EXPERIMENT :Estimation of Dissolved OxygenI want Result & Analysis for this experiment includes Precautions Observations:Table 1: Estimation of DO.SINO Volume of water Initialsample (ml)burette Finalburette VolumeofNazS203reading (ml)reading (ml)consumed (V2 ml)100.00.07.47.41100.07.414.87.42100.014.822.17.3Calculations:Normality of water sample:N¡V1(water sample) =N2V2(Na2S2O3)1N2 XV2x 7.440N1 =1.85 x 10-3V1100N1 = 1.85 x 10-3Amount of Dissolved oxygen in water sample:Amount of free chlorine in water sample= N1 x equivalent weight of oxygen1.85 х 10-3 х 8 д/L%3D3D 1.85 х 10-3 х 8х 103 рpm= 14.8 PPMAmount of free chlorine in water sample= 14.8 PPM3.

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

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter7: Statistical Data Treatment And Evaluation

Section: Chapter Questions

Problem 7.16QAP

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