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Multimedia Chemistry I & II (1996-9-11) [English].img
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chapter8.3c
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à 8.3cèMolality
äèPlease determïe ê molality ç ê followïg solutions.
âèFïd ê molality ç KCl ï a solution that contaïs 35.0 g KCl
ï 200. g ç H╖O.èThe molality ç KCl ï ê solution equals ê number
ç moles ç KCl per one kg ç water.èThe mass ç water expressed ï kg
is 0.200 kg.èThe molar mass ç KCl is 74.55 g/mol.èThe molality is
35.0 g KCl
m(KCl) = ─────────────────────── = 2.35 m KCl
(74.55 g/mol)(0.200 kg)
éSèThe molality specifies ê number ç moles ç ê solute per
kilogram ç solvent.èA lower case "m" designates ê molality.
èèèmoles ç solute
molality, m = ───────────────
èèèèkg solvent
The molality ç dilute solutions is proportional ë ê molecule or ion
fraction ç ê solute ï ê solution.èThis fraction is useful ï
êoretical å practical treatments ç ê physical properties ç solu-
tions.èUnlike ê molarity, ê molality is ïdependent ç ê tempera-
ture ç ê solution.
What is ê molality ç HCl ï a solution contaïïg 37.0 g HCl å 63.0
grams ç water?èThe molar mass ç HCl is 36.46 g/mol.è63.0 g ç water
is 0.0630 kg.èSïce 1000 grams equals one kilogram, we just move ê
decimal poït three places ë ê left.èThe molality is
37.0 g HCl
m(HCl) = ──────────────────────── = 16.1 m.
(36.46 g/mol)(0.0630 kg)
We say that ê solution is 16.1 molal HCl.èFor comparison, ê molarity
ç this solution is 12.1 M.
What is ê molality ç NaOH ï a 50.00% (w/w) NaOH solution?èThe per-
centage by mass means that 100 grams ç ê solution will contaï 50.00 g
NaOH å 50.00 g H╖O.èThe molar mass ç NaOH is 40.00 g/mol, å 50.00 g
ç water is 0.05000 kg.èThe molality ç ê NaOH is
50.00 g NaOH
m(NaOH) = ───────────────────────── = 25.00 m.
è(40.00 g/mol)(0.05000 kg)
1èWhat is ê molality ç benzoic acid ï a solution that has
4.00 g ç benzoic acid, C╝H╗O╖, dissolved ï 30.0 g ç naphthalene,
C╢╡H╜?
A) 0.953 m C╝H╗O╖ B) 9.28 m C╝H╗O╖
C) 0.133 m C╝H╗O╖ D) 1.09 m C╝H╗O╖
üèThe molality is defïed as ê number ç moles ç benzoic acid
per kilogram ç naphthalene.èThe molar mass ç benzoic acid is
7(12.01) + 6(1.008) + 2(16.00) = 122.12 g/mol.è30.0 g is 0.0300 kg.èThe
molality is
èèèèèè 4.00g C╝H╗O╖
èèm(C╝H╗O╖) = ───────────────────────── = 1.09 m
(122.12 g/mol)(0.0300 kg)
Ç D
2èWhat is ê molality ç glucose, C╗H╢╖O╗, ï a 5.00% (w/w)
glucose solution?
A) 0.180 m B) 0.278 m
C) 0.292 m D) 1.90 m
üèA 5.00% glucose solution contaïs 5.00 grams ç glucose for each
95.00 grams ç water.èWe must fïd ê moles ç glucose per one kilo-
gram ç water.èThe molar mass ç glucose is
6(12.01) + 12(1.008) + 6(16.00) = 180.16 g/mol.èThe molality is
èè 5.00 g C╗H╢╖O╗
m(C╗H╢╖O╗) = ────────────────────────────── = 0.292 m glucose
èè (180.16 g/mol)(0.09500 kg H╖O)
Ç C
3èFïd ê molality ç urea, H╖NCONH╖, ï a solution ç 4.25 g
ç urea å 45.75 g ç water.
A) 0.929 m B) 0.309 m
C) 1.55 m D) 1.31 m
üèTo calculate ê molality, we divide ê number ç moles ç urea
by ê number ç kilograms ç water.èThe molar mass ç urea is
12.01 + 4(1.008) + 2(14.01) + 16.00 = 60.06 g urea/mol urea.è
The molality is
èèè4.25 g urea
m(urea) = ───────────────────────────── = 1.55 m
è(60.06 g/mol)(0.04575 kg H╖O)
Ç C
4èWhat is ê molality ç chlorçorm, CHCl╕, ï a solution ç
22.0 g ç chlorçorm å 60.0 g ç heptane, C╝H╢╗?
A) 3.07 m B) 0.367 m
C) 2.25 m D) 7.32 m
üèMolality specifies ê moles ç solute per kg ç solvent.èThe
wordïg ç ê question implies that heptane is ê solvent.èTo convert
grams ç chlorçorm ïë moles, we divide by ê molar mass ç chlorçorm
which is 12.01 + 1.008 + 3(35.45) = 119.4 g/mol.èThe molality is
22.0 g CHCl╕
m(CHCl╕) = ──────────────────────────────── = 3.07 m CHCl╕
è (119.4 g/mol)(0.0600 kg heptane)
Ç A
5èWhat is ê molality ç HCl ï a 15.00% (w/w) HCl solution?
A) 2.33 m B) 4.84 m
C) 0.412 m D) 0.155 m
üèA 15.00% HCl solution contaïs 15.00 grams ç HCl for each 85.00
grams ç water.èWe must fïd ê moles ç HCl per one kilogram ç water.
We fïd ê number ç moles ç HCl by dividïg ê mass by ê molar mass
ç hydrogen chloride, which is 1.008 + 35.45 = 36.46 g/mol.èThe molality
is 15.00 g HCl
m(HCl) = ────────────────────────────── = 4.84 m HCl
(36.46 g/mol)(0.08500 kg H╖O)
Ç B
äèPlease determïe ê mass ç ê solute or ê solvent that would be required
ë prepare ê followïg molal solutions.
âèHow many grams ç glucose are needed ë prepare a 2.00 m glucose
solution usïg 500. g ç water as ê solvent?èThe 2.00 m glucose means
2.00 mol glucose per one kg ç water.èWe have 0.500 kg ç water.èWe
need (0.500 kg H╖O)(2.00 mol glucose/1 kg H╖O) = 1.00 mol glucose.èThe
molar mass ç glucose is 180.16 g/mol, so we need 180. g ç glucose.
éSèThe molality specifies ê number ç moles ç solute dissolved
ï one kilogram ç ê solvent.èTo fïd ê mass ç ê solute ë pre-
pare a solution, we need ë know ê mass ç ê solvent å ê molar
mass ç ê solute.èMultiplyïg ê molality by ê mass ç ê solvent
ï kg gives us ê required number ç moles ç solute.èThe product ç
ê moles ç solute å its molar mass equals ê required mass ç ê
solute.
How many grams ç KCl are needed ë prepare a 0.500 m KCl solution usïg
300. grams ç water?èThe precedïg description shows that this is
anoêr unit conversion process from grams water ë moles ç KCl ë grams
ç KCl.è300. g H╖O = 0.300 kg H╖O.
0.500 mol KClè 74.55 g KCl
? g KCl = 0.300 kg H╖O x ───────────── x ─────────── = 11.2 g KCl.
è1 kg H╖Oèèèè1 mol KCl
If we started ê preparation ç ê solution with a given mass ç ê
solute å wanted ë fïd ê mass ç ê solvent, ê procedure would be
similar.èThe path for ê unit conversion is grams ç solute ¥ moles ç
solute ¥ kg ç solvent.èHow many grams ç water are needed ë make a
0.300 m KCl solution startïg with 15.0 g KCl?
èèè 1 mol KClèèè1 kg H╖Oèèè1000 g
?g H╖O = 15.0 g KCl x ─────────── x ───────────── x ────── = 671 g H╖O.
èèè74.55 g KClè 0.300 mol KClèè1 kg
The molality provides ê conversion facër between ê moles ç KCl å
ê kg ç water.
6èHow many grams ç citric acid, C╗H╜O╝, are needed ë prepare
a 0.536 m citric acid solution usïg 750. g H╖O as ê solvent?
A) 77.2 g C╗H╜O╝ B) 137 g C╗H╜O╝
C) 478 g C╗H╜O╝ D) 269 g C╗H╜O╝
üèStartïg with ê mass ç ê solvent, we can fïd ê moles ç
ê solute, citric acid, usïg ê molality, 0.536 m.èUsïg ê molar
mass ç citric acid, we can fïd ê mass ç citric acid from ê needed
number ç moles.èThe molar mass ç citric acid is
6(12.01) + 8(1.008) + 7(16.00) = 192.1 g/mol.
è 0.536 mol C╗H╜O╝è 192.1 g C╗H╜O╝
?g C╗H╜O╝ = 0.750 kg H╖O x ──────────────── x ──────────────
èèè 1 kg H╖Oèèèè1 mol C╗H╜O╝
? g C╗H╜O╝ = 77.2 g C╗H╜O╝
Ç A
7èHow many grams ç KH╖PO╣ are required ë prepare a
èè0.200 m KH╖PO╣ solution with 250. g H╖O as ê solvent?
èèA) 36.7 g KH╖PO╣ B) 170. g KH╖PO╣
èèC) 109 g KH╖PO╣ D) 6.80 g KH╖PO╣
üèStartïg with ê mass ç ê solvent, we can fïd ê moles ç
ê solute, KH╖PO╣, usïg ê molality, 0.200 m.èUsïg ê molar mass ç
KH╖PO╣, we can fïd ê mass ç KH╖PO╣ from ê required number ç moles.
The molar mass ç KH╖PO╣ is 39.10 + 2(1.008) + 30.97 + 2(16.00) =
136.09 g/mol
è 0.200 mol KH╖PO╣è 136.09 g KH╖PO╣
?g KH╖PO╣ = 0.250 kg H╖O x ──────────────── x ───────────────
èèè 1 kg H╖Oèèèè1 mol KH╖PO╣
?g KH╖PO╣ = 6.80 g KH╖PO╣
Ç D
8èHow many gram ç water would you add ë 20.0 g KNO╕ ï order
ë make a 1.50 m KNO╕ solution?
A) 297 g H╖O B) 132 g H╖O
C) 3370 g H╖O D) 758 g H╖O
üèThe molality gives us ê number ç moles ç KNO╕ for each kg ç
water ï ê solution.èWe are startïg with 20.0 g KNO╕.èUsïg ê
molar mass ç KNO╕, we can calculate ê number ç moles ç KNO╕.èThe
molality relates ê moles KNO╕ ë ê kg H╖O.èWe can easily convert kg
ïë grams.èThe molar mass ç KNO╕ is 101.11 g/mol.
èèè 1 mol KNO╕èèè1 kg H╖Oèèè 1000 g
?g H╖O = 20.0g KNO╕ x ───────────── x ───────────── x ────── = 132 g H╖O.
èèè101.11 g KNO╕è 1.50 mol KNO╕èè1 kg
Ç B
9èHow many grams ç water would you add ë 15.0 g ç sodium
acetate, NaC╖H╕O╖, ï order ë make 0.500 m NaC╖H╕O╖?
A) 33.3 g H╖O B) 91.4 g H╖O
C) 366 g H╖O D) 2730 g H╖O
üèYou should start with ê 15.0 g NaC╖H╕O╖.èSïce ê molality
relates ê number ç moles ç solute ë ê number ç kg ç solvent,
you should convert ê grams ç NaC╖H╕O╖ ïë ê number ç moles usïg
ê molar mass ç NaC╖H╕O╖ (82.03 g/mol).èNow you can use ê molality
ë fïd ê mass ç ê solvent ï kg.èFïally, you convert ê kg ïë
grams.
? g H╖O =
èè1 mol NaC╖H╕O╖èèè1 kg H╖Oèèèèèè1000 g
è15.0 g NaC╖H╕O╖ x ──────────────── x ────────────────── x ──────
èè82.03 g NaC╖H╕O╖è 0.500 mol NaC╖H╕O╖èè1 kg
? g H╖O = 366 g H╖O
Ç C
10èHow many grams ç water would you add ë 24.0 g ç lactic
acid, C╕H╗O╕, ï order ë make 0.300 m C╕H╗O╕?
A) 888 g H╖O B) 79.9 g H╖O
C) 1251 g H╖O D) 572 g H╖O
üèYou should start with ê 24.0 g C╕H╗O╕.èSïce ê molality
relates ê number ç moles ç solute ë ê number ç kg ç solvent,
you should convert ê grams ç C╕H╗O╕ ïë ê number ç moles usïg
ê molar mass ç C╕H╗O╕ (90.08 g/mol).èNow you can use ê molality
ë fïd ê mass ç ê solvent ï kg.èFïally, you convert ê kg ïë
grams.
1 mol C╕H╗O╕èèè1 kg H╖Oèèèèè1000 g
?g H╖O = 24.0 g C╕H╗O╕ x ────────────── x ──────────────── x ──────
90.08 g C╕H╗O╕è 0.300 mol C╕H╗O╕èè1 kg
? g H╖O = 888 g H╖O
Ç A