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How to Solve Equivalent Weight and Normality?, Summaries of Analytical Chemistry

How to solve the number of equivalents of a solute divided by the volume of the solution in liters

Typology: Summaries

2023/2024

Available from 06/11/2024

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~ equivalent weight uniike Wokomar waght 1S PHORENTOWA moGs yf reWcal eynbies WhO Gambine or displave ster chemical antics Mo ¥eq chon. Molecular wtight _ Equivalent weight = E> volente {autor ee x Nalence Fadi - We determination of equivalent weignt 1S ony posse by determining we valence fautar v& re Mewical etry - Equivaleny Waght ok wemrrai zation toned: Trample: (in Mhe gwen neutralization YeauKion the eymuthent wright 6 gud ard Wade, are reapertivety Ay? + 2Wa dH — voryon, + 2-09 Stiwkiow: wilay brass. He a) eg.wt. (add) * 15 gin} 4 os 14) = U4 ae w aah “pe | (Wolay Mags: N= 29 0: Ww we 4 4ygina 4g. Wt (hase) > 4° ghmol Wn the reauiion 2VAON 4 My 804 —7 Noe HPA + 220 the equivorenr weight of the ouid is: Soluron: Molar Mass: Hy P04 a 40) yt (acid): 4 3 . ia aq 9S gt eo) & Tquivorc waght & Medex Reaction. Trxowgle: © Cry Ce” (reduced) Caine) oxidation Reaughon Solution’. Wlar Mass: C09 Cr. 5207-104 Oz _vday We 2A% ghvwol eq. wt (oxidation) = ne (0 ry = AG: +1 wh Eo -2 mm > 2414 ea x2 Key we Civitan \ exidation Gore) ~ (oxidation 4 see Vx swsnler of (« 3) x () 4) wright of some, WY: ot. of Sour, oes = (eS (ta Example: JO) What is the wormality of 0 soution grepand bby Aiccolveg WBe Knmdy, in ewugh Water W prepare, Stome sole? we wolor mass L Kwnog is Iss glmot, and the Perhtent veattion It underqpes 16 MnO —> Mat Given! weight of olute: 15.69 Volume of @burion: SO) mt = 1600 est) Molar tase: ISSgluayl ew (on) Mn > ~ Ke (ay) = - 149 --1 Y= -1ty fo 3]