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(Vi x Nx Eх 1000) c(O2) (mg/L) = [V¼(V2 – V3)/V2] Here, V = Volume of titrant Na2 S2O3 (ml) V2 = Total volume of water sample (ml) V3 Volume of (MnSO4 + KI solution) added (ml) V4 = Volume of analyte taken in titration E = Equivalent weight of O2 N = Normality of Na2S2O3 solution

(Vi x Nx Eх 1000) c(O2) (mg/L) = [V¼(V2 – V3)/V2] Here, V = Volume of titrant Na2 S2O3 (ml) V2 = Total volume of water sample (ml) V3 Volume of (MnSO4 + KI solution) added (ml) V4 = Volume of analyte taken in titration E = Equivalent weight of O2 N = Normality of Na2S2O3 solution

Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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Show the dimensional analysis or show each unit in the equation to obtain mg/L

(Vi × N × E x 1000) [V¼(V2 – V3)/V2] c(O2) (mg/L) = - Here, V1 Volume of titrant Na2 S2O3 (ml) V2 = Total volume of water sample (ml) V3 = Volume of (MNSO4 + KI solution) added (ml) V4 Volume of analyte taken in titration E Equivalent weight of O2 N = Normality of Na2 S2O3 solution
Transcribed Image Text:(Vi × N × E x 1000) [V¼(V2 – V3)/V2] c(O2) (mg/L) = - Here, V1 Volume of titrant Na2 S2O3 (ml) V2 = Total volume of water sample (ml) V3 = Volume of (MNSO4 + KI solution) added (ml) V4 Volume of analyte taken in titration E Equivalent weight of O2 N = Normality of Na2 S2O3 solution
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