g.com/ibiscms/modi bis/view.php?id 3224084 Saplinglea Nettlx f Facebook Secure c

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g.com/ibiscms/modi bis/view.php?id 3224084 Saplinglea Nettlx f Facebook Secure coud Access D Tower af God scanl N Pastsecret Acompleto guide to N Malta o dag breeder Kids Rides & Rdes to Account summary by Sapling Learning 5apling Learning I Indlana State University-Chem 32 7-WOLF Act huities and Due Dates I HV12 My Assignment Resources U s/5/2017 09:00AM A 26/100 4/25/2017 0702 PM Gradebook O ent Information Assig Print Calculatar Paradi Table Available From 00 AM 4 of 16 5/5/2017 09:00 AM Points possibl Map GA Sapling Learning Grade Category: Default An unknown amount of a compound with a molecular mass of 281.64 gmol is dissolved in a 10-mL Description flask. A aliquot of solution is transferred to a 25-mL volumetric fask and enough water is added to dilute to the mark. The absorbance of this diluted solution at 333 nm 0.439 in 1.000 Policies: cm cuvette. The molar absorptivity for this compound at 333 nm is ea3 *6387 M cm 1 (a) What is the concentration of the compound in the cuvette? any g Number eep trying to you get it right or give up. You lose 10% of the points available to (b) What is the concentration of the oompound in the 10-mL flask? Number eTextbook O Help with This Top (c) How many milligrams of compound were used to make the 10-mL solution? o Web Help & Videos OTechnical Support and Bug Reports nng AO 8 Gi Up & view Solution 2 Check Answer Next JExit Previous Hint 2017 Sapling Learning, about us careers privacy polcy tems of use contact us help O Logout Help

Explanation / Answer

A = e*l*M

a)

in cuvette:

0.439 = 6387*1*M

M = 0.439 /6387 = 0.00006873 M

b)

in the 10 mL flask:

C1V1 = C2V2

0.00006873*25 = C2*1

C2 = 0.0017182

then

C = 0.0017182 M in the 10 mL as well

c)

mg in V = 10 mL

mol = MV = (0.0017182)(10) = 0.017182 mmol

MW = 281.64 mg/mmol

mass = mol*MW = 0.017182*281.64  = 4.8391 mg

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