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Table 1: Estimation of DO. burette Final SLNO Volume of water Initial sample (ml) burette Volume of NazS203 reading (ml) reading (ml) consumed (V2 ml) 100.0 0.0 7.4 7.4 1 100.0 7.4 14.8 7.4 2 100.0 14.8 22.1 7.3 3 Calculations: Normality of water sample: N¡Vi(water sample) =N2V2(N22S2O3) -x 7.4 40 N2 XV2 N1 = 1.85 x 10-3 v1 100 N1 = 1.85 x 10-3 %3D Amount of Dissolved oxygen in water sample: Amount of free chlorine in water sample= N1 × equivalent weight of oxygen = 1.85 x 10-3 x 8 g/L 3D 1.85 х 10-3 х 8х 103 рpm 14.8 РPM %3D Amount of free chlorine in water sample= 14.8 PPM

Table 1: Estimation of DO. burette Final SLNO Volume of water Initial sample (ml) burette Volume of NazS203 reading (ml) reading (ml) consumed (V2 ml) 100.0 0.0 7.4 7.4 1 100.0 7.4 14.8 7.4 2 100.0 14.8 22.1 7.3 3 Calculations: Normality of water sample: N¡Vi(water sample) =N2V2(N22S2O3) -x 7.4 40 N2 XV2 N1 = 1.85 x 10-3 v1 100 N1 = 1.85 x 10-3 %3D Amount of Dissolved oxygen in water sample: Amount of free chlorine in water sample= N1 × equivalent weight of oxygen = 1.85 x 10-3 x 8 g/L 3D 1.85 х 10-3 х 8х 103 рpm 14.8 РPM %3D Amount of free chlorine in water sample= 14.8 PPM

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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EXPERIMENT :Estimation of Dissolved Oxygen

I want Result & Analysis for this experiment includes Precautions

Observations: Table 1: Estimation of DO. SINO Volume of water Initial sample (ml) burette Final burette Volume of NazS203 reading (ml) reading (ml) consumed (V2 ml) 100.0 0.0 7.4 7.4 1 100.0 7.4 14.8 7.4 2 100.0 14.8 22.1 7.3 Calculations: Normality of water sample: N¡V1(water sample) =N2V2(Na2S2O3) 1 N2 XV2 x 7.4 40 N1 = 1.85 x 10-3 V1 100 N1 = 1.85 x 10-3 Amount of Dissolved oxygen in water sample: Amount of free chlorine in water sample= N1 x equivalent weight of oxygen 1.85 х 10-3 х 8 д/L %3D 3D 1.85 х 10-3 х 8х 103 рpm = 14.8 PPM Amount of free chlorine in water sample= 14.8 PPM 3.
Transcribed Image Text:Observations: Table 1: Estimation of DO. SINO Volume of water Initial sample (ml) burette Final burette Volume of NazS203 reading (ml) reading (ml) consumed (V2 ml) 100.0 0.0 7.4 7.4 1 100.0 7.4 14.8 7.4 2 100.0 14.8 22.1 7.3 Calculations: Normality of water sample: N¡V1(water sample) =N2V2(Na2S2O3) 1 N2 XV2 x 7.4 40 N1 = 1.85 x 10-3 V1 100 N1 = 1.85 x 10-3 Amount of Dissolved oxygen in water sample: Amount of free chlorine in water sample= N1 x equivalent weight of oxygen 1.85 х 10-3 х 8 д/L %3D 3D 1.85 х 10-3 х 8х 103 рpm = 14.8 PPM Amount of free chlorine in water sample= 14.8 PPM 3.
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