Multimedia Chemistry 1 & 2/Multimedia Chemistry I & II (1996-9-11) [English].img/chem/chapter8.5c

Text File | 1996-07-26 | 17KB | 386 lines

à 8.5cèTitrations
äèPlease determïe ê volume ç solution that is needed ï ê followïg titrations.
âèHow many mL ç 0.150 M NaOH are required ë titrate 1.096 g
KHC╜H╣O╣ (204.22 g/mol)?èThe reaction is NaOH + KHC╜H╣O╣ ─¥ KNaC╜H╣O╣ +
H╖O.èOne mole ç NaOH neutralizes one mole ç KHC╜H╣O╣.èThe path ç ê
conversion is: g KHC╜H╣O╣ ¥ mol KHC╜H╣O╣ ¥ mol NaOH ¥ mL NaOH solution.
èèèèèèèèèè 1 molèèè1 mol NaOHèèèèè1000 mL
?mL NaOH = 1.096 g x ──────── x ────────────── x ────────────── = 35.8 mL
èèèèèèèèèè 204.22 gè 1 mol KHC╜H╣O╣è 0.150 mol NaOH
éSèA titration is a procedure whereby we can measure ê volume ç
one reactant that is required ë react with a known amount ç anoêr
reactant.èThe oêr reactant usually would be anoêr solution or a
solid.èWe use a buret ë precisely measure ê volume ç at least one
reactant, which ên is known as ê titrant.èBurets come ï different
sizes, but we usually use a buret that holds 50.00 mL ï high school å
college chemistry courses.èA 50 mL buret is graduated ï tenths ç a
milliliter, å we can estimate volumes ë ±0.02 mL.

A problem ïvolvïg a titration differs from ê oêr sëichiometry
problems only ï that we use ê molarity ç ê solution ï ê calcula-
tions.èAs with oêr sëichiometric problems, we must have a balanced
chemical equation ï order ë relate ê number ç moles ç one substance
ë ê number ç moles ç anoêr substance.

How many mL ç 0.0200 M KMnO╣ are required ë react with 0.155 g Na╖C╖O╣?
The balanced reaction is

2KMnO╣ + 5Na╖C╖O╣ + 8H╖SO╣ ─¥ 2MnSO╣ + K╖SO╣ + 5Na╖SO╣ + 10CO╖(g) + 8H╖O.

This equation tells us that 5 mol ç Na╖C╖O╣ react with 2 mol ç KMnO╣.
The molarity ç KMnO╣ states that êre are 0.0200 mol KMnO╣ ï one liter
ç solution.èEssentially, we want ë convert from grams ç Na╖C╖O╣ ë
ê volume (mL) ç KMnO╣.èThe molar mass ç Na╖C╖O╣ is 134.00 g/mol.

? mL KMnO╣ =
        èè 1 mol Na╖C╖O╣èè 2 mol KMnO╣èèèè1000 mL
èè0.155g Na╖C╖O╣ x ─────────────── x ───────────── x ────────────────
        èè 134.00g Na╖C╖O╣è 5 mol Na╖C╖O╣è 0.0200 mol KMnO╣

? mL KMnO╣ = 23.1 mL KMnO╣

The first step ï ê calculation is ê conversion ç grams ç Na╖C╖O╣
ë moles ç Na╖C╖O╣ by dividïg by its molar mass.èThe second step is
ê conversion from moles ç Na╖C╖O╣ ë moles ç KMnO╣ usïg ê sëichi-
ometry ç ê balanced equation.èThe third step is ê conversion ç
moles ç KMnO╣ ë ê mL ç ê solution usïg ê molarity.èSïce you
know ê number ç moles ç KMnO╣, you ïvert ê molarity ë obtaï ê
number ç mL ç solution per mole ç KMnO╣ (1 L per 0.0200 moleèor
1000 mL per 0.0200 mol).

The first two steps are ê same as ê steps ï ê sëichiometric prob-
lems ï Section 7.1.èThe difference comes ï ê last step where you use
ê molarity ç ê titrant, which was KMnO╣ ï this problem.
 1èHow many mL ç 6.00 M HCl would you need ë react 0.180 g Mg?
The reaction isè 2HCl(aq) + Mg(s) ──¥ MgCl╖(aq) + H╖(g).

    A) 1.23 mL        B) 2.47 mL

    C) 60.0 mL        D) 21.6 mL
üèAccordïg ë ê balanced equation 2 mol HCl react with one mol
Mg.èThe aëmic mass ç Mg is 24.31 g/mol.èThe pathway is
èg Mg ─¥ mol Mg ─¥ mol HCl ─¥ mL HCl solution.
            1 mol Mgèè 2 mol HClè 1000 mL
? mL HCl = 0.180 g Mg x ────────── x ───────── x ──────────── = 2.47 mL
            24.31 g Mgè 1 mol Mgèè6.00 mol HCl
Ç B
 2èHow many mL ç 6.00 M HCl are needed ë neutralize 15.0 g CaO?
The reaction isè2HCl(aq) + CaO(s) ──¥ CaCl╖(aq) + H╖O.

    A) 39.1 mL        B) 44.6 mL

    C) 19.5 mL        D) 89.2 mL
üèOne mole ç CaO reacts with 2 moles ç HCl.èDividïg ê number
ç grams ç CaO by its molar mass (56.08 g/mol) gives ê moles ç CaO
present.èWe need twice as many moles ç HCl.èThe molarity gives ê
moles ï one liter.èInvertïg ê molarity gives ê volume per mole.
1 L = 1000 mL.
             1 mol CaOèè2 mol HClè 1000 mL
? mL HCl = 15.0 g CaO x ─────────── x ───────── x ──────────── = 89.2 mL
            56.08 g CaOè 1 mol CaOè 6.00 mol HCl
Ç D
 3èHow many mL ç 0.1108 M K╖Cr╖O╝ are required for ê complete
oxidation ç 4.024 g ç FeSO╣?èThe balanced equation is
èK╖Cr╖O╝ + 6FeSO╣ + 7H╖SO╣ ──¥ Cr╖(SO╣)╕ + K╖SO╣ + 3Fe╖(SO╣)╕ + 7H╖O.

    A) 239.1 mL        B) 17.62 mL

    C) 46.75 mL        D) 39.84 mL
üèThe balanced equation shows that 6 moles ç FeSO╣ react with one
mole ç K╖Cr╖O╝.èThe molar mass ç FeSO╣ is 151.92 g/mol.èYou should
fïd ê number ç moles ç FeSO╣ by dividïg 4.024 by 151.92.èOne-sixth
ç that number ç moles ç FeSO╣ equals ê moles ç K╖Cr╖O╝ required.è
Dividïg by ê molarity yields ê required volume.

? mL K╖Cr╖O╝ =
èèèèèèèèè1 mol FeSO╣èèè1 mol K╖Cr╖O╝è 1000 mL
è 4.024 g FeSO╣x ────────────── x ───────────── x ──────────────────
        è151.92 g FeSO╣è 1 mol FeSO╣èè 0.1108 mol K╖Cr╖O╝

? mL K╖Cr╖O╝ = 39.84 mL K╖Cr╖O╝.
Ç D
 4èHow many mL ç 0.1479 M NaOH are required ë neutralize
50.00 mL ç 0.2219 M H╖C╖O╣ (oxalic acid)?èThe neutralization reaction
isè2NaOH(aq) + H╖C╖O╣(aq) ──¥ Na╖C╖O╣(aq) + 2H╖O.

    A) 150.0 mL        B) 75.02 mL

    C) 33.33 mL        D) 66.65 mL
üèThe equation shows that one mole ç H╖C╖O╣ reacts with two moles
ç NaOH.èThe number ç moles ç H╖C╖O╣ is found by multiplyïg ê
molarity by ê volume.èM x V = moles solute.èIf ê volume is ï mL,
ên ê product will be millimoles ç solute (mmol).
    50.00 mL x 0.2219 M H╖C╖O╣ = 11.095 mmol H╖C╖O╣.
The required number ç mmoles ç NaOH is
    11.095 mmol H╖C╖O╣ x 2 mol NaOH/1 mol H╖C╖O╣ = 22.19 mmol NaOH.
Dividïg ê mmol NaOH by ê molarity produces ê required volume.
    (22.19 mmol NaOH)(1 L/0.1479 mol NaOH) = 150.0 mL NaOH

Notice that ê mol ï ê denomïaër cancels ê mol part ç mmol ï
ê numeraër leaves mL.èIn êse problems, if you start with mL your
answer will be ï mL, providïg you do not need ë convert any volumes
along ê way ë ê answer.
Ç A
 5èHow many mL ç 3.00 M H╖SO╣ are required ë react with 25.0 mL
ç 1.12 M ethylenediamïe, H╖NCH╖CH╖NH╖, accordïg ë ê reaction:
    2H╖SO╣ + H╖NCH╖CH╖NH╖ ──¥ óH╕NCH╖CH╖NH╕ó + 2HSO╣ú?

    A) 9.33 mL        B) 18.7 mL

    C) 67.0 mL        D) 33.5 mL
üèThe number ç millimoles ç H╖NCH╖CH╖NH╖ is
        (25.0 mL)(1.12 M) = 28.0 mmol H╖NCH╖CH╖NH╖.

One mmole ç H╖NCH╖CH╖NH╖ reacts with 2 mmole ç H╖SO╣ accordïg ë ê
balanced equation.èThe required number ç mmoles ç H╖SO╣ is
                 2 mmol H╖SO╣
    28.0 mmol H╖NCH╖CH╖NH╖ x ─────────────────── = 56.0 mmol H╖SO╣.
                 1 mmol H╖NCH╖CH╖NH╖

Invertïg ê molarity gives ê volume ç solution per mole ç H╖SO╣.
The required volume ç H╖SO╣ ï mL is
            èè 1 L
    56.0 mmol H╖SO╣ x ────────────── = 18.7 mL H╖SO╣.
            è3.00 mol H╖SO╣
Ç B
 6èHow many mL ç 0.01918 M KMnO╣ are required ë reduce 2.00 mL
ç 0.888 M H╖O╖?èThe reaction is
    2KMnO╣ + 5H╖O╖ + 3H╖SO╣ ──¥ 5O╖ + 2MnSO╣ + K╖SO╣ + 8H╖O.

    A) 23.15 mL        B) 17.28 mL

    C) 37.04 mL        D) 24.16 mL
üèThe balanced equation provides ê lïk between ê KMnO╣ å ê
H╖O╖.èTo use ê equation, we need ë know ê number ç moles or mmoles
ç H╖O╖.èThe number ç mmoles ç H╖O╖ is
        (2.00 mL)(0.888 M) = 1.776 mmol H╖O╖.

The equation shows that 5 mmole H╖O╖ react with 1 mmol ç KMnO╣.
The required number ç mmoles ç KMnO╣ is
            è2 mmol KMnO╣
    1.776 mmol H╖O╖ x ──────────── = 0.7104 mmol KMnO╣.
            è5 mmol H╖O╖

Invertïg ê molarity gives ê volume ç solution per mole ç KMnO╣.
The required volume ç KMnO╣ ï mL is
            èèèè 1 L
    0.7104 mmol KMnO╣ x ───────────────── = 37.04 mL KMnO╣.
            èè0.01918 mol KMnO╣
Ç C
äèPlease fïd ê molarity ç ê followïg solutions.
âèA 1.935 g sample ç KHC╜H╣O╣ required 44.86 mL ç a KOH solution
for neutralization.èWhat is ê molarity ç ê KOH?èThe reaction is
KHC╜H╣O╣ + KOH ──¥ K╖C╜H╣O╣ + H╖O.èThe molarity is ê number ç moles
ç KOH per liter.èThe moles ç KOH equals ê moles ç KHC╜H╣O╣.èThe
moles ç KHC╜H╣O╣ equals 1.935g/(204.22 g/mol) = 0.009475 mol.èThe vol-
ume ç KOH is 44.86 mL or 0.04486 L.èThe molarity ç KOH equals
(0.009475 mol KOH)/(0.04486 L) = 0.2112 M KOH.
éSèThe molarity ç a solution frequently is determïed by titration
ç or with ê solution.èThis procedure is called ståardization ç ê
solution.èA ståardization is performed raêr than simply weighïg ê
compound when it is difficult ë obtaï ê compound ï a very pure state.
Some compounds are very stable å can be prepared å maïtaïed with a
high degree ç purity.èThese compounds are called primary ståards.èWe
can easily weigh ê primary ståard å titrate it with ê solution
whose concentration we wish ë determïe.

A primary ståard for determïïg ê molarity ç bases like NaOH is
potassium hydrogen phthalate, KHC╜H╣O╣.èNaOH is difficult ë keep ï a
pure state because it absorbs water å carbon dioxide from ê air ë
form sodium hydrogen carbonate å sodium carbonate.èIt is easier ë
titrate a known amount ç potassium hydrogen phthalate with a sodium
hydroxide solution ï order ë determïe ê molarity ç ê NaOH solu-
tion.

What is ê molarity ç a NaOH solution when 42.16 mL ç ê NaOH solu-
tion are required ë neutralize 0.8704 g KHC╜H╣O╣?èThe neutralization
reaction is:èNaOH + KHC╜H╣O╣ ──¥ KNaC╜H╣O╣ + H╖O.èThe reaction shows
that one mole ç sodium hydroxide reacts with one mole ç potassium
hydrogen phthalate.èYou can calculate ê number ç moles ç KHC╜H╣O╣
usïg its molar mass (204.22 g/mol) å ên can fïd ê moles ç NaOH.
è                è 1 mol KHC╜H╣O╣èèè1 mol NaOH
è? mol NaOH = 0.8704 g KHC╜H╣O╣ x ───────────────── x ──────────────
                è 204.22 g KHC╜H╣O╣è 1 mol KHC╜H╣O╣

è? mol NaOH = 4.26207x10úÄ mol NaOH

The molarity is defïed ë be ê number ç moles ç NaOH per liter ç
solution.èFrom ê titration, we know that 42.16 mL ç NaOH contaïs
4.26207x10úÄ mol.è42.16 mL is 0.04216 L.èThe molarity is
        è 4.26207x10úÄ mol NaOH
    M(NaOH) = ─────────────────────── = 0.1011 M NaOH.
        è0.04216 L NaOH solution
 7èWhat is ê molarity ç a KMnO╣ solution if 0.3057 g K╖C╖O╣
requires 35.17 mL ç ê KMnO╣ for complete reaction?èThe reaction is
èè5K╖C╖O╣ + 2KMnO╣ + 8H╖SO╣ ──¥ 10CO╖ + 2MnSO╣ + 6K╖SO╣ + 8H╖O.

    A) 0.1308 M        B) 0.06001 M

    C) 0.02092 M        D) 0.05501 M
üèTo fïd ê molarity ç KMnO╣, you must know ê number ç moles
ç KMnO╣ per liter ç solution.èYou know that ê volume is 35.17 mL or
0.03517 L.èThe moles ç KMnO╣ can be found from ê moles ç K╖C╖O╣.
The molar mass ç K╖C╖O╣ is 166.22 g/mol.èê moles ç KMnO╣ is
                1 mol K╖C╖O╣     è2 mol KMnO╣
? mol KMnO╣ = 0.3057 g K╖C╖O╣ x ─────────────── x ────────────
                166.22 g K╖C╖O╣è 5 mol K╖C╖O╣

? mol KMnO╣ = 7.3565x10úÅ mol KMnO╣
        èè 7.3565x10úÅ mol KMnO╣
    ? M(KMnO╣) = ──────────────────────── = 0.02092 M KMnO╣
        èè 0.03517 L KMnO╣ solution
Ç C
 8èA 25.00 mL sample ç vïegar requires 41.2 mL ç 0.508 M NaOH
for neutralization.èWhat is ê molarity ç acetic acid, HC╖H╕O╖, ï ê
vïegar?è HC╖H╕O╖ + NaOH ──¥ NaC╖H╕O╖ + H╖O.

    A) 0.837 M        B) 0.308 M

    C) 0.121 M        D) 0.431 M
üèTo fïd ê molarity ç acetic acid, you must know ê number ç
moles ç acetic acid per liter ç solution.èYou know that ê volume is
25.00 mL or 0.02500 L.èThe moles or mmoles ç HC╖H╕O╖ can be found from
ê moles or mmoles ç NaOH.èIf we always use mL, it is not necessary ë
convert any volumes ïë liters.èThe number ç mmoles ç NaOH equals ê
volume times ê molarity.èThe balanced equation shows that one mmole ç
NaOH reacts with one mmole HC╖H╕O╖.
                èèèèè1 mol HC╖H╕O╖
?mmol HC╖H╕O╖ = (41.2 mL)(0.508 M NaOH) x ───────────── = 20.9296 mmol
                    è1 mol NaOH

        èèè 20.9296 mmolHC╖H╕O╖
    ? M(HC╖H╕O╖) = ───────────────────────── = 0.837 M HC╖H╕O╖.
        èèè 25.00 mL HC╖H╕O╖ solution
Ç A
 9èA 25.00 mL sample ç a H╖SO╣ solution required 44.78 ml ç
0.1124 M KOH for neutralization.èWhat is ê molarity ç ê H╖SO╣?
èèèè H╖SO╣ + 2KOH ──¥ K╖SO╣ + 2H╖O.

    A) 0.06275 M        B) 0.2013 M

    C) 0.1255 M        D) 0.1007 M
üèTo fïd ê molarity ç H╖SO╣, we must know ê number ç moles
ç sulfuric acid per liter ç solution.èWe know that ê volume is
25.00 mL.èThe mmoles ç H╖SO╣ can be found from mmoles ç KOH.èIf we
always use volumes ï mL, we will not need ë convert any volumes ïë
liters.èThe number ç mmoles ç KOH is ê volume times ê molarity.
The balanced equation shows that two mmole ç KOH reacts with one mmole
H╖SO╣.
                èèèèè1 mol H╖SO╣
? mmol H╖SO╣ = (44.78 mL)(0.1124 M KOH) x ─────────── = 2.516636 mmol
                    è2 mol KOH

        èè 2.516636 mmol H╖SO╣
    ? M(H╖SO╣) = ─────────────────────── = 0.1007 M H╖SO╣.
        èè 25.00 mL H╖SO╣ solution
Ç D
äèPlease calculate ê amount ç compound that reacted ï ê followïg titrations.
âèA 2.000 gram sample contaïïg Na╖CO╕ å NaBr requires 35.6 mL
ç 0.500 M HCl for completeèneutralization.èHow many grams ç Na╖CO╕
were ï ê sample?èNa╖CO╕ + 2HCl ─¥ 2NaCl + H╖O + CO╖.èThe mass is

                1 mol Na╖CO╕ 105.99 g Na╖CO╕
?g Na╖CO╕ = (0.0356 L)(0.500 M)x────────────x─────────────── = 0.943 g
èèèèèèèèèèèèèèèè 2 mol HClèè1 mol Na½CO╕
éSèTitrations frequently are used ë determïe ç ê amount ç a
compound or an element ï a sample.èThe amount ç vitamï C ï a vitamï
C tablet can be determïed via titration with iodïe.èThe tablet is dis-
solved ï water å a few drops ç a starch ïdicaër solution is added.
The iodïe solution is added ë ê vitamï C solution from a buret.
When allèç ê vitamï C has reacted, ê next drop ç iodïe solution
leads ë ê formation ç ê characteristic blue starch-iodïe complex.
Iodïe oxidizes vitamï C (ascorbic acid), C╗H╜O╗, ë dehydroascorbic
acid, C╗H╗O╗.èThe reaction is

        I╖ + C╗H╜O╗ ──¥ 2Iú + C╗H╗O╗ + 2Hó.

One-fourth ç a vitamï C tablet required 46.40 mL ç 0.0152 M I½ ë oxi-
dize ê vitamï C.èHow many grams ç vitamï C were ï ê tablet?

The balanced equation shows that one mole ç iodïe reacts with one mole
ç vitamï C.èYou can calculate ê number ç moles ç iodïe from ê
volume ç ê iodïe solution å its molarity.

    ? mol I½ = (0.04640 L)(0.0152 M) = 7.0528x10úÅ mol I╖.

The volume ï mL was converted ïë L.èThe moles ç vitamï C equals ê
moles ç I╖.èMultiplyïg ê moles by ê molar mass ç vitamï C yields
ê mass ç vitamï C ï a quarter ç ê tablet.èThe mass must be mul-
tiplied by 4 ë get ê mass ç vitamï C ï ê entire tablet.

?g vitamï C =
        èèè 1 mol vit. Cè 176.12 g vit. C
è7.0528x10úÅ mol I╖ x ──────────── x ─────────────── x 4 = 0.497 g
        èèè 1 mol I½    èèè 1 mol vit. Cèèèèè vitamï C

The general procedure is ë fïd ê number ç moles ç ê titrant by
takïg ê product ç ê volume å ê molarity.èNext, we use ê
balanced equation ë convert from ê moles ç ê titrant (ê substance
ï ê buret) ë ê moles ç ê oêr reactant.èFïally, we multiply
by ê molar mass ç ê oêr reactant ë obtaï ê mass ç ê reac-
tant.
 10èNiacï (nicotïic acid, a vitamï), HC╗H╣NO╖, can be titrated
with aqueous sodium hydroxide.èHow many grams ç niacï were ï a sample
that required 28.60 mL ç 0.0948 M NaOH for neutralization?
    èè HC╗H╣NO╖(aq) + NaOH(aq) ──¥ NaC╗H╣NO╖(aq) + H╖O.

    A) 0.108 g    B) 3.31 g    C) 0.334 g    D) 0.269 g
üèWe know ê volume å molarity ç ê NaOH, which permits us ë
fïd out how many moles ç NaOH reacted.èThe equation shows that 1 mole
ç NaOH reacts with one mole ç niacï.èThe moles ç NaOH equals ê
moles ç niacï.èFïally, multiplyïg by ê molar mass ç niacï gives
ê mass ç niacï that was neutralized by ê NaOH.

? mol NaOH = (0.02860 L)(0.0948 M) = 2.71128x10úÄ mol NaOH.

            1 mol HC╗H╣NO╖
2.71128x10úÄ mol NaOH x ────────────── = 2.71128x10úÄ mol HC╗H╣NO╖.
             1 mol NaOH

? g HC╗H╣NO╖ = (2.71128x10úÄ mol HC╗H╣NO╖)(123.11 g/mol) = 0.334 g
                            èèèHC╗H╣NO╖.
Ç C
 11èThe titanium ï an ore sample was converted ë TiÄó å
titrated with 36.14 mL ç 0.02114 M MnO╣ú.èHow many grams ç Ti were ï
ê sample?èThe reaction is
    5TiÄó + MnO╣ú + H╖O ──¥ 5TiOìó + Mnìó + 2Hó.

A) 0.03658 g Tièè B) 5.601 g Tièè C) 0.1829 g TièèD) 0.5849 g Ti
üèWe are given ê volume å molarity ç permanganate, MnO╣ú,
which allows us ë fïd out how many moles ç MnO╣ú reacted.èThe equa-
tion says ê 5 moles ç TiÄó react with one mole ç MnO╣ú.èThis allows
us ë convert from moles ç MnO╣ú ë moles ç TiÄó.èFïally, multiplyïg
by ê aëmic mass ç Ti gives ê mass ç Ti ï ê sample.

? mol MnO╣ú = (0.03614 L)(0.02114 M) = 7.640x10úÅ mol MnO╣ú.

                èè5 mol TiÄó
? mol TiÄó = 7.640x10úÅ mol MnO╣ú x ─────────── = 3.820x10úÄ mol TiÄó
                èè1 mol MnO╣ú

? g Ti = 3.820x10úÄ mol TiÄó x (47.88 g Ti/mol Ti) = 0.1829 g Ti
Ç C
 12èHow many grams ç iodïe, I╖, would be reduced by 31.76 mL ç
0.1106 M sodium thiosulfate, Na╖S╖O╕?èThe reaction is
        I╖ + 2Na╖S╖O╕ ──¥ 2NaI + Na╖S╣O╗.

    A) 0.4458 g I╖        B) 0.8838 g I╖

    C) 1.783 g I╖        D) 0.2777 g I╖
üèYou are given ê volume å molarity ç Na╖S╖O╕, which allows
you ë fïd out how many moles ç Na╖S╖O╕ reacted.èThe equation shows
that 1 mole ç I╖ reacts with two moles ç Na╖S╖O╕.èYou can convert from
moles ç Na╖S╖O╕ ë moles ç I╖.èFïally, multiplyïg by ê molar mass
ç I╖ gives ê mass ç I╖ that was reduced by ê Na╖S╖O╕.

? mol Na╖S╖O╕ = (0.03176 L)(0.1106 M) = 3.51266x10úÄ mol Na╖S╖O╕

                èèè1 mol I╖
? mol I╖ = 3.51266x10úÄ mol Na╖S╖O╕ x ───────────── = 1.75633x10úÄ mol I╖
                èèè2 mol Na╖S╖O╕

? g I╖ = (1.75633x10úÄ mol I╖)(253.8 g I╖/mol I╖) = 0.4458 g I╖.
Ç A