Download Unit V The Mole

The Mole Concept
A.
Atomic Masses and Avogadro’s Hypothesis
1.
We have learned that compounds are _________________________________________
________________________________________________________________________
Therefore, compounds must be composed of ___________________________________
________________________________________________________________________
During a chemical reaction, the atoms that make up the starting materials ____________
________________________________________________________________________
The question that arises however, is how many atoms and molecules are involved in the
reaction or how much of one element will combine with another element? In addition,
since atoms are so small, how can we count them?
Early experimental work by English chemist _________ ______________ (1766–1844)
was concerned with how much of one element could combine with a given amount of
another element. He proposed the following hypotheses:

_____________________________________________________________________

If compound “B” contains twice the mass of element “X” as does compound “A”,
then compound “B must contain twice as many atoms of “X”.

Simple compounds are made up of only ____________________________________
_____________________________________________________________________
_____________________________________________________________________
VERSION: October 18, 2000
2
2.
CHEMISTRY 11
Dalton did not attempt to figure out the mass of an individual atom of any element.
Instead, he assigned an ____________________ __________ to each element. He made
the assumption that _______________________________________________________
________________________________________________________________________
________________________________________________________________________
Carbon was found to be 6 times heavier than hydrogen so it was assigned a mass of 6.
Oxygen was found to be 16 times heavier than hydrogen so it was assigned a mass of 16.
e.g.
The reaction between 2.74 g of hydrogen gas and 97.26 g of chlorine gas makes
100 g of hydrogen chloride gas. If we assume that hydrogen chloride contains
one atom each of hydrogen and chlorine, the relative mass of chlorine is
97.26 g
2.74 g = __________ times heavier than ____________________
Since hydrogen is assigned a mass of “1”, chlorine has a mass of “__________”.
If 46.0 g of sodium react with 71.0 g of chlorine, the relative mass of sodium is
46.0 g
71.0 g = __________ times the mass of ____________________
Since chlorine is assigned the mass of “__________”, the mass of sodium is
__________ x __________ = __________
In this way, Dalton was able to calculate the “____________________ _____________”
for several elements.
UNIT V THE MOLE CONCEPT
3.
3
Dalton’s atomic mass scale was partly in error because ___________________________
________________________________________________________________________
During the time that Dalton’s mass scale was just being introduced, the French chemist
______________________________ began to study how gases reacted. When Gay–
Lussac reacted pairs of gases at the same temperature and pressure, he found that
________________________________________________________________________
1 L of hydrogen gas reacts with 1 L of chlorine gas to make 2 L of HCl(g)
1 L of nitrogen reacts with 3 L of hydrogen gas to make 2 L of NH3(g)
2 L of carbon monoxide gas react with 1 L of oxygen gas to make 2 L of CO2(g)
By itself, Gay-Lussac’s findings did not seem to be related to atomic mass but then the
Italian chemist____________________ ____________________ proposed the following
explanation for Gay-Lussac’s data.
AVOGADRO’S HYPOTHESIS
______________________________________________________________________
______________________________________________________________________
______________________________________________________________________
In other words, ___________________________________________________________
________________________________________________________________________
Therefore, the molecule formed by reacting A with B is__________
Similarly, if 2 L of gas A reacts with 1 L of gas B, the molecules formed have the
formula __________
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CHEMISTRY 11
B.
The Mole
1.
Avogadro’s hypothesis _____________________________________________________
________________________________________________________________________
________________________________________________________________________
e.g.
If 1 L of nitrogen reacts with 3 L of hydrogen to form ammonia, then its formula
is __________
If 2 L of hydrogen reacts with 1 L of oxygen to form water, then its formula is
__________
So if we want to make a particular compound, all we need to do is react volumes of gases
in the ratio given by their formulas BUT how do we determine how much of one element
reacts with another element when they are not gases?
e.g.
How much iron is required to react with sulphur to produce iron (II) sulphide,
FeS, so that neither element is left over?
Since the easiest way to measure solids is to measure their mass, we need to relate mass
to the number of atoms.
2.
The periodic table shows us the ______________________________________________
Its units are “_____” which stands for “___________________________________”.
Unlike Dalton’s mass scale, the present day scale is not based on hydrogen.
Instead, _________________________________________________________________
________________________________________________________________________
UNIT V THE MOLE CONCEPT
5
A _______________ is the number of carbon atoms in exactly
_________________________
Molar Mass is the _____________________________________________
6
Atomic Number
C
Symbol
Atomic Mass
12.0
one “C” atom has a mass of __________
one MOLE of “C” atoms has a mass of __________
The periodic table gives us _________________________________________________
________________________________________________________________________
8
O
16.0
20
26
Ca Fe
40.1
55.8
16
17
S
Cl
32.1
35.5
Finding the molar mass of a compound simply involves __________________________
________________________________________________________________________
________________________________________________________________________
(remember the units of molar mass are grams).
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CHEMISTRY 11
EXAMPLE V.1
CALCULATING MOLAR MASSES
Problem:
What is the molar mass of iron (III) sulphate?
Solution:
Iron (III) sulphate is Fe2(SO4)3
To calculate the molar mass we need to add the masses of
2 Fe + 3 S + 12 O
2 (__________) + 3 (__________) + 12 (__________) = __________
Alternatively,
2 Fe + 3 (SO4)
2 (__________) + 3 (__________ + 3 (__________))
2 (__________) + 3 (__________) = __________
SAMPLE PROBLEM V.1
Problem:
CALCULATING MOLAR MASSES
Calculate the molar masses of
(a) Na2Cr2O7
Solution:
(b) Ag2SO4
(c) Pb3(PO4)4
(a)
__________________________________________________
(b)
__________________________________________________
(c)
__________________________________________________
UNIT V THE MOLE CONCEPT
7
C.
Relating Moles to Mass, Volume of Gas, and Number of Particles
1.
The molar mass of a compound ______________________________________________
________________________________________________________________________
________________________________________________________________________
Since we know that
__________ of “X” has a mass of ______________________________
we have two conversion factors:
1 mol
(molar mass of "X") g
(molar mass of "X") g or
1 mol
EXAMPLE V.2
Problem:
Solution:
RELATING MASS AND MOLES
(a)
What is the mass of 3.25 mol of CO2?
(b)
What is the mass of 1.36 x 10-3 mol of SO3?
(c)
How many moles of N2 are there in 50.0 g of N2?
(d)
How many moles of CH3OH are there in 0.250 g of CH3OH?
(a)
1 mol CO2 = __________
mass CO2 = __________ x __________ = __________
(b)
1 mol SO3 = __________
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CHEMISTRY 11
mass SO3 = ____________________ x __________ = __________
(c)
1 mol N2 = __________
mol N2 = __________ x __________ = __________
(d)
1 mol CH3OH = __________
mol N2 = __________ x __________ = _______________
SAMPLE PROBLEMS V.2
Problem:
Solution:
RELATING MASS AND MOLES
(a)
What is the mass of 0.834 mol of FeSO4?
(b)
What is the mass of 2.84 x 10-2 mol of Na3N?
(c)
How many moles of CH4 are there in 27.5 g of CH4?
(d)
How many moles of Ca(NO3)2 are there in 35.0 g of Ca(NO3)2?
(a)
1 mol FeSO4 = __________
mass FeSO4 = __________ x __________ = __________
(b)
1 mol Na3N = __________
mass Na3N = ____________________ x __________ = __________
UNIT V THE MOLE CONCEPT
9
(c)
1 mol CH4 = __________
mol CH4 = __________ x __________ = __________
(d)
1 mol Ca(NO3)2 = __________
mol Ca(NO3)2 = __________ x __________ = __________
2.
Calculations involving gas volumes are simplified by Avogadro’s hypothesis. Recall that
AVOGADRO’S HYPOTHESIS
______________________________________________________________________
______________________________________________________________________
______________________________________________________________________
The MOLAR VOLUME of a gas is the _____________________________________
______________________________________________________________________
Since the volume of a gas is _________________________________________________
________________________________________________________________________
________________________________________________________________________
STANDARD TEMPERATURE AND PRESSURE
(_________) = _________ and _________
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CHEMISTRY 11
Avogadro’s hypothesis can be interpreted to mean that all gas samples with the same
temperature, pressure, and numbers of particles occupy the same volume. This can be
re–stated as ______________________________________________________________
________________________________________________________________________
Experimentally, it is determined that
__________ of any gas _______________ has a volume of __________
In other words, the ________________________________________________________
We can obtain two conversion factors:
1 mol
22.4 L
or
22.4 L
1 mol
EXAMPLE V.3
Problem:
Solution:
3.
RELATING VOLUME OF A GAS AND MOLES
(a)
How many moles of gas are contained in a balloon with a
volume of 10.0 L at STP?
(b)
What volume will 0.250 mol of CO2 occupy at STP?
(a)
mol of gas = __________ x __________ = __________
(b)
volume of CO2 = __________ x __________ = __________
The mole is the fundamental unit in chemistry for measuring the ___________________
________________________________________________________________________
UNIT V THE MOLE CONCEPT
11
In a sense, the mole is simply a ____________________ ____________________. Just
as a
dozen = 12
experimentally,
__________ = ____________________
This value, 6.02 x 1023, is called ____________________ number. Notice that there are
no units in the same way a “dozen” stands for “12”.
Conversion factors:
1 mol particles
6.02 x 1023 particles
or
1 mol particles
6.02 x 1023 particles
EXAMPLE V.4
Problem:
RELATING NUMBER OF PARTICLES AND MOLES
(a)
How many molecules are there in 0.125 mol of molecules?
(b)
How many moles of N are there in 5.00 x 1017 N atoms?
(c)
How many atoms are in 5 molecules of CuSO4•5H2O?
(a)
molecules = __________ x
Solution:
23
6.02 x 10 molecules
1 mol molecules
= ____________________ molecules
(b)
moles N2 = ____________________ x
1 mol atoms
23
6.02 x 10 atoms
= ____________________
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CHEMISTRY 11
(c)
atoms
atoms = ____________________ x 21 molecule
= __________
SAMPLE PROBLEMS V.3
Problem:
Solution:
RELATING VOLUME OF GAS / NUMBER OF PARTICLES AND
MOLES
(a)
How many moles of gas are contained in a balloon with a
volume of 17.5 L at STP?
(b)
What volume of gas will 0.074 mol of gas occupy at STP?
(c)
How many atoms are there in 0.0185 mol of atoms?
(d)
How many moles of Fe2O3 are there in 8.75 x 1020 Fe2O3
molecules?
(e)
How many atoms of H are in 30 molecules of Ca(H2PO4)2?
(a)
mol of gas = __________ x __________ = __________
(b)
volume of gas = __________ x __________ = __________
(c)
atoms = _______________ x ________________________
= ____________________
(d)
moles Fe2O3 = _______________ x ____________________
= ____________________
(e)
H atoms = ___________________ x ___________________
= _______________
UNIT V THE MOLE CONCEPT
4.
13
All of the previous problems have involved single–step conversions between moles,
mass, volume, or number of particles. The following is a summary of the conversion
factors needed:
CONVERSION
CONVERSION FACTOR
MOLES  NUMBER OF PARTICLES
23
6.02 x 10 particles
1 mol
(molar mass) g
1 mol
MOLES  MASS
22.4 L
1 mol
MOLES  VOLUME (gases @ STP)
MOLECULES  ATOMS
(atom count) atoms
1 mol
or
1 mol
23
6.02 x 10 particles
1 mol
or (molar mass)
1 mol
or 22.4 L
1 mol
or (atom count) atoms
The following flow chart will help to simply calculations that involve multiple
conversions.
PARTICLES
(atoms/molecules)
6.02 x 10
23
MOLES
molar mass
MASS
(in grams)
22.4 L/mol
VOLUME
(at STP)
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CHEMISTRY 11
EXAMPLE V.5
Problem:
MOLE CALCULATIONS INVOLVING MULTIPLE CONVERSIONS
(a) What is the volume occupied by 50.0 g of NH3(g) at STP?
(b) What is the mass of 1.00 x 10
(c)
12
atoms of Cl?
How many oxygen atoms are contained in 75.0 L of SO3(g) at
STP?
Solution:
(a) MASS  MOLES  VOLUME
? L = __________ x __________ x
__________ = __________
(b) ATOMS  MOLES  MASS
? g = _____________ atoms g x
=
(c)
1 mol
23
6.02 x 10
atoms
x ________
_______________
VOLUME  MOLES  MOLECULES ATOMS of O
23
? O’s = _________ x _________x
6.02 x 10
1 mol
= ____________________
SO3
x _________
UNIT V THE MOLE CONCEPT
4.
15
So far, all the volumes have been of gases at STP. If ____________________ is
mentioned at any point in a problem, recall that d = m/V
•
__________________________________________________, calculate the volume
(
from V = m/d
) . (Note you cannot use the molar volume of a gas, 22.4 L, when
calculating the volume of a solid or liquid.)
•
______________________________, you will need both mass and volume to
(
calculate density from d = m/V
•
).
________________________________________, use the density and volume to
calculate mass from (m = d • V) and then convert mass to moles.
•
________________________________________, it can be calculated by the
(molar mass) 

equation: d = (molar volume)


EXAMPLE V.6
Problem:
DENSITY AND MOLE CALCULATIONS
(a)
What is the volume occupied by 3.00 mol of ethanol,
C2H5OH? (d = 0.790 g/mL)
(b)
How many moles of Hg(l) are contained in 100 mL of Hg(l)?
(d = 13.6 g/mL)
Solution:
(c)
What is the density of O2(g) at STP?
(a)
We want volume of liquid so V =
m
/d , and mass can be
related to moles, so
? mL = __________ x __________x __________= __________
(b)
To get moles, we need to calculate mass first and then
convert to moles.
? mol = __________ x __________x __________= __________
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CHEMISTRY 11
(c)
There are no numbers given but the term STP suggests a
volume of 22.4 L for one mole of gas and the mass of one
mole of O2(g) is 32.0 g.
m
density = V
mass of 1 mol (molar mass)
= volume of 1 mol (molar volume)
= __________= __________
EXAMPLE V.7
Problem:
MORE DENSITY CALCULATIONS
(a)
A 2.50 L bulb contains 4.91 g of a gas at STP. What is the
molar mass of the gas?
(b)
Al2O3(s) has a density of 3.97 g/mL. How many atoms of Al
are in 100 mL of Al2O3?
Solution:
(a)
Density can be found from d =
m
/V , so
d = __________= __________
(b)
Change volume to mass, then to moles, then to molecules,
and finally atoms.
6.02 x 10
? Al =_______ x _______x _______x
23
Al2O3
1 mol
= ____________________ atoms Al
x
2 atoms Al
Al2O3
UNIT V THE MOLE CONCEPT
17
C.
Percentage Composition
1.
The PERCENTAGE COMPOSITION is _____________________________________
________________________________________________________________________
EXAMPLE V.8
Problem:
Solution:
PERCENTAGE COMPOSITION
(a)
What is the percentage composition of H2SO4?
(b)
What is the percentage composition of water in CuSO 4 •
5H2O?
(a)
Assume that there is 1 mole of the compound.
Molar mass = __________
Total mass of H = _______________ = __________
Total mass of S = _______________ = __________
Total mass of O = _______________ = __________
% H = __________ x 100% = __________
% S = __________ x 100% = __________
% O = __________ x 100% = __________
(b)
Assume that there is 1 mole of the compound.
Molar mass = __________
Total mass of H2O = _______________ = __________
% H2O = __________ x 100% = __________
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CHEMISTRY 11
D.
Empirical and Molecular Formulas
1.
The empirical formula is called the ___________________________________________
________________________________________________________________________
________________________________________________________________________
CH2, C2H4, C3H6, C4H8, and C5H10 all contain twice as many H’s as there are C’s. The
empirical formula (simplest formula) for all of these molecules is __________
Finding empirical formula is essentially the opposite of determining percentage
composition.
EXAMPLE V.9
Problem:
Solution:
EMPIRICAL FORMULAS
(a)
What is the empirical formula of a compound consisting of
80.0% C and 20.0% H?
(b)
A compound contains 58.5 % C, 7.3% H, and 34.1 % N. What
is the empirical formula of the compound?
(a)
Assume __________ of the compound
mass of C = ____________________= __________
mass of H = ____________________ = __________
Use the mass to determine the moles of each element
mole C = __________ x __________ = __________
mole H = __________ x __________ = __________
UNIT V THE MOLE CONCEPT
19
Determine smallest ratio by
dividing by smallest number of moles (÷ 6.67)
C = _______________ = __________
H = _______________ = __________
Empirical Formula is __________
(b)
Assume 100.0 g of the compound
mass of C = __________, mass of H = __________,
mass of N = __________
Use the mass to determine the moles of each element
1 mol
mole C = 58.5 g x 12.0 g = __________
1 mol
mole H = 7.3 g x 1.0 g = __________
1 mol
mole N = 34.1 g x 14.0 g = __________
Determine smallest ratio by
dividing by smallest number of moles (÷ 2.44)
4.88 mol
C = 2.44 mol = __________
7.3 mol
H = 2.44 mol = 2.99  __________
2.44 mol
N = 2.44 mol = __________
Empirical Formula is __________
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CHEMISTRY 11
EXAMPLE V.10
Problem:
MORE EMPIRICAL FORMULAS
What is the empirical formula of a compound consisting of 81.1% C
and 18.2% H?
Solution:
Assume 100.0 g of the compound
mass of C = __________, mass of H = __________
Use the mass to determine the moles of each element
1 mol
mole C = 81.1 g x 12.0 g = __________
1 mol
mole H = 18.2 g x 1.0 g = __________
Determine smallest ratio by
dividing by smallest number of moles (÷ 6.82)
6.82 mol
C = 6.82 mol = __________
18.2 mol
H = 6.82 mol = __________
DO NOT just round off ratio, you must multiply both number by 2, 3, 4,
or 5 until both are whole numbers
Multiplying both numbers by _____ gives whole numbers
_____ : _____
Empirical Formula is __________
ALWAYS ______________________________________________________________
________________________________________________________________________
Improper round-off calculations will cause you to multiply by the wrong number when
trying to obtain whole numbers.
UNIT V THE MOLE CONCEPT
2.
21
The ____________________ ____________________ can be found be using the molar
mass of the empirical formula; that is, the EMPIRICAL MASS.
The molecular formula is ___________________________________________________
________________________________________________________________________
CH2, C2H4, C3H6, C4H8, and C5H10 all have the same empirical formula CH2.
The whole number multiple (N) is given by the formula
molar mass
Multiple = N = empirical mass
_________________________ = _________________________
3.
It may be necessary to calculate the molar mass from information that is given in the
question.
(a)
Finding molar mass from density of a gas at STP
If:
density of gas “X” = 1.43 g/L (at STP)
Then: molar mass of “X” = 1.43 g/L x 22.4 L/mol = __________
(b)
Finding molar mass from mass and volume of a gas at STP
If you are told: “0.0425 L of gas ‘X’ at STP has a mass of 0.135 g”
0.135 g
Then: density of gas “X” = 0.0425 L = 3.176 g/L
And:
molar mass of “X” = 3.176 g/L x 22.4 L/mol = __________
22
CHEMISTRY 11
(c)
Finding molar mass from mass and a given number of moles
If you are told: “0.0250 mol of ‘X’ has a mass of 1.775 g”
1.775 g
Then: molar mass = 0.0250 mol = __________
(d)
Finding molar mass from of the molar mass if given as a multiple of a known
molar mass
If you are told: “X” has a molar mass which is 1.64 times that of CO2
Then: molar mass of CO2 = __________
And:
molar mass of “X” = 1.64 x 44.0 g/mol = __________
EXAMPLE V.11
MOLECULAR FORMULA
Problem:
The empirical formula of a compound is SiH3. If 0.0275 mol of the
compound has a mass of 1.71 g, what is the compound’s molecular
formula?
Solution:
Empirical mass
Molar mass of SiH3 = __________ + __________ = __________
Molar mass of compound
Molar mass = _______________ = __________
Molecular formula
N = __________ = _____
_____ x __________ = __________
UNIT V THE MOLE CONCEPT
23
E.
Molar Concentrations
1.
SOLUTIONS are ________________________________________________________
________________________________________________________________________
________________________________________________________________________
Most solutions contain a solid (_______________) dissolved in a liquid
(_______________); however, there are solutions of gases as well.
The CONCENTRATION of a substance in solution _____________________________
________________________________________________________________________
________________________________________________________________________
Chemists use the “mole” to describe the amount of substance in a solution.
_______________ ____________________ or ____________________ of a substance is
the _____________________________________________________________________
e.g.
If 2.0 L of solution contain 5.0 mol of NaCl, what is the molarity of the NaCl?
molar concentration = _______________ = __________
mol
The unit symbol for L is “M”
When expressed in words, the unit symbol “M” is written as “_______________”.
The short-hand symbol for “______________________________” is a set of brackets:
[…], so [NaCl] means the “________________________________________”
24
2.
CHEMISTRY 11
The definition of molar concentration leads to the following equation:
moles
n
molar concentration = volume or M = V
mol
M
mol
V
M
mol
V
M
mol
V
M
V
c = molar concentration, in mol/L, n = number of moles, V = volume, in litres
EXAMPLE V.12
Problem:
Solution:
CALCULATING CONCENTRATION
What is the [NaCl] in a solution containing 5.12 g of NaCl in 250.0 mL
of solution?
In order to find molarity (c), the moles (n), and volume are needed.
The volume is given and the mass must be converted to moles.
moles of NaCl = __________ x __________ = __________
n
[NaCl] = M = V = _______________ = __________
EXAMPLE V.13
Problem:
Solution:
CALCULATING MASS CONTAINED IN SOLUTIONS
What mass of NaOH is contained in 3.50 L of 0.200 M NaOH?
The molarity (c) and volume (V) are given so moles can be found.
Moles can then be converted to mass.
n
solving M = V for n gives M = c • V
moles of NaOH = _______________ x __________ = __________
mass of NaOH = _______________ x __________ = __________
UNIT V THE MOLE CONCEPT
EXAMPLE V.14
25
CALCULATING CONCENTRATION FROM DENSITY
Problem:
What is the molarity of pure sulphuric acid, H2SO4, having a density of
1.839 g/mL?
Solution:
Since both density and molarity both have units of amount/volume,
density =
amount (as mass)
volume
and
molarity =
amount (as moles)
volume
therefore all we need to do is convert from mass to moles using molar
mass.
[H2SO4] = _______________ x __________ = __________
F.
Dilution Calculations
1.
When two solutions are mixed, ______________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
(total moles of chemical)
molarity of mixture = (total volume of mixture)
Consider the following dilution,
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CHEMISTRY 11
Mole (in itia l) = Mi x V i
H 2O
Mole (fi nal) = Mf x V f
Diluted Solution
Stock Solution
since the ________________________________________________________________
________________________________________________________________________
________________________________________________________________________
M1V1 = M2V2
this equation can be rearranged to give the following dilution equation
M2 =
EXAMPLE V.15
Problem:
M1 x V1
V2
SIMPLE DILUTION CALCULATIONS
If 200.0 mL of 0.500 M NaCl is added to 300.0 mL of water, what is the
resulting [NaCl] in the mixture.
Solution:
M2 =
M1 x V1
V2
[NaCl] = __________ x _______________________ = __________
EXAMPLE V.16
MAKING DILUTE SOLUTIONS FROM CONCENTRATED
SOLUTIONS
Problem:
What volume of 6.00 M HCl is needed to make 2.00 L of 0.125 M HCl?
Solution:
Since
UNIT V THE MOLE CONCEPT
27
M1V1 = M2V2
then,
VHCl = _________________________ = __________
2.
When two solutions having different concentrations of the same chemical are mixed,
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
EXAMPLE V.17
Problem:
Solution:
MIXING SOLUTIONS OF DIFFERENT CONCENTRATION
If 300.0 mL of 0.250 M NaCl is added to 500.0 mL of 0.100 M NaCl,
what is the resulting [NaCl] in the mixture?
For each solution, the diluted concentration is given by
M2 =
M1 x V1
V2
300.0 mL
[NaCl]first solution = __________ x (300.0 + 500.0) mL = __________
500.0 mL
[NaCl]second solution = __________ x (300.0 + 500.0) mL = __________
[NaCl]total = [NaCl]first solution + [NaCl]second solution
[NaCl]total = _______________ + _______________ = __________