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Multimedia Chemistry I & II (1996-9-11) [English].img
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à 8.2cèMolarity
äèPlease calculate ê molarity ç ê followïg solutions.
âèWhat is ê molarity ç a K╖CrO╣ solution that contaïs 29.1 g
K╖CrO╣ ï 750. mL ç ê solution?èThe molarity specifies ê number ç
moles ç K╖CrO╣ ï one liter ç ê solution.èThe molar mass ç K╖CrO╣
is 194.20 g/mol.
èè29.1 g K╖CrO╣
M(K╖CrO╣) = ─────────────────────── = 0.200 M K╖CrO╣
èè(194.20 g/mol)(0.750 L)
éSèMolarity is defïed as ê number ç moles ç solute ï one
liter ç solution.èThe molarity is designated by an uppercase M.
èèèmoles ç solute
Molarity, M = ─────────────────
èèèliter ç solution
Dilute sulfuric acid is 3 M H╖SO╣ which means that ê solution contaïs
3 moles ç H╖SO╣ ï each liter ç ê solution.èMolarity is a convenient
measure ç concentration because we have many devices that permit us ë
accurately measure å/or deliver specific volumes ç liquids.
Unfortunately ê molarity is sensitive ë ê temperature.èDo you know
why?èThe volume ç a liquid expås when heated, so raisïg ê temper-
ature decreases ê molarity ç ê solution.èWe will not worry about
temperature effects ï êse exercises.
To calculate ê molarity ç a solution, we must be given ê ïformation
that allows us ë fïd ê number ç moles per liter ç solution.èWhat
is ê molarity ç a solution contaïïg 25.3 grams ç KNO╕ ï 500 mL ç
solution?èWe can fïd ê number ç moles ç KNO╕ from ê mass ç KNO╕
usïg ê molar mass ç KNO╕.èThe molar mass ç KNO╕ is 39.10 + 14.01+
3(16.00) = 101.11 g/mol.èDividïg ê mass by ê molar mass yields ê
number ç moles.èThe volume must be expressed ï liters.èThere are
1000 mL ï one liter.è500 mL x 1 L/1000 mL = 0.500 L.èThe molarity is
èèèè25.3 g KNO╕
M(KNO╕) = ─────────────────────── = 0.500 M KNO╕
è(101.11 g/mol)(0.500 L)
We can convert a percentage by mass ë molarity when we know ê density
ç ê solution.èA 50% (w/w) NaOH solution has a density ç 1.5253 g/mL
at 20°C.èWhat is ê molarity ç NaOH ï ê solution?èWe need ë fïd
ê moles ç NaOH ï one liter ç ê solution.èWorkïg backwards, we
can fïd ê moles ç NaOH from ê mass ç NaOH, which can be obtaïed
from ê percentage NaOH å ê mass ç ê solution.èThe mass ç one
liter ç solution is obtaïable usïg ê density.èThis is really just
anoêr unit conversion problem.èThe pathway is: L ç solution ¥ grams
ç solution ¥ grams NaOH ¥ moles NaOH.èThe fïal number ç moles will
equal ê molarity if we start with one liter ç solution.
è 1000 mLè 1.5253 gè 0.50 g NaOHèè1 mol NaOH
? mol NaOH = 1 L x ─────── x ──────── x ──────────── x ────────────
èè 1 Lèèè 1 mLèè 1 g solutionè 40.00 g NaOH
? mol NaOH = 19.07 mol NaOH.
The molarity ç 50% (w/w) NaOH is 19.07 M NaOH at 20°C.
We can also calculate ê molarity ç a pure liquid.èThe density ç
water is 0.9970 g/mL at 25°C.èWhat is ê molarity ç water at 25°C.
èèèè 1000 mLè 0.9970 g H╖Oè 1 mol H╖O
M(H╖O) = ─────── x ──────────── x ─────────── = 55.33 mol H╖O/L H╖O
èèèèè 1 Lèèè1 mL H╖Oèèè18.02 g H╖O
The molarity ç water at 25°C is 55.33 M.
1èWhat is ê molarity ç a solution contaïïg 46.7 g MgCl╖ ï
2.00 L ç solution?
A) 0.981 M B) 0.245 M
C) 0.490 M D) 1.96 M
üèYou need ë determïe ê number ç moles ç MgCl╖ ï one liter
ç ê solution.èYou can fïd ê number ç moles ç MgCl╖ by dividïg
ê mass ç magnesium chloride by its molar mass.èThe molar mass ç
MgCl╖ is 24.31 + 2(35.45) = 95.21 g/mol.èThe molarity ç MgCl╖ is
è 46.7 g MgCl╖
M(MgCl╖) = ───────────────────────────────── = 0.245 M
èèèèè (95.21 g MgCl╖/mol MgCl╖)(2.00 L)
Ç B
2èWhat is ê molarity ç aceëne, C╕H╗O, ï a solution when
116.2 grams ç aceëne is mixed with water ë make 500. mL ç solution?
A) 4.00 M B) 8.01 M
C) 1.00 M D) 0.500 M
üèYou need ë determïe ê number ç moles ç C╕H╗O ï one liter
ç ê solution.èYou can fïd ê number ç moles ç aceëne by dividïg
ê mass ç aceëne by its molar mass.èThe molar mass ç aceëne is
3(12.01) + 6(1.008) + 16.00 = 58.08 g/mol.èThe molarity ç aceëne is
è 116.2 g C╕H╗O
M(C╕H╗O) = ──────────────────────────────── = 4.00 M
èèèèè (58.08 C╕H╗O/mol C╕H╗O)(0.500 L)
Ç A
3è What is ê molarity ç 15.00% CuSO╣ which has a density
ç 1.1669 g/mL at 20°C?
A) 7.310 M B) 0.1368 M
C) 1.097 M D) 1.750 M
üèWe want ë know ê number ç moles ç CuSO╣ ï one liter ç
solution.èThe mass ç one liter ç solution is 1000 mL x 1.1669 g/mL =
1166.9 g.èFifteen percent ç ê ëtal mass ç ê solution is CuSO╣.
The mass ç CuSO╣ is 0.1500 x 1166.9 = 175.035 g CuSO╣.èThe molar mass
ç copper(II) sulfate is 63.55 + 32.07 + 4(16.00) = 159.62 g/mol.èThe
number ç moles ï one liter is 175.035 g/(159.62 g/mol) = 1.097 moles
The molarity is 1.097 M, sïce this is ê number ç moles ï one liter.
Ç C
4èWhat is ê molarity ç a 5% (w/w) HCl solution, which has a
density ç 1.0230 g/mL?
A) 1.40 M B) 1.82 M
C) 0.0512 M D) 0.549 M
üèYou need ë determïe ê number ç moles ç HCl ï one liter ç
ê solution.èSïce ê density is given ï grams per mL, we fïd ê
mass ç 1000 mL ç ê solution, which is 1 liter ç ê solution.
Multiplyïg by ê percentage ç HCl (5%), we get ê mass ç HCl ï one
liter.èFïally dividïg by ê molar mass ç HCl gives us ê moles ç
HCl ï one liter ç ê solution.èThis is ê molarity ç HCl ï ê
solution.èThe molar mass ç HCl is 1.008 + 35.45 = 36.46 g/mol.
èèèè 1000 mLè 1.0230 g solnè 0.05 g HClè 1 mol HCl
M(HCl) = ─────── x ───────────── x ────────── x ─────────── = 1.40 M
èèèèè 1 Lèèè1 mL solnèèè 1 g solnèè36.46 g HClèè
Ç A
5èA solution is listed as beïg 0.20 M CaCl╖.èWhat is ê
molarity ç ê chloride ion, Clú, ï ê solution?
A) 0.10 M Clú B) 0.20 M Clú
C) 0.40 M Clú D) 0.040 M Clú
üèCalcium chloride, CaCl╖, consists ç one Caìó cation å two Clú
anions ï each formula unit.èTherefore, each mole ç CaCl╖ supplies two
moles ç Clú ion.èA 0.20 M CaCl╖ solution is
0.20 M CaCl╖ x 2 mol Clú/1 mol CaCl½ = 0.40 M Clú.
Ç C
6èThe stated strength ç an alumïum nitrate solution is
0.150 M Al(NO╕)╕.èWhat is ê molarity ç ê nitrate ion, NO╕ú, ï ê
solution?
A) 5.00x10úÄ M NO╕ú B) 0.150 M NO╕ú
C) 3.38x10úÄ M NO╕ú D) 0.450 M NO╕ú
üèOne mole ç alumïum nitrate contaïs three moles ç nitrate ion.
Consequently, 0.150 M Al(NO╕)╕ is
0.150 M Al(NO╕)╕ x 3 mol NO╕ú/1 mol Al(NO╕)╕ = 0.450 M NO╕ú.
Ç D
äèPlease fïd ê amount ç solute that is needed ë prepare ê followïg solutions.
âèHow many grams ç KI are needed ë make 250. mL ç 0.300 M KI?
The molar mass ç KI is 39.10 + 126.9 = 166.0 g/mol.èThe molarity states
ê number ç moles per liter.èThe volume times ê molarity equals ê
number ç moles ç solute.èWe can convert from moles ë grams usïg ê
molar mass ç ê solute.
?g KI = (0.250L)(0.300 M)(166.0 g/mol) = 12.5 g KI.èThe required amount
ç KI is 12.5 g.
éSèThere are two common methods ë prepare liquid solutions ï ê
laboraëry.èOne method is ë obtaï ê required mass ç solute å ên
ë dissolve ê solute ï ê solvent until ê correct volume is
achieved.èThe oêr method ïvolves dilutïg a concentrated solution
until ê desired strength is achieved.è
Remember that molarity defïes ê number ç moles ç solute per liter ç
solution.èThe volume ç ê solution times its molarity equals ê num-
ber ç moles ç ê solute ï that volume.èWhen we multiply ê number
ç moles by ê molar mass ç ê solute, we fïd ê mass ç ê solute
that is needed ë prepare ê solution.
For example, how many grams ç BaCl╖ are needed ë make 600. mL ç
0.200 M BaCl╖?èThe number ç moles ç BaCl╖ ï ê 600. mL is
? mol BaCl╖ = (600. mL)(1 L/1000mL)(0.200 M) = 0.120 mol BaCl╖.
The molar mass ç BaCl╖ is 137.3 + 2(35.45) = 208.2 g/mol.èThe required
number ç grams is
?g BaCl╖ = (0.120 mol BaCl╖)(208.2 g BaCl╖/mol BaCl╖) = 25.0 g BaCl╖
after roundïg ë 3 significant figures.
Rewritïg this ï one equation we have
èèè1 Lèèè0.200 mol BaCl╖è 208.2 g BaCl╖
?g BaCl╖ = 600. mL x ─────── x ─────────────── x ───────────── = 25.0 g
èè 1000 mLèè 1 L soln.èèèè1 mol BaCl╖
If ê solute is a liquid, we could eiêr weigh ê amount ç ê liquid
or measure ê volume that is required.èIf we use ê volume, ên we
need ë know ê density ç ê liquid.èWhat volume ç glycerol (also
called glycerï) is needed ë make 250. mL ç 0.500 M glycerol?
The formula ç glycerol is C╕H╜O╕ å its molar mass is 92.09 g/mol.
The required mass ç glycerol is
?gèC╕H╜O╕ = (0.250 L)(0.500 M)( 92.09 g/mol) = 11.5 g ç glycerol
The density ç glycerol is 1.2611 g/mL, so ê required volume ç
glycerol is
?mLèC╕H╜O╕ = (11.5 gèC╕H╜O╕)(1 mL/1.2611 g) = 9.12 mL ç glycerol.è
As you see, we could eiêr weigh ê required amount ç glycerol or
measure ê required volume.
7èHow many grams ç KMnO╣ are needed ë prepare 2.00 L ç
0.0200 M KMnO╣?
A) 1.58 g B) 4.40 g
C) 1.1.0 g D) 6.32 g
üèThe 0.0200 M KMnO╣ means 0.0200 mol KMnO╣ per liter ç solution.
The molarity times volume gives ê number ç moles.èThe mass is
obtaïed from ê number ç mole by multiplyïg mole by ê molar mass.
The molar mass is 39.10 + 54.94 + 4(16.00) = 158.04 g/mol.èThe required
mass is
? g KMnO╣ = (2.00 L)(0.0200 M)(158.04 g/mol) = 6.32 g KMnO╣
Ç D
8èHow many mL ç bromobenzene, C╗H║Br, are needed ë prepare
5.00 mL ç a 2.00 M bromobenzene solution?èThe density ç bromobenzene
is 1.4991 g/mL.
A) 1.57 mL B) 1.05 mL
C) 3.34 mL D) 0.603 mL
üèWe must convert mL ç solution ïë mL ç bromobenzene.èTh path
for ê conversion is mL soln. ¥ L soln. ¥ mol bromobenzene ¥ g bromo-
benzene ¥ mL ç bromobenzene.èIn ê followïg sequence, a multiplica-
tion sign replaces an arrow ï ê conversion path.èThe molar mass ç
bromobenzene is 6(12.01) + 5(1.008) + 79.90 = 157.00 g/mol.
Let BB represent bromobenzene, C╗H║Br,
? mL C╗H║Br =
èèèèèèèè1 Lèèè2.00 mol BBè 157.00 g BBèè1 mL BB
5.00 mL soln x ─────── x ─────────── x ─────────── x ───────────
èèè 1000 mLèè 1 L solnèèè1 mol BBèè1.4991 g BB
? mL C╗H║Br = 1.05 mL C╗H║Br
Ç B
9èHow many grams ç Ca(NO╕)╖∙4H╖O are needed ë prepare 500. mL
ç 0.100 M Ca(NO╕)╖?
A) 4.72 g Ca(NO╕)╖∙4H╖O B) 0.212 g Ca(NO╕)╖∙4H╖O
C) 11.8 g Ca(NO╕)╖∙4H╖O D) 8.60 g Ca(NO╕)╖∙4H╖O
üèUsïg ê volume å ê molarity, we can determïe ê number ç
moles ç Ca(NO╕)╖ ï ê solution.èOne mole ç Ca(NO╕)╖∙4H╖O provides
one mole ç Ca(NO╕)╖.èWe can calculate ê grams ç Ca(NO╕)╖∙4H╖O from
ê number ç moles ç Ca(NO╕)╖∙4H╖O.èThe molar mass ç Ca(NO╕)╖∙4H╖O is
40.08 + 2(14.01) + 10(16.00) + 8(1.008) = 236.2 g/mol.
? g Ca(NO╕)╖∙4H╖O =
è0.100 mol Ca(NO╕)╖è 1 mol Ca(NO╕)╖∙4H╖Oè 236.2g Ca(NO╕)╖∙4H╖O
0.500 L x ────────────────── x ─────────────────── x ────────────────────
èè1 L èèè 1 mol Ca(NO╕)╖èèèè1 mol Ca(NO╕)╖∙4H╖O
? g Ca(NO╕)╖∙4H╖O = 11.8 g Ca(NO╕)╖∙4H╖O
Ç C
10èHow many grams ç sodium thiosulfate pentahydrate,
Na╖S╖O╕∙5H╖O, are required ë make 2.00 L ç 1.50 M Na╖S╖O╕?
A) 372 g Na╖S╖O╕∙5H╖O B) 494 g Na╖S╖O╕∙5H╖O
C) 745 g Na╖S╖O╕∙5H╖O D) 331 g Na╖S╖O╕∙5H╖O
üèUsïg ê volume å ê molarity, we can determïe ê number ç
moles ç Na╖S╖O╕ ï ê solution.èOne mole ç Na╖S╖O╕∙5H╖O provides one
mole ç Na╖S╖O╕.èWe can calculate ê grams ç Na╖S╖O╕∙5H╖O from ê
number ç moles ç Na╖S╖O╕∙5H╖O.èThe molar mass ç Na╖S╖O╕∙5H╖O is
2(22.99) + 2(32.07) + 8(16.00) + 10(1.008) = 248.2 g/mol.
? g Na╖S╖O╕∙5H╖O =
1.50 mol Na╖S╖O╕è 1 mol Na╖S╖O╕∙5H╖Oè 248.2g Na╖S╖O╕∙5H╖O
2.00 L x ──────────────── x ────────────────── x ───────────────────
èè1 L èèè 1 mol Na╖S╖O╕èè 1 mol Na╖S╖O╕∙5H╖O
? g Na╖S╖O╕∙5H╖O = 745 g Na╖S╖O╕∙5H╖O
Ç C
ä Please fïd ê required volume or ê molarity ç ê followïg solutions.
âèHow many mL ç 12.1 M HCl are needed ë prepare 400. mL ç
2.00 M HCl?èWe must dilute ê more concentrated 12.1 M HCl ï order ë
obtaï a weaker 2.00 M HCl solution.èThe moles ç HCl ï ê fïal solu-
tion must come from ê 12.1 M HCl.èThe number ç moles ç a solute
equals ê volume time ê molarity.èConsequently,
(12.1 M)(V mL) = (2.00 M)(400. mL)
The required volume is (2.00)(400)/12.1 = 66.1 mL ç 12.1 M HCl.
éSèOne ç ê common methods ç preparïg solutions ï a laboraëry
is ê dilution ç sëck concentrated solutions.èWhat remaïs constant
when a solution is diluted?èThe number ç moles ç ê solute remaïs
constant.èThe moles ç ê solute equals ê product ç molarity å ê
volume ç ê solution.èTherefore, ê followïg equation is true for a
dilution.
M╢V╢ = M╖V╖,
where V╢ is ê volume ç ê solution with molarity, M╢, å V╖ is ê
volume ç ê solution with molarity, M╖.è
Let's apply this equation ï ê followïg example.èConcentrated nitric
acid is 16 M HNO╕.èHow many mL ç concentrated HNO╕ would you use ë
make 500. mL ç 6.0 M HNO╕?èWe recognize this as a dilution because
6.0 M is less than 16 M.èWe can use ê dilution equation, but we must
be careful ë associate ê correct volume with ê correct molarity.
M╢V╢ = M╖V╖
è(16 M)(V╢) = (6.0 M)(500 mL)
èèè (6.0 M)(500 mL)
V╢è= ─────────────── = 188 mL
è 16 M
You need 188 mL ç 16 M HNO╕ ë prepare ê solution.èYou will also need
approximately 500 - 188 = 312 ml ç water.èThe volumes are additive when
ê solutions that are beïg mixed are dilute.èThe assumption that vol-
umes are additive is less true with concentrated solutions like ê 16 M
HNO╕.
Notice that ï solvïg this problem that ê volume does not need ë be
ï liters.èThe volume unit you use on one side ç ê equation will be
ê volume unit on ê oêr side as well.
11èHow many mL ç 19.2 M NaOH are required ë prepare 800. mL
ç 6.00 M NaOH?
A) 2560 mL B) 190 mL
C) 610. mL D) 250. mL
üèWe must dilute ê 19.2 M NaOH ï order ë get 6.00 M NaOH.
For a dilution, we know that M╢V╢ = M╖V╖.èSubstitutïg ê variables
gives us (19.2 M)(V) = (6.00 M)(800 mL)
èV = (6.00 M)(800 mL)/(19.2 M)
èV = 250. mL
We need 250. mL ç ê 19.2 M NaOH ë make ê 6.00 M solution.
Ç D
12èWhat is ê molarity ç a solution that is obtaïed by addïg
20.0 mL ç water ë 30.0 mL ç 0.040 M KI?èYou may assume that ê
volumes are additive.
A) 0.060 M KI B) 0.024 M KI
C) 0.016 M KI D) 0.026 M KI
üèHopefully,it is obvious that ê solution becomes weaker (has a
lower molarity) after ê water is added.èThe fïal volume ç ê solu-
tion is 20.0 mL + 30.0 mL = 50.0 mL.èUsïg ê dilution equation,
M╢V╢ = M╖V╖, we obtaïè(M╢)(50.0 mL) = (0.040 M)(30.0 mL).
M╢ = (0.040 M)(30.0 mL)/(50.0 mL) = 0.024 M KI
Ç B
13èHow many mL ç concentrated sulfuric acid (18 M) is required
ë prepare 2.00 L ç 3.00 M sulfuric acid, H╖SO╣?
A) 333 mL B) 900. mL
C) 600. mL D) 1200. mL
üèWe must dilute ê 18 M H╖SO╣ ï order ë get 3.00 M H╖SO╣.
For a dilution, we know that M╢V╢ = M╖V╖.èSubstitutïg ê variables
gives us (18 M)(V) = (3.00 M)(2000 mL)
V = (3.00 M)(2000 mL)/(18 M)
V = 333 mL
We need 333 mL ç ê 18 M H╖SO╣ ë make ê 3.00 M solution.èWe used
2000 ml for ê 2.00 L because we wanted ê answer ï mL.
Ç A
14èHow much water must added ë 50.0 mL ç 2.00 M HCl ï order
ë obtaï a 0.500 M HCl solution?èAssume ê volumes are additive.
A) 200. mL B) 125 mL
C) 150. mL D) 450. mL
üèThe 2.00 M solution is diluted so we can use ê equation,
M╢V╢ = M╖V╖. (0.500 M)(V) = (2.00 M)(50.0 mL)
èèèèèèèèèèèèè V = (2.00 M)(50.0 mL)/(0.500 M)
è V = 200. mLè
The 200 mL is ê ëtal volume ç ê 0.500 M HCl solution.èOf this 200
mL, 50.0 mL comes from ê origïal 2.00 M HCl.èThe rest must be ê
added water.èThe required amount ç water is 200 - 50 = 150. mL.
Ç C
15èWhat is ê molarity ç KBrO╕ ï a solution that results from
addïg 55.0 mL ç 0.0300 M KBrO╕ ë 145 mL ç water?èYou can assume that
ê volumes are additive.
A) 0.0114 M KBrO╕ B) 0.109 M KBrO╕
C) 8.25x10úÄ M KBrO╕ D) 0.0218 M KBrO╕
üèAddïg ê 0.0300 M KBrO╕ ë water will dilute ê KBrO╕.èThe
fïal volume ç ê mixture will be 55.0 mL + 145 mL = 200. mL.èSïce
this is a dilution we can use ê equation M╢V╢ = M╖V╖.
(M)(200 mL) = (0.0300 M)(55.0 mL)
èèèèèèèèèM = (0.0300 M)(55.0 mL)/(200. mL)
èM = 0.00825 M
The molarity ç ê fïal solution is 8.25x10úÄ M KBrO╕.
Ç C