Multimedia Chemistry 1 & 2/Multimedia Chemistry I & II (1996-9-11) [English].img/chem/chapter8.1c

Text File | 1996-07-26 | 10KB | 232 lines

à 8.1cèPercentage Concentrations
äèPlease fïd ê percentage concentration ç or amount ç solute for ê followïg
solutions.
âèHow many grams ç MgSO╣ are required ë prepare 500. mL ç
10% (w/w) MgSO╣, which has a density ç 1.1034 g/mL?èThe volume times
ê density yields ê mass ç ê required solution.èMultiplyïg by
percentage gives ê required mass.
            è1.1034 g soln.è10 g MgSO╣
?g MgSO╣ = 500 mL soln. x ────────────── x ─────────── = 55.2 g MgSO╣
            èè1 mL soln.èè100 g soln.
éSèA solution is a homogeneous mixture ç several possible combïa-
tions ç solids, liquids, å gases.èIn general chemistry at ê high
school å ê first year or two ç college level chemistry, we normally
are concerned with solids or liquids dissolved ï oêr liquids.èIn par-
ticular, we are ïterested ï aqueous (water) solutions.èThe component
ç ê solution that makes up ê majority ç ê solution is called ê
solvent.èWater is ê solvent ï aqueous solutions.èThe mïor component
ç ê solution is called ê solute.èA solution can contaï several
solutes.

When usïg solutions, we need ë know ê strength or concentration ç
ê solute(s) ï ê solution.èThere are various ways ç specifyïg ê
strength.èIn this section, we will look at percentage by mass or
percentage by volume ë ïdicate ê strength.èA German white wïe might
be 7.5% alcohol by volume.èOne hundred milliliters (100 mL) ç ê wïe
contaïs 7.5 mL ç alcohol.èThe hydrogen peroxide solution sold ï a
drugsëre is 3% hydrogen peroxide.èThis solution is 3% hydrogen peroxide
by mass, which means that 100 g ç ê solution contaïs 3 grams ç
hydrogen peroxide.èThe percentage by mass is ïdicated by "% (w/w)".
The defïition is
èèèèèèèèègrams ç solute
% solute (w/w) = ───────────────── x 100.èA 50% (w/w) solution ç sodium
èèèèèèèè grams ç solution
hydroxide contaïs 50 grams ç sodium hydroxide ï 100 grams ç ê
sodium hydroxide solution.

There are two ways ë prepare a solution.èWe can weigh ê appropriate
amount ç solute å ê amount ç ê solvent.èWe can start with a more
concentrated solution å add more solvent ë obtaï a weaker solution.
The second procedure is called a dilution.èHow many grams ç sodium
hydroxide are needed ë prepare 500 g ç a 2% sodium hydroxide solution?
The 2% means that 100 g ç solution contaïs 2 g ç NaOH or movïg ê
decimal two places ë ê left we could say 0.02 g NaOH/g solution.èThe
required mass ç ê sodium hydroxide is
èèè(500 g solution)(0.02 g NaOH/g solution) = 10 g NaOH.
This also means that we would need 490 g ç water ë prepare ê solution.

We could also prepare ê 2% solution from ê sëck 50% solution by add-
ïg water.èWe must dilute ê sëck solution.èThe mass ç ê sodium
hydroxide is constant ï ê dilution.èAs ï ê previous example, ê
mass ç ê solution times ê percentage equals ê mass ç ê solute.
The equation for a dilution is
èèèèèèèèèè┌è massèè ┐èèèèèèèèèèè┌è massèè ┐
 [% concentration 1]│ solution 1 │ = [% concentration 2]│ solution 2 │
         èè└èèèèèè┘èèèèèèèèèèè└èèèèèè┘
Substitutïg ïë this equation, we fïd
        è(2% NaOH)(500 g) = (50% NaOH)(X grams)

         (2%)(500 g)/(50%) = X grams

            èèè20 g = X grams
This shows that we would need 20 grams ç ê 50% NaOH solution å also
480 g ç water (500 - 20).

Problems ïvolvïg % by volume are treated ï ê same manner with ê
followïg difference.èYou would use ê ëtal volume ç ê solution
ïstead ç ê mass ç ê solution.èIf ê problem mixes ê volume å
mass, ên we must know ê density ç ê solution or ê solute depend-
ïg on ê wordïg ç ê question.è For example, how many grams ç pure
acetic acid are needed ë make 750. mL ç a 15.0% (w/w) acetic acid solu-
tion.èThe density ç 15.0% acetic acid is 1.0195 g/mL.èThe poït ë
notice ï this problem is that you are given ê density ç ê solution,
not acetic acid.èThe mass ç ê solution is ê volume times ê dens-
ity.èThe mass ç acetic acid is ê percentage ç acetic acid ï decimal
form times ê mass ç ê solution.

            è1.0195 g solnè 0.150 g acetic acid
?g acetic acid = 750 mL x ───────────── x ───────────────────
            è1 mL solnèèèè1 g soln

?g acetic acid = 115 g acetic acid
We would mix 115 g ç acetic acid with enough water ë make 750 mL ç
solution.
 1èWhat is ê % by mass ç NaCl ï a solution ç 42 g NaCl å
228 g water?

    A) 16% NaCl        B) 11% NaCl

    C) 23% NaCl        D) 35% NaCl
ü The percent by mass ç NaCl ï ê solution is ê mass ç sodium
chlorideèdivided by ê ëtal mass ç ê solution times 100 ë convert
ë percent.
        èèèè42 g NaCl
    ?% NaCl = ───────────────────── x 100 = 16%èNaCl
        è42 g NaCl + 228 g H╖O
Ç A
 2èWhat is ê % by mass ç ZnCl╖ ï a solution contaïïg 35 g
ZnCl╖ å 140. g H╖O?

    A) 4.0% ZnCl╖        B) 5.0% ZnCl╖

    C) 20.% ZnCl╖        D) 25% ZnCl╖
ü The percent by mass ç ZnCl╖ ï ê solution is ê mass ç zïc
chlorideèdivided by ê ëtal mass ç ê solution times 100 ë convert
ë percent.
        èèèè35 g ZnCl╖
    ?% ZnCl╖ = ───────────────────── x 100 = 20.%èZnCl╖
        è35 g ZnCl╖ + 140 g H╖O
Ç C
 3èHow many grams ç concentrated HCl, 37% (w/w) HCl, are needed
ë prepare 500. grams ç a 10% HCl solution?

    A) 1850 g        B) 106 g

    C) 245 g        D) 135 g
üèWe can tell that this is a dilution, because ê sëck solution
is more concentrated (37%) than ê fïal solution (10%).èWe use ê
equation
èèèèèèèèèè┌è massèè ┐èèèèèèèèèèè┌è massèè ┐
 [% concentration 1]│ solution 1 │ = [% concentration 2]│ solution 2 │
         èè└èèèèèè┘èèèèèèèèèèè└èèèèèè┘
 (37% HCl)(X grams)= (10% HCl)(500. g)

    è X grams = 10x500/37 = 135 g ç concentrated HCl are required.
Ç D
 4èHow many grams ç concentrated HCl, 37% (w/w) HCl, do you need
ë prepare 500. mL ç 20.% (w/w) HCl?
The density ç 20% HCl is 1.0980 g/mL

    A) 356 g        B) 1016 g

    C) 193 g        D) 297 g
üèTo prepare this solution we must dilute ê sëck 37% solution.
We need ë know ê mass ç ê 20% solution ë use with ê mass per-
cent.èThe mass ç ê fïal solution is (500 mL)(1.0980 g/mL) or
549 g ç solution.èNow we can use ê equation for dilutions:
èèèèèèèèèè┌è massèè ┐èèèèèèèèèèè┌è massèè ┐
 [% concentration 1]│ solution 1 │ = [% concentration 2]│ solution 2 │
         èè└èèèèèè┘èèèèèèèèèèè└èèèèèè┘

 (37%)(X grams) = (20%)(549 g)

èèèèX grams = 20x549/37 = 297 g ç concentrated HCl are required ë
prepare ê 20% solution.
Ç D
 5èHow many grams ç sucrose (table sugar), C╢╖H╖╖O╢╢, 342 g/mol,
are required ë make 25.0 g ç a 30.0% (w/w) sucrose solution?èThe dens-
ity ç a 30.0% sucrose solution is 1.1270 g/mL.

    A) 103 g sucrose    B) 7.50 g sucrose

    C) 0.243 g sucrose    D) 4.10 g sucrose
üèA lot ç extraneous ïformation was provided ï this question.
The 30.0% (w/w) means 30.0 g ç sucrose ï 100 g ç solution.èWe are
makïg 25.0 g ç solution.èThe required mass ç sucrose is simply,
? g sucrose = (25.0 g soln)(0.300g sucrose/1 g soln) = 7.50 g sucrose.
Ç B
 6èHow many mL ç water are needed ë prepare 1.50 L ç a
70.% (v/v) isopropyl alcohol solution (rubbïg alcohol)?

    A) 1050 mL        B) 200 mL

    C) 450 mL        D) 500 mL
üèA 70% (v/v) isopropyl alcohol solution is 30% water by volume.
The ëtal volume ç ê desired solution is 1.50 L or 1500 mL.èThirty
percent ç ê 1500 mL will be water.
    ?mL H╖O = 1500 mL soln. x 0.30 mL H╖O/1 mL soln = 450. mL H╖O
Ç C
 7èHydrogen peroxide is available as a 50% (w/w) solution.èHow
many grams ç 50% H╖O╖ would you use ë make 400. mL ç 3% (w/w) H╖O╖?
The density ç 3% (w/w) H╖O╖ is 1.00 g/mL.

    A) 24 g        B) 267 g

    C) 12 g        D) 133 g
üèFirst you must determïe how many grams ç hydrogen peroxide you
need for ê 3% solution.è

è ?g H╖O╖ needed = (400 mL)(1.00 g/mL)(0.03 g H╖O╖/g soln.) = 12 g H╖O╖

Now you need ë fïd out how many grams ç ê 50% solution will supply
12 grams ç H╖O╖.è 50% (w/w) H╖O╖èmeans that are 50 g H╖O╖ ï 100 g ç
ê solution.èConversely, we could say that êre are 100 g ç solution
per 50 g H╖O╖.
        èèèèèèè 100 g 50% H╖O╖ soln.
è ? g 50% H╖O╖è= 12 g H╖O╖ x ──────────────────── = 24 g ç 50% H╖O╖
            èèèè50 g H╖O╖
Ç A
 8èHow many mL ç a 50% (v/v) glycerol solution would you use ë
prepare 350. mL ç 15% (v/v) glycerol solution?

    A) 81 mL         B) 105 mL

    C) 269 mL        D) 1167 mL
üèYou must dilute a 50% solution ï order ë make a 15% solution.
In this situation, ê volume ç ê glycerol is constant.èWe obtaï ê
volume by multiplyïg ê percentage by ê volume ç ê solution.
èèèèèèèèèè┌èvolumeèè┐èèèèèèèèèèè┌èvolumeèè┐
 [% concentration 1]│ solution 1 │ = [% concentration 2]│ solution 2 │
         èè└èèèèèè┘èèèèèèèèèèè└èèèèèè┘
(50%)(V mL) = (15%)(350 mL).èV = 15x350/50 = 105 mL ç ê 50% solution.
Ç B
 9èWhat is ê percentage by mass ç a Cu(NO╕)╖ solution when
300. g ç a 25% (w/w) Cu(NO╕)╖ is added ë 500. g ç water?

    A) 15% (w/w)        B) 3.1% (w/w)

    C) 9.4% (w/w)        D) 5.0% (w/w)
üèThe percentage by mass if ê mass ç ê Cu(NO╕)╖ ï 100 grams
ç ê solution.èYou can calculate ê mass ç Cu(NO╕)╖ ï ê 300. g
ç ê 25% solution by multiplication.
è?g Cu(NO╕)╖ = (300. g soln)(25 g Cu(NO╕)╖/100 g soln) = 75 g Cu(NO╕)╖
The ëtal mass ç ê fïal solution is 300 g + 500 g = 800 g.
The % by mass equals (75 g Cu(NO╕)╖/800. g) x 100 = 9.4% (w/w)
Ç C
 10èAn isoënic salïe solution for ïtravenous ïjection is
0.9% (w/w) NaCl.èHow many grams ç NaCl are required ë prepare 20.0 L
ç this solution?èThe density ç 0.9% NaCl is 1.0046 g/mL at 20°C.
(Assume that ê 0.9% has an ïfïite number ç significant figures.)

èA) 181 g NaClèè B) 105 g NaClèè C) 30.9 g NaClèè D) 64.9 g NaCl
üèUsïg ê density ç ê solution, we can fïd ê mass ç 20.0
liters ç ê solution.èFrom ê solution mass we can determïe ê
required mass ç sodium chloride.
        èèèè 1000 mlè 1.0046 gè 0.9 g NaCl
?g NaCl = 20.0 L soln. x ─────── x ──────── x ────────── = 181 g NaCl
        èèèèè1 Lèèè 1 mlèè 100 g soln.
Ç A