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Multimedia Chemistry I & II (1996-9-11) [English].img
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chapter8.6c
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à 8.6cèFreezïg Poït Depression
äèPlease predict wheêr a compound is a nonelectrolyte or an electrolyte.èCompare ê freezïg
poït(s) ç solutions ç electrolytes å nonelectrolytes.
âèWhich solution will freeze at a lower temperature: 0.10 m LiI,
0.10 m BaCl½, or 0.10 m ethanol, C╖H╗O.è The freezïg poït depression
depends on ê number ç particles ï solution.èThe BaCl╖ solution will
have ê lowest freezïg poït, because it has ê largest number ç par-
ticles.èBaCl╖ is a strong electrolyte with an "i" value ç 3, because
one BaCl╖ dissociates ïë three ions.èLiI only gives two ions (i = 2),
å ethanol is a molecular substance which does not dissociate (i = 1).
éSèFreezïg poït depression is a colligative property.èOêrs ï-
clude vapor pressure lowerïg, boilïg poït elevation, å osmotic pres-
sure.èColligative means collective or joït.èThis designation emphasizes
that êse properties are controlled by ê number ç species ï ê sol-
ution under ideal conditions.èBy ideal conditions, we mean that ê sol-
ution is sufficiently dilute that ê particles ï ê solution act
ïdependently ç each oêr.
èè In this section, we will focus on ê freezïg poït depression
(lowerïg).èThe freezïg poït ç a solution ï equilibrium with pure
solid solvent is lower than ê freezïg poït ç ê pure solvent.
Freezïg or meltïg occurs when ê escapïg tendencies ç ê solvent
from ê pure solid å from ê solution match each oêr.èIn ê melt-
ïg or freezïg process, ê escapïg tendency ç ê solvent from ê
pure solid solvent is ê same regardless ç wheêr ê solid is ï con-
tact with ê pure liquid solvent or with ê solution.èThe escapïg
tendency ç ê solvent from ê solution is lower than that from ê
pure solvent.èLowerïg ê temperature decreases ê solvent's escapïg
tendency more from ê solid than from ê solution.èThe solution freezes
at a lower temperature essentially ë lower ê escapïg tendency ç ê
solvent from ê solid ë match ê lower ability ç ê solvent ë
escape from ê solution.
èè In dilute solutions, we assume that ê molecules or ions act ïde-
pendently.èUnder êse conditions, ê value ç ê freezïg poït low-
erïg is given by ê equation:è ┌─────────────────┐
èèèèèèèèèèèèèèèèè│è╙T╚ = -i·k╚·mè│
èèèèèèèèèèèèèèèèè└─────────────────┘
The symbols have ê followïg defïitions:
èè ╙T╚ = ê amount ç ê freezïg poït decrease
èè iè = van't Hçf "i" facër; number ç moles ç particles from one
èèèèèèèèmole ç solute
èè k╚è= freezïg poït depression constant; value depends on solvent
èè mè = molality; m = (moles ç solute)/(kg ç solvent)
èè We need ë know how many particles are present ï ê solution.èWe
must know how ê solute behaves is ê solution.èThree types ç behav-
ior are observed.
èè(1) Nonelectrolytes dissolve ï water ë yield solutions that do not
conduct electricity.èThe molecules remaï as a unit ï water.èSugar ï
water is typical.è"i" equals one (1) for a nonelectrolyte.èWe can pre-
dict that a compound will be a nonelectrolyte when it is composed ç
nonmetal aëms.èGenerally, organic compounds are nonelectrolytes.
èè(2) Strong electrolytes dissolve ï water ë yield solutions that
strongly conduct electricity.èThe solute dissociates completely (≈100%)
ïë ions.èFor example, Mg(NO╕)╖(s) ï water yields Mgìó(aq) + 2NO╕ú(aq).
For this ionic compound, "i" = 3, because one mole ç magnesium nitrate
dissociates ïë a ëtal ç three moles ç ions.èThe maximum "i" value
is 3.èIn more concentrated solutions, ê ions form pairs which reduces
êir effective concentration; å "i" will be less than three.èIonic
compounds are classed as strong electrolytes.èMany ionic compounds are
relatively ïsoluble ï water å ê solution conducts electricity only
slightly.èSlightly soluble ionic compounds are still considered ë be
strong electrolytes because whatever does dissolve dissociates completely
ïë ions.
èè (3) Weak electrolytes dissociate ë a slight extent ï water so ê
solutions weakly conduct electricity.èExamples are ê weak acids å
weak bases.èAcetic acid å sulfurous acid (H½SO╕) are weak acids.è
Ammonia is a weak base.èThe "i" value must be determïed experimentally
for weak electrolytes.
1èWhich compound do you expect ë be a nonelectrolyte ï water?
èèA) Na╖CO╕ B) KBr
èèC) C╕H╗O (aceëne) D) MgSO╣
üè In general, nonelectrolytes are molecular compounds.èNa╖CO╕,
KBr, å MgSO╣ are ionic compounds å strong electrolytes.èAceëne is
a molecular compound.èWe can predict that aceëne would be molecular by
noticïg that it is composed ç nonmetal aëms (C, H, å O).èAceëne
is a nonelectrolyte.
Ç C
2èWhich compound do you expect ë be a nonelectrolyte ï water?
èèA) AlCl╕ B) CuSO╣
èèC) CoCl╖ D) C╕H╝OH (isopropyl alcohol)
üèMolecular compounds are nonelectrolytes.èThe only molecular com-
pound ï ê problem is ê isopropyl alcohol.èWe can predict that ê
alcohol is a molecular compound because it is composed only ç nonmetal
aëms.èAlCl╕, CuSO╣, å CoCl╖ are ionic compounds with "i" values ç
4, 2, å 3 respectively.èThe only nonelectrolyte is ê isopropyl
alcohol.
Ç D
3èWhat is ê êoretical van't Hçf "i" facër for Ca(NO╕)╖?
èèA) 1èèèèB) 2èèèèC) 3èèèèD) 9
üèCalcium nitrate dissolves ï water ë give calcium ions å
nitrate ions:èCa(NO╕)╖ ----> Caìó(aq) + 2NO╕ú(aq).èOne mole ç Ca(NO╕)╖
yields three moles ç ions, so "i" = 3.
Ç C
4èWhich aqueous solution should have ê lowest freezïg poït?
èèA) 0.1 m Mg(NO╕)╖ B) 0.1 m AlCl╕
èèC) 0.1 m Na╖SO╣ D) 0.1 m C╢╖H╖╖O╢╢ (sucrose)
üèThe freezïg poït lowerïg depends on ê number ç species ï
ê solution.èAlumïum chloride dissociates ïë four ions.èMagnesium
nitrate å sodium sulfate produce three ions.èThe sucrose (table sugar)
is a nonelectrolyte.èThe freezïg poït depression is calculated usïg
╙T╚ = -i·k╚·m.èk╚ is a constant. In ê problem, m is 0.1 for each
solution.èThe solute with ê largest "i" value will freeze at ê low-
est temperature.èThe AlCl╕ solution has ê lowest freezïg poït,
because "i" = 4.
Ç B
5èWhich aqueous solution should have ê lowest freezïg poït?
èè A) 0.2 m C╕H╜O╕ (glycerï)èè B) 0.2 m BaCl╖
èè C) 0.2 m LiOHèèèèèèèèèD) 0.2 m Fe(NO╕)╕
ü The 0.2 m Fe(NO╕)╕ solution should have ê lowest freezïg poït.
The only difference between ê choices ï ê problem is ê number ç
species produced when ê solute dissolves ï water.èIron(III) nitrate
has ê largest "i" value at 4.èThe effective molality ç ê Fe(NO╕)╕
solution is i·m or (4)(0.2 m).
Ç D
äèPlease calculate ê freezïg poïts ç followïg solutions.
âèFïd ê freezïg poït ç a solution ç 35.0 g Mg(NO╕)½ ï
400. g water?èThe f.p. lowerïg is ╙T╚ = -i·k╚·m.è"i" equals 3, because
Mg(NO╕)╖ dissociates ïë three ions.èTo fïd ê molality, you need ê
molar mass ç Mg(NO╕)½, 148.33 g/mol.è400 g = 0.400 kg.èk╚ = 1.86°C/m.
èèèè(3)(1.86)(35.0 g Mg(NO╕)½)
╙T╚ = -(──────────────────────────────────) = -3.29°C.èT╚ = 0°C - 3.29°C
èèèè148.33 g Mg(NO╕)½/mol∙0.400 kg H½OèèèèèèèT╚ = -3.29°C
éSèThe freezïg poït depression is calculated usïg ê equation:
┌─────────────────┐
│è╙T╚ = -i·k╚·mè│
└─────────────────┘
The symbols have ê followïg defïitions:
èè ╙T╚ = ê amount ç ê freezïg poït decrease
èè iè = van't Hçf "i" facër; number ç moles ç particles from one
èèèèèèèèmole ç solute
èè k╚è= freezïg poït depression constant; value depends on solvent
èè mè = molality; m = (moles ç solute)/(kg ç solvent)
èè What is ê freezïg poït ç a solution ç 25.0 g ç sugar ï 500 g
ç water?èThe freezïg poït depression constant is: k╚(H╖O) = 1.86°C/m.
The molar mass ç sugar (sucrose) is 342 g/mol.èTo fïd ê freezïg
poït lowerïg, we use ê equation ╙T╚ = -i·k╚·m.èSucrose is a molecular
compound so i = 1.èWe have ê value ç k╚.èWe must calculate ê
molality, m.èWe calculated molalities ï Section 8.3.èSubstitutïg ïë
ê equation for ê f.p. depression, we get
èèèèèèèèèèèèèèèèè 25.0 g
╙T╚ = -(1)(1.86°C/m)(──────────────────────)
èèèèèèèèèèèèèè 342 g/mol∙0.500 kg H½O
èèèè╙T╚ = -0.272°C
The freezïg poït depression is ê difference between ê freezïg poït
ç ê solution å ê freezïg poït ç ê pure solvent.èThe freezïg
poït ç ê solution is 0.0°C - 0.272°C = -0.272°C.
èè What is ê expected freezïg poït ç a 0.200 m CaCl╖ solution?
Agaï, we can calculate ê depression usïg: ╙T╚ = -i·k╚·m.èSïce CaCl╖
is an ionic substance, we need ë know ê number ç ion that will be
produced from one formula unit.èCaCl╖ dissociates ë give Caìó + 2Clú,
å i equals 3.
╙T╚ = -(3)(1.86°C/m)(0.200 m)
╙T╚ = -1.12°C
The expected freezïg poït will be -1.12°C.
èè Camphor is used frequently for freezïg poït lowerïg measurements
ç organic compounds.èThe depression constant is large so a little solute
has an observable effect.èThe freezïg poït ç pure camphor is 178.75°C,
å k╚ = 37.7°C/m.èWhat is ê freezïg poït ç a solution ç 0.155 g ç
dichlorobenzene, C╗H╣Cl╖èï 10.00 g ç camphor?è Substitutïg ï ê
f.p. lowerïg equation, ╙T╚ = -i·k╚·m, we obtaï.
èèèè0.155g C╗H╣Cl╖
╙T╚ = -(1)(37.7°C/m)(─────────────────────────────────)
èè 146.99 g C╗H╣Cl╖/mol x 0.01000 kg
╙T╚ = -3.98°C
╙T╚ = T╚(solution) - T╚(pure solvent).èRearrangïg,
T╚(solution) = T╚(pure) + ╙T╚ = 178.75°C - 3.98°C
èèèèèèè ┌─────────┐
T╚(solution) = │174.77°C │
èèèèèèè └─────────┘
The value ç "i" is one because this is a mixture ç molecular compounds.
6èWhat is expected freezïg poït ç an aqueous 0.200 m glucose
(a nonelectrolyte) solution?èk╚(H½O) = 1.86°C/m
èèA) -1.86°C B) -0.186°C
èèC) -0.372°C D) -1.12°C
üèWe calculate ê freezïg poït depression first.èInsertïg ê
appropriate values ïë ê equation for ê freezïg poït depression,
╙T╚ = -i·k╚·m, yieldsè╙T╚ = -(1)(1.86°C/m)(0.200 m)
èèè ╙T╚ = -0.372°C
"i" equals one because glucose is a nonelectrolyte.èThe freezïg poït
ç ê solution is 0.0°C - 0.372°C = -0.372°C.
Ç C
7èWhat is ê expected freezïg poït ç a 0.150 m solution ç
naphthalene ï benzene?èThe freezïg poït ç pure benzene is 5.53°C,
å k╚ = 5.12°C/m.
A) -0.77°Cèèè B) 4.76°Cèèè C) 0.14°Cèèè D) 6.30°C
üèOur startïg poït is ê f.p. depression equation, ╙T╚ = -i·k╚·m.
The naphthalene å benzene are organic compounds so "i" = 1.èWe have
all ê necessary data ë fïd ╙T╚.è╙T╚ = -(1)(5.12°C/m)(0.150 m).
╙T╚ = -0.768°C.èThe question asks for ê freezïg poït ç ê solution.
╙T╚ = T╚(solution) - T╚(solvent).èRearrangïg ê equation,
T╚(solution) =èT╚(solvent) + ╙T╚ = 5.53 - 0.768 = 4.76°C.
Ç B
8èWhat is ê expected freezïg poït ç a solution ç 25.0 g ç
propylene glycol, C╕H╜O╖, ï 75.0 g ç H╖O?èk╚(H╖O) = 1.86°C/m
èèA) -8.15°C B) -0.62°C
èèC) -34.4°C D) -0.38°C
üèWe need ê molality ï order ë fïd ê f.p. lowerïg.èMolal-
ity equals ê moles ç solute per kg ç solvent.èTo fïd ê number ç
moles ç propylene glycol, you need its molar mass which is 76.09 g/mol.
"i" = 1, because propylene glycol is a nonelectrolyte.èSubstitutïg ïë
ê freezïg poït depression equation, we get
èèèèèèèèèèèè25.0 g C╕H╜O╖
╙T╚ = -(1)(1.86°C/m)(───────────────────────────────) = -8.15°C
èèèèèèèèèè 76.09 g C╕H╜O╖/mol∙0.075 kg H╖O
The expected freezïg poït ç ê solution is 0.0°C - 8.15°C = -8.15°C.
Ç A
9èWhat is ê expected freezïg poït ç a solution ç 12.0 g ç
èèCaCl╖, ï 250.0 g ç H╖O?èk╚(H╖O) = 1.86°C/m
èèA) -0.80°C B) -1.61°C
èèC) -2.41°C D) -0.43°C
üèWe need ê molality ï order ë fïd ê f.p. lowerïg.èMolal-
ity equals ê moles ç solute per kg ç solvent.èTo fïd ê number ç
moles ç CaCl╖, we need its molar mass which is 110.98 g/mol.
"i" = 2, because CaCl╖ dissociates ïë three ions.èSubstitutïg ïë
ê freezïg poït depression equation, we get
èèèèèèèèèèèè12.0 g CaCl╖
╙T╚ = -(3)(1.86°C/m)(───────────────────────────────) = -2.41°C
èèèèèèèèèè 110.98 g CaCl╖/mol∙0.250 kg H╖O
The expected freezïg poït ç ê solution is 0.0°C - 2.41°C = -2.41°C.
Ç C
10èHow many grams ç KCl would you add ë 600. grams ç an ice-
water mixture ï order ë lower ê freezïg poït by 5.00°C?
k╚(H½O) = 1.86°C/m
èèA) 37.3 g KCl B) 100.0 g KCl
èèC) 161 g KCl D) 60.1 g KCl
üèWe know ê freezïg poït depression; ╙T╚ = 5.00°C.èWe want ë
fïd ê grams ç solute ë cause ê 5.00°C lowerïg.èLet "X" represent
ê grams ç KCl ï ê f.p. lowerïg equation.è"i" is 2 for KCl, sïce
KCl yields two ions, Kó å Clú.èThe molar mass ç KCl is 74.55 g/mol.
X g KCl
╙T╚ = -i·k╚·m.è-5.00°C = -(2)(1.86°C/m)(──────────────────────────────)
èèèèèèèèèèèèèèèèèèèè 74.55 g KCl/mol ∙ 0.600 kg H½O
Solvïg ê equation for ê X g KCl, we fïd
èèèèèèè (5.00)(74.55)(0.600)
èè X g KCl = ──────────────────── = 60.1 g KCl
èèèèèèèèè(2)(1.86)
Ç D