Observations: Table 1: Estimation of DO. SI No 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 7.4 100.0 14.8 22.1 7.3 3 Calculations: Normality of water sample: N¡V1(water sample) =N2V2(Na2S2O3) Ex 7.4 40 N2 XV2 N13 = 1.85 x 10-3 V1 100 N1 = 1.85 × 10-3 Amount of Dissolved oxygen in water sample: Amount of free chlorine in water sample= N1 x equivalent weight of oxygen = 1.85 x 10-3 x 8 g/L = 1.85 x 10-3 x 8 × 103 ppm = 14.8 PPM Amount of free chlorine in water sample= 14.8 PPM
Observations: Table 1: Estimation of DO. SI No 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 7.4 100.0 14.8 22.1 7.3 3 Calculations: Normality of water sample: N¡V1(water sample) =N2V2(Na2S2O3) Ex 7.4 40 N2 XV2 N13 = 1.85 x 10-3 V1 100 N1 = 1.85 × 10-3 Amount of Dissolved oxygen in water sample: Amount of free chlorine in water sample= N1 x equivalent weight of oxygen = 1.85 x 10-3 x 8 g/L = 1.85 x 10-3 x 8 × 103 ppm = 14.8 PPM Amount of free chlorine in water sample= 14.8 PPM
Chapter7: Statistical Data Treatment And Evaluation
Section: Chapter Questions
Problem 7.16QAP
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EXPERIMENT :Estimation of Dissolved Oxygen
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