Experiment-10: Salt water Analysis
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Solution chemistry
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Solubility and factors affecting solubility
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Evaporation
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Molarity and % Calculations
In our chemistry laboratory, a solution is defined as a homogenous mixture of two or
more pure substances, with each substance retaining its chemical identity. The process of
mixing is a physical change. No new chemicals are made and no change in chemical
composition of the components. A simple solution has two components - a solute and a
solvent. Solute is the minor portion and solvent is the major portion of a solution.
Any solution in which water is the solvent is termed as an aqueous solution and other
solutions are known as nonaqueous solution. Other common solvents are alcohol,
acetone, hexane etc. Some solutions are so common to us that we give them a unique
name. A solution of water and sugar is called syrup. A solution of sodium chloride
(common table salt) in water is called brine. A sterilized specific concentration (0.15 M)
of sodium chloride in water is called saline. A solution of carbon dioxide in water is
called seltzer, and a solution of ammonia gas in water is called ammonia. Some common
properties of true liquid solutions are:
1. The particles of solute are the size of individual small atoms, molecules or ions.
One nanometer (1x 10-9 m) is about the maximum diameter for a solute particle.
2. The mixture components do not separate on standing. Gravity has no effect.
3. The mixture can’t be separated by common filtrations because entire solution will
pass through the filters.
4. Once it is completely mixed, the mixture is homogeneous. Properties of the
solution will be the uniform throughout.
5. The mixture appears clear rather than cloudy. It may have some color to it, but it
seems to be transparent otherwise. A beam of light will pass through.
6. The solution of an ionic material into water will result in an electrolyte solution,
which can carry electrical current.
7. A solution has a higher boiling point as compared to the pure solvent. The boiling
point elevation is directly proportional to the amount of solute present.
8. A solution has a lower freezing point as compared to the pure solvent. The
freezing point depression is directly proportional to the amount of solute present.
A simple definition of Solubility: maximum number of grams of solute that will dissolve
o
in 100 g of solvent at a given temperature (usually 25 C). A saturated solution is a
solution that contains the maximum amount of solute that can be dissolved under a set of
experimental conditions at which the solution exists. It means that you cannot dissolve
anymore solute in a saturated solution. An unsaturated solution contains less solute than
the maximum amount that could dissolve. In this case, it is possible for you to dissolve
more solute in solution. In some cases, using special experimental procedures, it is
possible to make supersaturated solutions, which are unstable. A supersaturated
solution is a solution that contains more dissolved solute than present in a saturated
solution. Usually, solubility of solids and liquids increases with increasing temperature
of solvent. But, solubility of a gas as a solute into a liquid solvent decreases with
increasing temperature of solvent. For example, one can dissolve a greater amount of
salt or sugar if the water is hot. However, less amount of carbon dioxide gas will dissolve
into water it is hot. Therefore, to maximize the tingling taste of seltzer it is kept cold
because at a higher temperature carbon dioxide gas will be less soluble and will escape
seltzer thereby making it flat in taste. Pressure has little or no effect on the solubility of
solids and liquids in water. However, higher pressure increases the solubility of a gas in
water. All carbonated drinks contain carbon dioxide gas at a higher pressure so that it can
be more soluble. A dilute solution contains small amount of solute in solution relative to
that could dissolve. A concentrated solution contains a large amount of solute in
solution relative to the amount that could dissolve. Dilution is achieved by carefully
adding the concentrated solution into a calculated amount of solvent. The like dissolves
like rule describes the general principle for dissolving a solute in a solvent. Commonly,
ionic and polar solutes tend to dissolve in polar solvents such as water. Nonpolar
compounds (mostly organic chemicals) tend to dissolve in nonpolar solvents such as
hexane. Liquids that dissolve in one another are known as miscible.
Solution Concentrations
A concentration is number, which reflects the amount of solute present in a specified
amount of. solvent or solution. Four most common expressions for concentrations are:
 Moalrity (M)
 Percent by mass or mass-mass percent (% m/m)
 Percent by volume or volume-volume percent (% v/v)
 Mass-volume percent (% m/v)
Example-1: Calculate the molar concentration of a sodium chloride solution, which
contains 4.256 g NaCl per 100.0 mL solution. The molar mass of NaCl is 58.44 g mol.
Molarity (M) is the number of moles of solute in a liter of solution. Mathematically,
M = (# mol solute/ volume solution in L)
Convert 3.05 g NaCl into mol NaCl (# mol solute).
# mol NaCl = ( grams of NaCl / Molar mass) = (4.256 g / 58.44 g/mol) = 0.07283 mol
Convert 100.0 mL solution into Liters. (1000 Ml = 1 L) ; 100.0 mL = 0.1000 L
M = (0.07283 mol/ 0.1000 L) = 0.7283 mol/L = 0.7283 M
Example-2: What is the percent by mass of a solution made by mixing 6.0 g of sugar
with 85.0 g water?
Percent by mass
= (mass of solute/ mass of solution) x 100 = % m/m
In this example, mass of solution = 6.0 g + 85.0 g = 91.0 g
= (6.0 g/91.0 g) x 100 = 6.6%
Example-3: 15.0 mL of isopropyl alcohol is mixed with 60.0 mL water. Calculate
Percent by volume or volume-volume percent (% v/v) of isopropyl alcohol.
Percent by volume
= (volume of solute/ volume of solution) x 100 =
% v/v
= (15.0 mL / 75.0 mL) x 100 = 20 %
Example-4: Calculate grams of acetic acid present in one teaspoon (5.0 mL) of a
commercial vinegar sample, which is a 4.0 % (m/v) aqueous solution of acetic acid.
Mass-volume percent = (mass of solute g/ volume of solution mL) x 100=(% m/v)
4.0 %
Xg
= (X g / 5.0 mL) x 100
= mass of solute g = grams of acetic acid present in one teaspoon (5.0 mL)
= (4.0 % x 5.0 mL) / 100 g
= 0.20 g acetic acid
Procedure:
1. Record your unknown salt solution number__________.
2. Find the mass of a dry and clean 250-mL beaker. Record all the digits throughout
the experiment. Use the same scale throughout this experiment.
3. Transfer exact 50.0 mL of unknown salt solution into the 250-mL beaker. Find
the mass of the 250-mL beaker with 50.0 mL of unknown salt solution.
4. Place this beaker on top of an electrical hot plate. Cover this beaker with a round
watchglass. It is necessary to cover the beaker because the dry salt tends to spatter
out especially at the end. Turn on the electrical hot plate. Use lower settings to
begin with then gradually increase the heat level. In few minutes, the unknown
salt solution inside the beaker will start boiling. Do not lean close to this assembly
to have a better look! Continue the process of heating and boiling until all water
has been evaporated and a dry white solid (NaCl) is left in the beaker. It is
necessary to make sure that the NaCl you recover at the end is completely dry.
Shut-off and unplug the electrical hot plate when you see completely dry NaCl.
Carefully remove the watchglass using crucible tongs/towel. Carefully remove the
hot beaker using tongs and allow it to cool.
5. Now, find the mass of the 250-mL beaker with dry residue (salt).
6. Dispose-off salt into sink and flush with water.
7. Calculate mass/mass % and molar concentration of your unknown salt solution.
Laboratory Report#10: Salt water Analysis
Last Name_____________________________, first name________________
Date of Experiment______________ Instructor’s Initials__________
Data and Calculations
1. Your Unknown sample #_____________
2. Mass of empty and dry 250-mL beaker
=_____________________g
3. Mass of (250-mL beaker +50.0 mL of unknown salt solution) =_________________g
4. Mass unknown salt solution used = #3-#2
=_______________g
5. Volume of unknown salt solution used
= _______________ L
= 50.0 mL
After evaporation, drying, and cooling
6. Mass of 250-mL beaker containing dry salt
=_____________________g
7. Mass salt recovered = #6-#2
=_____________________g
8. Mol of salt recovered = convert #7 into mol = (#7 g/ 58.44 g/mol)= ___________ mol
9. Calculate the Molarity of unknown solution
M
= (# mol solute/ volume solution in L)
= (#8 / #5)
=__________M
10. Calculate the percent by mass of unknown solution
% m/m = (mass of solute/ mass of solution) x 100
= (#7 g/ #4 g) x 100
= ________ %
Show all your work! Use proper units and significant figures !!
1. A solution weighs 44 g. This sample contains 12 g of salt. Calculate % m/m salt.
2. A 250.0 mL solution contains 7.305 g NaCl. Calculate its Molarity.
3. Define and give one example:
Saturated solution
Miscible liquids
Polar solvent
Molarity