New User Special Price Expires in

Let's log you in.

Sign in with Facebook


Don't have a StudySoup account? Create one here!


Create a StudySoup account

Be part of our community, it's free to join!

Sign up with Facebook


Create your account
By creating an account you agree to StudySoup's terms and conditions and privacy policy

Already have a StudySoup account? Login here


by: Heaven Wehner


Heaven Wehner
GPA 3.98


Almost Ready


These notes were just uploaded, and will be ready to view shortly.

Purchase these notes here, or revisit this page.

Either way, we'll remind you when they're ready :)

Preview These Notes for FREE

Get a free preview of these Notes, just enter your email below.

Unlock Preview
Unlock Preview

Preview these materials now for free

Why put in your email? Get access to more of this material and other relevant free materials for your school

View Preview

About this Document

Class Notes
25 ?




Popular in Course

Popular in Food Science & Technology

This 10 page Class Notes was uploaded by Heaven Wehner on Friday September 18, 2015. The Class Notes belongs to FOS 4321 at University of Florida taught by Marshall in Fall. Since its upload, it has received 9 views. For similar materials see /class/206858/fos-4321-university-of-florida in Food Science & Technology at University of Florida.

Popular in Food Science & Technology




Report this Material


What is Karma?


Karma is the currency of StudySoup.

You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 09/18/15
Fos4327 00d Anaysys Pan 111 Molarin and Normality Mnlarity xm 1a solution The umtxs mnlL and can be abbrevxated 11th a capxtal letter i MoleMw Mole i M L L is ch98 072 for H250 1 mole 98 072 g 9707Zg 1M 1 L A Examgle For me followmg Hasoa concentmon calculate the Molanty 1 5 g m 50 mL 2 1 mgmL Caleulauons s 47 1130 1 male 1000 mL X 9a 072 g X 1 L 50 mL 1 02 M 1mg H250 1g 1mole 10mm 7 lmL X1000 mgX9807ZgX 1L quot101M An important equauon to remember w gram M moleL x Mw grammole x V L n r n Note Nexther eby 39 w of Umts SI Ithas advantages when can39ymg outmxauon ea1eu1auons Use ofnormahty xs dumuraged by NIST and Amenean Chemical Society FOS4327rFood Anaysls egu valentw ght n e anurnber ofcalculatlons lnvolve repomng tlne results of analyses uslng equrvalents a revlew may help to understand how and why equrvalents are used eonesponds to a reacting unit 3 g a rnole ofhydrogen lnydronyl eleetron unrvalent ron or V2 rnole of oxygen Anytlung equal to l rnole ofreactmg unrtrs l equrvalent sq cannot answer egulvalent welght offhand wnlnout knowmg the reaeuon Eguivalents in Snllltinn s de ned already N or norrnalrty refers to tlne number of equrvalent sq of a component m one lrter ofsolunon orrnrllrequrvalent rneq m l rnL ofsolunon eq 7 Meg NT 7 W Mass of compound sq Equzualerltwezght A Acidrb as e tra on The reaeung unrt for acldrbase trtratron ls tlne proton H39 1 HO Na0H a NaCl Hao lH exchanged 2 H250o 2Na0H a NazSOo 2H20 2 H exchanged One H exchanged in reactlnn 1 For HCl m tlus reaetron molarmluj A g 36465 Egurvalentwergn MW of HC effl 1m 1 eq 36 46 g For 1 L soluuon eontarnrng 3o 46 g ofHCl N1 eqlL l N For NaOH FOS4327rFood Anaysts 39997 f5 39997 g Egmvalentwexght i or leq 39997g sq For 2 L soluuon eontammg 39 997 g ofNaOH N1eq2L05N Twn H is exchanged in martian 217quot Haso in this reaeu39tm 98 D78 quot1011217111255 7 mg g b H exchanged t Z Egurva1ertt werght e m L of mwwmm o 1 sq 49 039 g 2 sq Exa les ea1eu1ate the normahty of H2501 solution 5 g m 50 m1 Calculauons 2041v 1000 mL 1 L weakly romzedto tatrate So we ear tatrate one two or three protons m HgPO to obtam three separate and pomts as m reaetror 3 4 and 5 3 H3PO NaoH a NaHzPOo H20 1H exchanged 4 H3PO 2NaoH a NazHPO 2H20 2 H exchanged 5 H3PO 3NaoH a NagPO 3H20 3 H exchanged Fur ngoun 980 5 980 g Reacmon 3 Eqmvalentwexght i 1 eq98 0g lmole1 eq 930 TE 490 g Reacmon 4 Equxvalentwexght i 1 eq w 0g lmole2 eq 930 T5 325 g Reacmon 5 Eqmvalentwexght i 1 eq32 6g lmole3 eq Fos4327 00d Analysts Nnrmality is useful in the cnntzxt hr acidrh axe titratinn In sueh reaction 1 mL of 1 N soluu on of and would neutxahze 1 mL of 1 N solution of alkah Equation mLxN mszN For example 20 mL of 1 N ofNaOH ean neutrahze 100 mL of 0 2 NHCl 20mL x 1N100mLx 0 2N B Reduc oneoxida on reaction The reacting umthere is the electron gamed or 10 t 3 12 KNLnOo SHCI SF8C11 MnClz4HzOKCl Or m Ionized fonn ez MnOt39 8H39 SFe MnZ 4H20 It consists oftwo half nesmans owr 8H 5e e an 4HzO sFez e SFe Se Fwe electrons are exchanged mthls nesmon Each Fez loses one e1eetmn whtle Mnohgsnns ve e ee ons For the reducing agent FeCl molm MM hnnbm of elecnonscoef clmz Egmval ent wetght 12675 1267g E 5 1 mole 1 sq For the oxxdumg agent KMnm 316ns g 1 Egmvalent wetght 1mole5eq C Precipita on and complexel orma on FOS4327eFood Anaysls In these eases there ls no reactmg umt exchanged bunhe reactants merely eomblne The charge Welght dlvldedby the charge on the catlon Example agent for WW 1 cholz Pbclol The metal lan lstz and charge ls 2 molar lulu Egnlvalenl waght hangs on the canon metal ml 1942 g T 71 5 oh 1 eq 971 g 1 male 2 eq 01 o z N Examgle 2 IMEDTA for Al 7 EDTA abbrevlanon for ethylenedlamlnetetxaaceu aeld ls a ehelaang agent T ll quot01 a on n It elm complex Wllh metal ion ln 1 lrauo A 3 EDTA a AlEDTA3 The metal lan lsAl3 and charge ls 3 ma a mun Eguw em wage Oxng on the canon metal ml 2924 97414 or leq974lg sllmmaly nr Nnrma ty Calculatinn 2 me NL mL FOS4327rFood Anaysys Mass of compound gram Sq Equwalmt weight Nnrmality is ralzted m Mnlarity hythe fnlluwing equz nn N n x M ntota1 number ofreactmg umts n 1 for HCL HNO NaOH orKOH n 2 for H2502 It can be 1 2 or 3 foerPOo dependmg on the reaction wakes apart FOS4321Food Analysis Problems for Part III due on September 19th 25 point 1 N E 4 Determine the number of millimoles of solute contained in each of the following a 25 mL of 025 M PbNO3z b 5 mL of5 M NH4OH 2 points a 25 mL X 025 molL 25 mL X 025 mmolmL 625 mmol b 5 mL X 5 molL 5 mL X 5 mmolmL 25 mmol Determine the Molarity of a 10 wv of sucrose solution b 37 ww of HCl with speci c gravity of 1184 c 30 ww of KOH with speci c gravity of 129 gmL 3 points a 1000 mL X 10 wv 100 g 100 g 3423 gmol 0292 mol 0292 mol1000 mL 0292 M b 1000 mL X 1184 gmL 1184 g 1184 g X 37 ww 4381 g 4381g 365 gmol 120 mol 120 mol1000 mL 120 M c 1000 mL X 129 gmL 1290 g 1290 g X 30 ww 387 g 387 g 56 gmol 69 mol 69 mol1000 mL 69 M Find the number of equivalent of solute in a 25 mL of6 N H2804 b 15 L of 05 N NaOH and C 300 mL of 125 N NaZCO3 3 points a 25 mLX 6N00025LX6N0015 eg b 15LX05N075eg c 300 mL X 125N03 L X 125N0375 eg How many grams of solute are contained in each of the following a 25 mL of 6 N H2804 b 15 mL of 05 N of BaOHz for reaction BaOHz HzSO4 gtBaSO42HzO c 25 mL of 01 N KMnO4 as oxidizer in acid medium where it is reduced to Mn2 d 25 mL of 01 N KMnO4 as oxidizer in neutral medium where it is reduced to MnOz 3 points a No reaction is given for HzSO4 However because H2804 is a strong acid it almost always donates two H in acidbase titration Thus N2XM or MN23molL 25 mL X 3 mol L X 98072 gmol 736 gram b 15 mL of 05 N of BaOHz for reaction BaOHz HzSO4 gtBaSO42HzO Two H are exchanges in this reaction for one BaOHz thus n2 7 UI FOS4321Food Analysis N 2 X M or M 052 025 molL 15 mL X 025 mol L X 1713 gmol anhydrous 064 gram c 25 mL of 01 N KMn04 as oxidizer in acid medium where it is reduced to Mn2 Mn7 was reduced to Mn2 Five 6 was exchanged for each KMn04 N 5 X M or M N5 002 molL 25 mL X 002 mol L X 158 gmol 0079 gram c 25 mL of 01 N KMn04 as oxidizer in neutral medium where it is reduced to Mn02 Mn7 was reduced to Mn4 Three 6 was exchanged for each KMn04 N 3 X M or M N3 0033 molL 25 mL X 0033 mol L X 158 gmol 0132 gram Determine the normality of the following a 21 gram of KOH in 25 L solution b 220 g H2804 in a 5 L solution and C 306 g K2Cr207 in 250 mL of solution for reaction Cr20727 14H 6e gt 2Cr3 7H2O 4 points a 21 gram of KOH in 25 L solution Each KOH require one H in acidbase titration n1 21 g 561 gmol 25 L 015 molL NnX M 1X015 015 N b 220 g H2804 in a 5 L solution Each H2804 donates two H in acidbase titration n2 220 g 9807 gmol 5 L 045 molL N nX M 2X045 09 N c 306 g K2Cr207 in 250 mL of solution for reaction Cr20727 14H 6e7 gt 2Cr3 7H2O 6e was exchanged in this reaction for each K2Cr207 n6 306 g 2942 gmol 250 mL 00416 molL N 6X M 6X00416 0250 N Find the molarity and normality of the following a 167 gram of H3P04 in 250 mL for reaction H3P04 Na0H gt NaH2P04 H20 b 6425 g of Ba0H2 in 500 mL for reaction Ba0H2 H2804 gtBa8042H20 5 points gt1 FOS4321Food Analysis a Each H3PO4 donates one H in this reaction n1 167 g 98 gmol 250 mL 686 molL N nX M 1X685 m b Each BaOHz requires two H in this reaction n2 6425 g 1713 gmol anhydrous 500 mL 0075 molL N 2X M 2X0075 w Calculate the molarity and normality of the following stock solutions used in chemical laboraoties a HCl speci c gravity 1189 38 ww b HzSO4 96 ww speci c gravity 1835 gmL and c HNO3 71 ww speci c gravity 1418 5 points a HCl speci c gravity 1189 38 ww HCl is a strong acid One HCl will donate one H in acidbase titration n1 1000 mL X 1189 gmL 1189 g 1189 g X 38 ww 4518 g ofHCl 4518 g 365 gmol 124 mol M 124 mol1000 ml M N nX M 1X124 M b H2804 96 ww speci c gravity 1835 gmL H2804 is a strong acid One HZSO4will donate two H in acidbase titration n2 1000 mL X 1835 gmL 1835 g 1835 g X 96 ww 17616 g oszSO4 17616 g 9807 gmol 1796 mol M 1796 mol1000 ml 1796 M N nX M 2X1796 m c HNO3 71 ww speci c gravity 1418 HN03 is a strong acid One HNngill donate one H in acidbase titration n1 1000 mL X 1418 gmL 1418 g 1418 g X 71 ww 10068 g of HNO3 10068 g 630 gmol 160 mol M 160 mol1000 ml M FOS4321Food Analysis NnX M1X16016


Buy Material

Are you sure you want to buy this material for

25 Karma

Buy Material

BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.


You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

Why people love StudySoup

Bentley McCaw University of Florida

"I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

Anthony Lee UC Santa Barbara

"I bought an awesome study guide, which helped me get an A in my Math 34B class this quarter!"

Steve Martinelli UC Los Angeles

"There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."


"Their 'Elite Notetakers' are making over $1,200/month in sales by creating high quality content that helps their classmates in a time of need."

Become an Elite Notetaker and start selling your notes online!

Refund Policy


All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email


StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here:

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.