:

When salts are dissolved in water, there is often a temperature

change associated with the process. Some salt dissolve, releasing

heat in the process. Others dissolve while absorbing heat. As we

may remember, processes that proceed with a release of heat

energy are called “exothermic” processes, while those that absorb

heat are called “endothermic” processes. The energy is released

to the surroundings in an exothermic process; the heat of

solution would be given a negative value because the energy of

the system is decreasing.

NaOH(s) Na+(aq) + OH-(aq) ΔHsolution = - 44.51 kJ

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On the other hand, energy is absorbed from the surroundings in

endothermic processes, the heat of solution would be given a

positive value because the energy of the system is increasing.

KNO3(s) K+(aq) + NO3-(aq) ΔHsolution = + 34.89 kJ

In this experiment students will start with a known mass of any

salt, and a known volume of water. Students will determine the

magnitude of the temperature change associated with the

dissolving process, and use the masses of the solute and solvent,

the temperature change (called ΔT), and the known heat capacity

(specific heat) of water, 4.18 J/(g⋅°C), to calculate the heat of

solution.

Procedure:

1. Obtain a “calorimeter” and add to it a known amount of

distilled water.

2. Secure a thermometer to stand up in the calorimeter, using

ring stand (your instructor will show you how.)

3. Weigh out 4.000 grams of potassium nitrate.

4. Record the initial temperature of the water in the

calorimeter.

5. Add the potassium nitrate to the water in the calorimeter, all

at once, and begin stirring the solution to dissolve the salt

as rapidly as possible. Stir by swirling the cup with your18


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hand. DO NOT use a glass stir rod, as it will affect your

results. Watch out for the fragile thermometer! Record the

lowest temperature achieved during the dissolving of the salt.

6. Cleanup : Rinse the solution down the sink with LOTS of water.

Rinse the Styrofoam cup, and return it to the side counter. DO

NOT throw the Styrofoam cup away – we re-use them. Rinse the

thermometer and return it to your lab drawer.

Results:

Observations and Data:

1. Mass potassium nitrate used ________________g

2. Volume of water used ________________mL

3. Initial temperature (T1) ________________°C

4. Final (lowest) temperature (T2) ________________°C

Calculations:

1. Calculate the amount of energy absorbed as the potassium

nitrate dissolved. In order to do this, we must assume that

the solution has the same specific heat (cp) as pure water,

and that the density of the water was 1.00 gram/mL. Express

your final answer in kilojoules, kJ!

q = cp x m x ΔT 19


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Here,

q is the energy that is absorbed or released,

cp is the specific heat or heat capacity ofwater,

m is the mass of solution and

ΔT is the temperature change of the system

Heat of Solution Lab Background and Terms

Enthalpy (H) is the total energy of a system. The change in

enthalpy (heat energy) of a process is symbolized by H.

An exothermic process is one that release excess energy to the

environment.

An endothermic process is one that absorbs energy from the

environment in order to happen.

The lattice energy of an ionic solid is a measure of the strength

of bonds (cation to anion attraction) in that ionic compound.

Thus, it requires energy (endothermic process) to break apart

those bonds.

Solvation describes the process of polar water molecules being

attracted to (and often surrounding) cations and anions of an

ionic salt that is being dissolved (this releases free energy).

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Dissolution (or dissolving) can be viewed as occurring in three

steps. If sum of all these energies determines the overall energy

properties of the process:

1. Breaking solute-solute attractions (endothermic), which is

known as lattice energy in salts (ionic compounds).

2. Breaking solvent-solvent attractions (endothermic), for

instance that of hydrogen bonding in water.

3. Forming solvent-solute attractions (exothermic), in

solvation.

CASE I Dissolving process is Exothermic Process: If the heat

given off in the dissolving process (solvation energy) is greater

than the heat required to break apart the ionic solid (lattice energy)

and separate the water molecules (H bonding energy) the net dissolving

reaction is exothermic (energy given off). Thus the solution gets

hotter. The addition of added external heat (increases temperature)

would only inhibit the dissolving reaction since excess heat is

already being produced by the reaction. This situation is not very

common where an increase in temperature produces a decrease in

solubility.

CASE II Dissolving process is Endothermic Process: If the heat given

off in the dissolving reaction (solvation energy) is less than the heat

required to break apart the solid (lattice enrgy) and separate the water21

http://en.wikipedia.org/wiki/Solvation

http://en.wikipedia.org/wiki/Hydrogen_bonding

http://en.wikipedia.org/wiki/Lattice_energy


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molecules (H bonding energy), the net dissolving reaction is endothermic

(energy required). When the dissolving reaction is endothermic – it

requires heat. Therefore the heat is drawn from the surroundings, the

solution feels cold. The addition of more heat facilitates the dissolving

reaction by providing energy to break bonds in the solid. This is the most

common situation where an increase in temperature produces an increase in

solubility for solids. The use of first-aid instant cold packs is an

application of this solubility principle.

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Experiment # 07.

Name of Experiment: Identification of Iron II and Iron

III Ions.

Process Objectives:

To design an experiment to determine if an unknown solution

contains iron II or iron III ions.

To organize the data collected into a chart.

Learning Objectives:

To list compounds useful in identifying the iron II and the

iron III ions.

To learn how to confirm the presence of the iron III ion.

Introduction:

In the identification tests for the Fe2+ and Fe3+ ions shall use

the complex ferrocyanide, Fe(CN)64 ion. The deep-blue precipitate

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results when the complex ion combines with Fe3+and light blue

precipitate results when the complex ion combines with Fe2+.

This provides us with the means of identifying either iron. If

the light-blue precipitate is formed on addition of Potassium

Ferrocyanide,K4[Fe(CN)6] the complex indicates the presence of

the iron II ion. Similarly, a deep-blue precipitate formed with

the Potassium Ferrocyanide, K4[Fe(CN)6] complex indicates the

presence of the iron (III) ion.

Chemical Reactions:

In the case of Iron (III) chloride the complex reaction is as

follows….

4FeCl3 + 3K4[Fe(CN)]6 Fe4[Fe(CN)6]3 + 12KCl

In the case of Iron (II) sulfate the complex reaction is as

follows….

2FeSO4 + K4[Fe(CN)]6 Fe2[Fe(CN)6] + 2K2SO4

Apparatus:

1. 3 beakers

2. 2 test tubes24

Prussian Blue

Pale/light


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3. 3 Stirring rods

4. 3

Spatulas

Materials:

1. Iron (III) chloride, FeCl3

2. Iron (II) sulfate, FeSO4

3. Postassium ferrocyanide, K4[Fe(CN)6]

4. Distilled water

Procedures:

Describe by yourself…..

Observation/ Results:


Safety:

Take the necessary safety precaution before beginning this

experiment.

1. Wear Safety goggles and apron.

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2. As you conduct this experiment, you are required to handle

various chemicals.

3. Do not touch the chemicals with your hands.

4. Carefully check the labels on the reagent bottles before

removing any of their contents.

5. Observe all safety precautions while conducting experiments.

6. It is important to use good safety techniques while

conducting experiments.

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Experiment # 08.

Name of Experiment: Conductometric Analysis of

Different Aqueous Solutions.

Process Objectives/Purposes of the Experiment:

To design an experiment to determine the amount of ions

present in different aqueous solutions.

Identification of strong and weak acid as well as strong and

weak base.

To characterize the solutions whether they are strong

electrolyte / weak electrolyte / non electrolyte.

To find out the relationship of solution’s concentration on

To organize the data collected into a chart.

Learning Objectives:

To establish a relationship between the concentration and

the conductivity of any solution.

To find out a relationship between the conductivity and the

temperature of a solution with a specific strength.

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To estimate the amount of dissolved solids in a given

solution.

Introduction:

Conductivity is a measure of the concentration of ions in

solution. By completing the circuit of a conductivity meter, the

conductivity of the solution can be measured. The conductivity is

proportional to the current that flows between the electrodes.

For current to flow, ions must be present in solution to carry

the charge from one electrode to another. Increasing the number

of ions in solution will increase the amount of charge that can

be carried between electrodes and will increase the conductivity.

The units microSiemens/cm (uS/cm) and milliSiemens/cm

(mS/cm) are most commonly used to describe the conductivity of

aqueous solutions. The Siemen was formerly called mho (pronounced

"mo"), which was derived as a unit of conductivity by reversing

the letters in "ohm," the unit of resistance.

Theory:

When a strong electrolyte such as NaCl is place in aqueous

solution, it is said that it dissociates into its ions

(essentially100%). However, when a weak electrolyte is placed in

aqueous solution, it is said that it dissociates only partially.

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This experiment considers the electrical conduction of

aqueous solution. Although “pure water” is a very poor conductor,

ionic species present in aqueous solutions of acids, bases, and

other electrolytes significantly improve its conduction. The

conductivity of a solution is proportional to the concentration

of ions present in solution. More specifically, the conduction of

various ions depends on there charge and mobility in solution.

Aqueous solutions containing electrolytes obey Ohm’s law in

the same way that metallic conductors do:

V = I R where V = Potential Difference (in

volts), I = current (in amps), and R = resistance (in

ohms Ω).

Conductance is defined as the reciprocal of resistance, of a

homogeneous body of uniform cross Section is proportional to the

cross-sectional area A and inversely proportional to the length

L:

1/R= κA/L where κ = the specific conductance ( Ω-

1cm-1).

Apparatus:

5. Conductivity meter

6. Beakers 29


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7. Stirring rods

8. Spatulas

Materials:

5. Tap water

6. Distill water

7. NaCl (aq)

8. NaOH(aq) [0.1 M, 0.5 M, 1.0 M, 2.0 M]

9. HCl(aq)

10. CH3COOH (aq)

11. Ethanol

Strong

Electrolytes

HCl, HBr, HI, HClO4, HNO3, H2SO4, KBr, NaCl,

NaOH, KOH, Other soluble ionic compounds

Weak Electrolytes

Acetic acid (CH3COOH), HF

Nonelectrolytes H2O, CH3OH (Methanol), C2H5OH (Ethanol), C12H22O11

(Sucrose), Most organic compounds

Procedures:


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Observation/ Results:

[Some experimental values of conductivity on 8th November

2013]

Substances Conductivity @ 29 °C in mS/cm

Tap water 0.27

Distilled water 0.0

NaCl (aq) 17.4

NaOH (aq) [0.1 M, 0.5 M,

1.0 M, 2.0 M]

26.7, 96.5, 156.5

HCl (aq) 32.90

CH3COOH (aq) 0.40

C2H5OH 0.04

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Discussions:

1. Draw a graph by plotting the concentration and the

conductivity of a given solution.

2. Show the relationship between the conductivity of a given

solution with the temperature of the system by plotting a

graph.

3. Describe the purposes of this experiment.

Temperature Compensation of different chemical species

Substance % change per °C

Acids 1.0 to 1.6

Bases 1.8 to 2.2

Salts 2.2 to 3.0

Neutral water 2.0

Safety:

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7. Wear Safety goggles and apron.

8. As you conduct this experiment, you are required to handle

various chemicals.

9. Do not touch the chemicals with your hands.

10. Carefully check the labels on the reagent bottles

before removing any of their contents.

11. Observe all safety precautions while conducting

experiments.

12. It is important to use good safety techniques while

conducting experiments.

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CHM 118_Sec_A_B_Spring_2015

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