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A student began experiment for determination of 5 -day, $20^{\circ} \mathrm{C} \mathrm{BOD}$ on Monday. Since the $5^{\text {th }}$ day fell on Saturday, the final DO readings were taken on next Monday. On calculation, BOD (i.e. 7 day, $20^{\circ} \mathrm{C}$ ) was found to be $150 \mathrm{mg} / \mathrm{L}$. What would be the5-day, $20^{\circ} \mathrm{C} \mathrm{BOD}$ (in $\left.\mathrm{mg} / \mathrm{L}\right)$ ? Assume value of $\mathrm{BOD}$ rate constant $(\mathrm{k})$ at standard temperature of $20^{\circ} \mathrm{C}$ as $0.23 /$ day $($ base $e) .$


Question ID - 155857 | Toppr Answer

A student began experiment for determination of 5 -day, $20^{\circ} \mathrm{C} \mathrm{BOD}$ on Monday. Since the $5^{\text {th }}$ day fell on Saturday, the final DO readings were taken on next Monday. On calculation, BOD (i.e. 7 day, $20^{\circ} \mathrm{C}$ ) was found to be $150 \mathrm{mg} / \mathrm{L}$. What would be the5-day, $20^{\circ} \mathrm{C} \mathrm{BOD}$ (in $\left.\mathrm{mg} / \mathrm{L}\right)$ ? Assume value of $\mathrm{BOD}$ rate constant $(\mathrm{k})$ at standard temperature of $20^{\circ} \mathrm{C}$ as $0.23 /$ day $($ base $e) .$

1 Answer - 5876 Votes

3537

Answer Key : (127 to 132) -

127 to 132



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