Chm151 exam 4 slides sp10

CHM 151 General Chemistry

EXAM 4

SLIDES

Electron Energy Levels

Electrons are arranged inspecific energy levels that are labeled n = 1, n = 2,

n = 3, and so on increase in energy as n

increases have the electrons with the

lowest energy in the first energy level (n = 1) closestto the nucleus

Energy Level Changes

An electron absorbsenergy to “jump” to a higher energy level.

When an electron falls to a lower energy level, energy is emitted.

In the visible range, the emitted energy appears as a color.

In each of the following energy level changes, indicateif energy is 1) absorbed, 2) emitted, or 3) not changed.

A. An electron moves from the first energy level (n = 1) to the third energy level (n = 3).

B. An electron falls from the third energy level to the second energy level.

C. An electron moves within the third energy level.

Learning Check

Sublevels

Sublevels contain electrons with the same energy are found within each energy level. are designated by the letters s, p, d, and f

The number of sublevels is equal to the value of the principal quantum number (n).

Number of Sublevels

Energy of Sublevels

In any energy level

the s sublevel has the lowest energy

the s sublevel is followed by the p, d, and f sublevels (in order of increasing energy)

Orbitals

An orbital is a three-dimensional space

around a nucleus where an electron is most likely to be found

has a shape that represents electron density (not a path the electron follows)

can hold up to 2 electrons contains two electrons that must

spin in opposite

s Orbitals

An s orbital has a spherical shape

around the nucleus increases in size around the

nucleus as the energy level n value increases

is a single orbital found in each s sublevel

p Orbitals

A p orbital has a two-lobed shape is one of three p orbitals that make up each p

sublevel increases in size as the value of n increases

Sublevels and Orbitals

Each sublevel consists of a specific number of

orbitals. An s sublevel contains one s orbital. A p sublevel contains three p orbitals. A d sublevel contains five d orbitals. An f sublevel contains seven f orbitals.

Electrons in Each Sublevel

• An orbital diagram represents each orbital with a box, with orbitals in the same subshell in connected boxes; electrons are shown as arrows in the boxes, pointing up or down to indicate their spins.

• Two electrons in the same orbital must have opposite spins.

Orbital Diagrams

↑↓

• An electron configuration lists the occupied subshells using the usual notation (1s, 2p, etc.). Each subshell is followed by a superscripted number giving the number of electrons present in that subshell.

• Two electrons in the 2s subshell would be 2s2 (spoken as “two-ess-two”).

• Four electrons in the 3p subshell would be 3p4 (“three-pea-four”).

Electron Configuration

• Hydrogen contains one electron in the 1s subshell.

1s1

• Helium has two electrons in the 1s subshell.

1s2

Electron Configurations of Elements

↑

↑↓

Electron Configurations of Elements

• Lithium has three electrons.1s2 2s1

• Beryllium has four electrons.1s2 2s2

• Boron has five electrons.1s2 2s2 2p1

↑↓ ↑

↑↓ ↑↓

↑↓ ↑↓ ↑

Orbital Diagram of Carbon

• Carbon, with six electrons, has the electron configuration of 1s2 2s2 2p2.

• The lowest energy arrangement of electrons in degenerate (same-energy) orbitals is given by Hund’s rule: one electron occupies each degenerate orbital with the same spin before a second electron is placed in an orbital.

↑↓ ↑↓ ↑ ↑

Other Elements in the Second Period

• N 1s2 2s2 2p3

• O 1s2 2s2 2p4

• F 1s2 2s2 2p5

• Ne 1s2 2s2 2p6

↑↓ ↑↓ ↑ ↑ ↑

↑↓ ↑↓ ↑↓ ↑ ↑

↑↓ ↑↓ ↑↓ ↑↓

↑

↑↓ ↑↓ ↑↓ ↑↓ ↑↓

• Because their electron configurations can get long, larger atoms can use an abbreviated electron configuration, using a noble gas to represent core electrons.

Fe: 1s2 2s2 2p6 3s2 3p6 4s2 3d6 → [Ar] 4s2 3d6

Ar

Electron Configurations of Heavier Atoms

Chapter 7 Visual Summary

Daniel L. RegerScott R. GoodeDavid W. Ball

www.cengage.com/chemistry/reger

Chapter 8The Periodic Table:

Structure and Trends

http://www.cengage.com/chemistry/reger

• The 4s orbital is lower in energy than the 3d orbital and fills first, starting the fourth period at potassium.

• The 3d orbitals fill after the 4s.

• Similar inversions occur in the remaining periods.

Electron Configurations

• For anions, the additional electrons fill orbitals following the same rules that applies to atoms.

Cl: [Ne] 3s2 3p5 Cl-: [Ne] 3s2 3p6

As: [Ar] 4s2 3d10 4p3 As3-: [Ar] 4s2 3d10 4p6

• Many stable anions have the same electron configuration as a noble gas atom.

Electron Configurations of Anions

Test Your Skill• Write the electron configurations of the

following ions: (a) N3- (b) Co3+ (c) K+

Test Your Skill• Write the electron configurations of the

following ions: (a) N3- (b) Co3+ (c) K+

• Answers:

(a) 1s2 2s2 2p6

(b) [Ar] 4s23d4

(c) [Ar]

Size Trends for an Isoelectronic Series

Sizes of the Atoms and Their Cations

• Atoms are always larger than their cations.

Sizes of the Atoms and Their Cations• If an atom makes more than one cation,

the higher-charged ion has a smaller size.

• Anions are always larger than their atoms.

Atomic and Ionic Radii

• Identify the larger species of each pair: (a) Mg or Mg2+ (b) Se or Se2-

Test Your Skill

• Identify the larger species of each pair: (a) Mg or Mg2+ (b) Se or Se2-

• Answer: (a) Mg is larger.(b) Se2- is larger.

Test Your Skill



Chapter 9Chemical Bonds


Chemical Bonds

• Chemical bonds are the forces that hold the atoms together in substances.

• This chapter discusses two limiting types of bonding.• Ionic bonding• Covalent bonding

Lewis Electron-dot Symbols

• A Lewis electron-dot symbol consists of the symbol for the element surrounded by dots, one for each valence electron.

Lewis Symbols for Cations

• Cations of most representative elements have no valence shell electrons shown in the Lewis symbol.

Na× ® Na+ + e-

×Ca× ® Ca2+ + 2e-

Lewis Symbols for Anions• The Lewis symbols

of most monatomic anions show eight valence electrons.

Cl + e- → Cl -

Se + 2e- → Se 2-

Ionic Bonding• Ionic bonding results from the electrostatic attraction

between cations and anions.

• Formation of an ionic bond can be viewed as a transfer of electrons.

Na + F → Na+ + F - (or NaF)

• A covalent bond result from the sharing of two electrons between two atoms, as shown here for H2.

Covalent Bonding

• Two hydrogen atoms become more stable as their orbitals, each containing one electron, overlap.

Orbital Overlap

Lewis Structures• Lewis structures represent covalent bonding by showing

how the valence electrons are present in a molecule.• Bonding pairs are shared between two atoms and are

represented by lines .• Lone pairs are entirely on one atom and are represented

by two dots.

H ClBonding Pair

Lone Pair

The number of Covalent BondsThe number of covalent bonds can be determined fromthe number of electrons needed to complete an octet.

Octet Rule

Octet Rule: atoms share electrons until each atom is surrounded by eight.

• Single Bond - sharing one pair of electrons

• Double Bond - sharing two pairs of electrons

• Triple Bond - sharing three pairs of electrons

1. Write the skeleton structure.

2. Sum the valence electrons.

3. Subtract two electrons for each bond in the skeleton structure.

4. Count the number of electrons needed to satisfy octet rule for each atom.• If the number of electrons needed equals the number remaining, go

to 5.• If fewer electron remain, add one bond for every two additional

electrons needed.

5. Place remaining electrons as lone pairs to satisfy the octet rule for each atom (not H).

Writing Lewis Structures

Writing Lewis Structures• Write the Lewis structure of fromaldehyde, H2CO. The

skeleton structure is

O

HHC

• The total number of valence electrons is

1(C) 1 x 4 = 4

1(O) 1 x 6 = 6

2(H) 2 x 1 = 2

12


O

HHC

• Remaining valence electrons = 6• 8 electrons needed to obey the octet rule

• Add one bond because 2 more electrons are needed than are available.


O

HHC

needs 6e- to complete octet

needs 2e- to complete octet

O

HHC

• Finish the structure by placing remaining electrons as lone pairs.

• Check that the final Lewis structure has the correct number of valence electrons (12) and each atom (not H) has 8 electrons.


O

HHC

Test Your Skill• Write the Lewis structure of N2H2. The skeleton structure is:

H N N H

Test Your Skill• Write the Lewis structure of N2H2.

• Answer:

H N N H

Polar BondPolar Bond –

A covalent bond in which the two atoms do not share the bonding electrons equally.

Note: The higher electronegative element takes on a partial negative charge and the lower electronegative element takes on a partial positive charge.

+ -d d

I—Br (a polar covalent bond)

arrow indicates direction of more electronegative element

• In I2 the sharing of the electrons in the covalent bond is equal; in ClF it is not.

• Dipole moment is a measure of the unequal sharing of electrons.

• The unequal sharing leads to a polar covalent bond that is indicated with the symbol d followed by a sign to show partial charges. + -d d

Cl-F

Bond Polarity

• Electronegativity is a measure of the ability of an atom to attract the shared electrons in a chemical bond.

Electronegativity

Electronegativity Trends

Example: Electronegativity• Select the most polar bond.

Cl-F O-F P-F

Compound Cl2 ICl NaI NaF

Electroneg.Difference

0 0.5 1.6 3.1

Polarity nonpolar polar ionic ionic

Phase (R.T.)

gas solid crystalline solid

crystalline solid

m.p. (oC) -103 27 661 988

b.p. (oC) -35 97 1304 1695

Bonding covalent polar covalent

ionic ionic

Properties of Compounds

• Formal charge is a charge assigned to atoms in Lewis structures by assuming the shared electrons are divided equally between the bonded atoms.

• Equation for Formal Charge:

(number of valence electrons in atom) –

(number of lone pair electrons) – ½ (number of shared electrons)

Formal Charges

Atom # B

onds

# B

onds

# B

onds

N

O

C

-1 0 +1

2

2

3

3

3

1

4

4

N N N

O O O

C C

_

_____

___

_

__

__

_ _

__

___

_

:: ::

::

:

:::

Formal Charges

Formal Charges• Add formal charges to the Lewis structure of HNO3 shown

below.

O

N O O

H

• The nitrogen atom has 4 bonds giving it a +1 formal charge.

• The oxygen atom on the bottom left only has 1 bond giving it a -1 formal charge.

• The sum of formal charges equals the charge of the species.

Formal Charges

ON O

O

H

Test Your Skill• Add formal charges to the Lewis structure of HNO3 shown

below.

ON O

O

H

• Answer:

Test Your Skill

ON O

O

H

• Lewis structures that show the smallest formal charges are favored.

• Lewis structures that have adjacent atoms with formal charges of the same sign are much less favorable.

• Lewis structures that place negative formal charges on the more electronegative atoms are favored.

• Formal charges of opposite sign are usually on adjacent atoms.

• Fewer formal charges overall.

Structure Stability (which structure is more favored)

Test Your Skill• Of the two structures shown for HNO3, use the stability rules

to predict which will be more favored.

ON O

O

HO

N OO

H

• Answer:

• The structure on the left is favored because

• it has fewer formal charges.

• it does not have adjacent atoms with the same formal charge.

Test Your Skill

ON O

O

HON O

O

H

Resonance in Lewis Structures• Resonance structures differ only in the distribution of the

valence electrons. • All resonance structures follow the rules for writing Lewis

structures.• Resonance structures are indicated by a double headed

arrow.

ON O

O

HO

N OO

H

↔

• Draw the third possible resonance structure for HNO3, the first two are below.

Drawing Resonance Structures

ON O

O

HO

N OO

H

↔

• First two:

• In the third, add the double bond between the nitrogen and the oxygen to the bottom left. Complete the octets with the remaining electrons and add formal charges.

Drawing Resonance Structures

ON O

O

HO

N OO

H

↔

ON O

O

H

HNO3 Resonance Forms

• A total of three resonance forms can be written for HNO3.

• The first and last structures are equally favored because of fewer formal charges; the middle structure less favored.

ON O

O

HO

N OO

H

↔ ↔O

N OO

H

Test Your Skill• Write all resonance structures, including formal charges, for

O3, O-O-O skeleton structure.

Test Your Skill• Write all resonance structures, including formal charges, for

O3, O-O-O skeleton structure.

• Answer:

O ↔ O

O

OO

O

• No resonance structure is correct by itself; the correct structure is an average of all resonance structures.

Average Structure

Contribution of Resonance Structures• Equivalent resonance structures, such as the two for O3,

contribute equally to the average structure. Bond order in O3 is the average of a double bond and a single bond = 1.5.

O ↔ O

O

OO

O

Test Your Skill• Draw the Lewis structure of IF3.

• Answer:

I

F

F

F



Chapter 10 Molecular Structure and

Bonding Theories


Valence-Shell Electron-Pair Repulsion Model (VSEPR) predicts shape from Lewis Structures.

• VSEPR Rule 1:

• A molecule has a shape that minimizes electrostatic repulsions between valence-shell electron pairs.

• Minimum repulsion results when the electron pairs are as far apart as possible.

VSEPR

• Steric number = • (number of lone pairs on central atom) +

(number of atoms bonded to central atom)• The steric number is determined from the

Lewis structure.

• Steric number determines the bonded-atom lone-pair arrangement, the shape that maximizes the distances between the valence-shell electron pairs.

Steric Number

Geometric Arrangements

• In the Lewis structure of BeCl2,

beryllium has two bonded atoms and no lone pairs,

steric number = 2.

• A linear geometry places the two pairs of electrons on the central beryllium atom as far apart as possible.

Steric Number = 2

BeCl Cl

• The Lewis structure of HCN (H-C º N:) shows that the carbon atom is bonded to two atoms and has no lone pairs, steric number = 2.

• The bonded-atom lone-pair arrangement is linear.

• The number of bonded atoms, not the number of bonds, determines the steric number.

Molecules with Multiple Bonds

• The Lewis structure of BF3

shows the boron atom has a steric number = 3; the bonded-atom lone-pair arrangement is trigonal planar.

Steric Number = 3

B F

F

F

• The Lewis structure of CH4

shows the carbon

atom has a steric number = 4; the bonded-atom lone-pair arrangement is tetrahedral.

Steric Number = 4

C H

H

H

H

• The phosphorus atom in PF5 has a steric number = 5; the bonded-atom lone-pair arrangement is trigonal bipyramidal.

Steric Number = 5

• The sulfur atom in SF6 has a steric number = 6; the bonded-atom lone-pair arrangement is octahedral.

Steric Number = 6

• The Lewis structure of H2O is

• Steric number = 4, 2 bonded atoms and 2 lone pairs.

• The bonded-atom lone-pair arrangement is tetrahedral.

Central Atoms with Lone Pairs

OHH

• Molecular shape is the arrangement of the atoms in a species.

• The bonded-atom lone-pair arrangement of H2O is tetrahedral (top); the molecular shape is bent or V-shaped (bottom).

Molecular Shape of H2O

• What is the electron pair geometry and molecular shape of NH3?

Molecular Shape of NH3

First, draw the Lewis structure.

• The nitrogen has 3 bonded atoms and 1 lone pair; the steric number = 4 and the bonded-atom lone-pair arrangement is tetrahedral.


NHH

H

• The bonded-atom lone-pair arrangement of NH3 is tetrahedral (top), molecular shape is a trigonal pyramidal (bottom).


• The measured bond angle in H2O (104.5o) is smaller than the predicted angle (109.5o)

• Explanation (VESPR Rule #2) Forces between electron pairs vary as: • lone pair-lone pair replusion > lone pair-bonding pair replusion >

bonding pair-bonding pair repulsion

Electron Pair Repulsions

• What is the steric number, the bonded-atom lone-pair arrangement, and the molecular shape of ClF3?

Test Your Skill

Answer: The steric number is 5, the bonded-atom lone-pair arrangement is trigonal bipyramidal and the molecule is “T” shaped with the two lone pairs in equatorial positions.

Test Your Skill

• The geometry of each central atom is determined separately.

• The CH3 carbon in CH3CN has tetrahedral geometry and the other carbon has linear geometry.

Multiple Central Atoms

• What are the bonded-atom lone-pair

arrangements and the shapes about each central atom in NH2SH?

• Draw the Lewis structure.

• The bonded-atom lone-pair arrangements of both are tetrahedral, the nitrogen shape is trigonal pyramidal and sulfur is “V” shaped.

Shapes of Molecules

N SH

H

H

Polar Bond

Polar Bond –

A covalent bond in which the two atoms do not share the bonding electrons equally.

Note: The higher electronegative element takes on a partial negative charge and the lower electronegative element takes on a partial positive charge.

+ -d d

I—Br (a polar covalent bond)

arrow indicates direction of more electronegative element

• In I2 the sharing of the electrons in the covalent bond is equal; in ClF it is not.

• Dipole moment is a measure of the unequal sharing of electrons.• Equals the magnitude of the separated charges X the distance

between them

• The unequal sharing leads to a polar covalent bond that is indicated with the symbol d followed by a sign to show partial charges. + -d d

Cl-F

Bond Polarity

• The bond dipoles in CO2 cancel because the linear shape orients the equal magnitude bond dipoles in exactly opposite directions.

• Bond dipole determined by difference in electronegativities of bonded atoms.

Polarity of Molecules

• The bond dipoles do not cancel in COSe; they are oriented in the same direction and are of unequal length. They do not cancel in OF2 because the V-shape of the molecule does not orient them in opposite directions.


• The bond dipoles in BCl3 and CCl4 cancel because of the regular shape and equal magnitude.


• The bond dipoles in BCl2F and CHCl3 do not cancel because they are not of the same magnitude.


• Are the following molecules polar or nonpolar: H2S, SiF4, CH2Cl2?

Test Your Skill

• Are the following molecules polar or nonpolar: H2S, SiF4, CH2Cl2?

• Answer: H2S and CH2Cl2 are polar, SiF4 is nonpolar.

Test Your Skill

• Hybrid orbitals are:

Orbitals obtained by mixing two or more atomic orbitals on the same central atom.

Hybrid Orbitals

• The bonds in BeCl2 arise from the overlap of two sp hybrid orbitals on the beryllium atom with the 3p orbitals on the two chlorine atoms.

Bonding in BeCl2

• The bonds in BF3 arise from the overlap of three sp2 hybrid orbitals on the boron atom with 2p orbitals on the three fluorine atoms.

Bonding in BF3

• The bonds in CH4 arise from the overlap of four sp3 hybrid orbitals on the carbon atom with 1s orbitals on the four hydrogen atoms.

Bonding in CH4

• Hybrid orbitals can hold lone pairs as well as make bonds.

Lone Pairs and Hybrid Orbitals

Steric Number

Electron pair geometry

Hybridization

2 linear sp 3 trigonal planar sp2 4 tetrahedral sp3 5 trigonal

bipyramid sp3d

6 octahedral sp3d2

Hybrid Orbitals

Test Your Skill• Identify the hybrid orbitals on the central

atoms in SiH4 and HCN.

• Identify the hybrid orbitals on the central atoms in SiH4 and HCN.

• Answer: sp3 hybrid orbitals on silicon; sp hybrid orbitals on carbon.

Test Your Skill

• Sigma bonds (s): the shared pair of electrons is symmetric about the line joining the two nuclei of the bonded atoms.

Types of Bonds: Sigma Bonds

• The C-C sigma bond in C2H4 arises from overlap of sp2 hybrid orbitals and the four C-H sigma bonds from overlap sp2 hybrid orbitals on C with 1s orbitals on H.

• The second C-C bond forms from sideways overlap of p orbitals.

Bonding in C2H4

• Pi bonds (p) places electron density above and below the line joining the bonded atoms – they form by sideways overlap of p orbitals.

Types of Bonds: Pi Bonds

• The double bond in C2H4 is one sigma bond and one pi bond – each bond is of similar strength.

Bonding in C2H4

Test Your Skill

• Describe the bonds made by the carbon atom in HCN.

• Describe the bonds made by the carbon atom in HCN.• Answer: The carbon is sp hybridized.

• C sp hybrid forms s bond with H 1s. • C sp hybrid forms s bond. N has s, p (along bonding

direction) or sp hybrid of the two available for this s bond.

• Two pi bonds form by sideways overlap of two p orbitals on both C and N.

Test Your Skill



Chapter 11Liquids and Solids


• Intermolecular forces are the attractions that hold molecules together in the liquid and solid states.

State Volume Shape of State Density Compressibility

gas assumes shape and volume of container

low easily compressed

liquid definite volume, assumes shape of container

high nearly incompressible

solid both definite shape and volume

high nearly incompressible

Characteristic Properties of Gases, Liquids, and Solids

Physical State Relation Between Energy of Attraction and Kinetic Energy of Molecules

solid kinetic energy << energy of attraction

liquid kinetic energy ≈ energy of attraction

gas kinetic energy >> energy of attraction

Physical State and Energy of Attraction

Boiling Point• The boiling point of a liquid is the

temperature at which the vapor pressure is equal to the external pressure.

• The normal boiling point of a liquid is the temperature at which its equilibrium vapor pressure is equal to 1 atmosphere.

• At the boiling point, bubbles filled with vapor form below the surface of the liquid.

• Electrostatic forces account for all types of intermolecular attractions. There are three types of attractions:

• Dipole-dipole attractions• London dispersion forces• Hydrogen bonding

Intermolecular Attractions

• Dipole-dipole attractions result from electronic forces between molecular dipoles:

Dipole-Dipole Attractions

• London dispersion forces arise from the attractions between instantaneous dipoles and induced dipoles.

London Dispersion Forces

Dispersion Forces and Periodic Trends

• Polarizability is the ease with which a charge distorts the electron cloud in a molecule.• Polarizability generally increases with the

number of electrons in the molecule.

• For related series of molecules, London dispersion forces increase going down any group in the periodic table.

• Hydrogen bonding occurs between a hydrogen atom bonded to N, O, or F, and a lone pair of electrons on a second N, O, or F.• Hydrogen bonds are

sometimes shown as dotted lines.

Hydrogen Bonding

• Hydrogen bonding causes ice to have a lower density than liquid water.

Structure of Solid Water

• Identify the kind of intermolecular forces:

(a) BF3, BBr3

(b) C2H5OH, C2H5Cl

Example: Intermolecular Forces

Answers

a) London dispersion forces for both.

b) C2H5OH is Hydrogen Bonding

C2H5Cl is Dipole-Dipole

Liquids: Capillary Action• Capillary action causes water to rise in

a small diameter glass tube.

• Capillary action is the result of a competition between:• cohesion: the attraction of molecules for

other molecules of the same substance.• adhesion: the attraction of molecules for

other molecules of a different substance.

Capillary Action• Water rises because adhesion is stronger

than cohesion.• Mercury is lowered because cohesion is

stronger than adhesion.

• Viscosity is the resistance of a fluid to flow.

• The stronger the intermolecular forces of attraction, the greater the viscosity.

• Other factors contribute to viscosity as well, like structure, size, and shape of molecules.

Liquids: Viscosity



Chapter 12Solutions


• There are a number of ways to express concentration:

• Molarity• mass percentage• ppm and ppb• molality

Solution Concentration

• All concentration units are fractions.• The numerator contains the quantity of solute.• The denominator is the quantity of either

solution or solvent.

• They differ in the units used to express these two quantities.

Solution Concentration

Units of Concentration Used Earlier

solution of liters

solute of moles

:Molarity

M

Mass Percent Composition

%100solution grams

solute gramspercent mass

• A solution is prepared by dissolving 3.00 g of NaCl (molar mass = 58.44 g/mol) in 150 g of water. Express its concentration as mass percent.

• Answer: 1.96 %

Example: Percent Composition

Molality• Molality (m or molal) is defined as

solvent of kilograms

solute of moles molality

Example: Calculate Molality

• What is the molality of a solution prepared by dissolving 3.00 g NaCl (molar mass = 58.44 g/mol) in 150 g of water?

• Answer: 0.342 m

ppm & ppb

• ppm (parts per million)

= (g solute / g solution) x 1,000,000

• ppb (parts per billion)

= (g solute / g solution) x 1,000,000,000

Units for Concentration Conversion

Concentration Unit Numerator Units Denominator

units

Mass % Grams 100 g solution

Molarity Moles 1 L solution

Molality Moles 1 kg solvent

Mole Fraction Moles 1 mol total of solution

• Express the concentration of a 3.00% H2O2 solution as molality.

Answers:

0.910 molal

Example: Concentration Conversion

Test Your Skill• Calculate (a) the molality, (C2H5OH;

molar mass = 46.07 g/mol) in a wine that has an alcohol concentration of 7.50 mass percent.

• Answers: 1.76 molal

• Conversion of most concentration units to molarity usually involve using the density of the solution to convert units mass to units of volume.

• The density of a 12.0% sulfuric acid (H2SO4; molar mass = 98.08 g/mol) is 1.080 g/mL. What is the molarity of this solution?

Answer = 1.32 M

Example: Conversion to Molarity

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