Fig. 1 shows a river downstream from a village sewage outlet. At point A, untreated sewage is released. At point B, 2 km downstream, aquatic plants grow alo...
Question 1 Report
Fig. 1 shows a river downstream from a village sewage outlet. At point A, untreated sewage is released. At point B, 2 km downstream, aquatic plants grow along the bank. A student collected water samples and used sterile bottles to test the number of bacteria and the concentration of dissolved oxygen.
Table 1 gives the results. The water temperature was 20 degrees C at both points.
| Sampling point | Bacteria / colonies cm-3 | Dissolved oxygen / mg dm-3 |
|---|---|---|
| Upstream of outlet | 18 | 8.7 |
| A | 760 | 2.9 |
| B | 210 | 6.1 |
(a) Describe the difference in dissolved oxygen between the upstream sample and point A. [2]
(b) Use the results to suggest why oxygen concentration is low at point A. [2]
(c) Suggest why dissolved oxygen is higher at point B than at point A. [2]
(d) Name one stage in a sewage-treatment plant that removes solid waste before water enters a river. [2]
Answer Details
(a) Dissolved oxygen was lower at point A than upstream. It fell from \(8.7\text{ mg dm}^{-3}\) to \(2.9\text{ mg dm}^{-3}\): \[8.7-2.9=5.8\text{ mg dm}^{-3}\] [2]
(b) Point A had many more bacteria, 760 colonies \(\text{cm}^{-3}\), than upstream. These bacteria decompose sewage and use oxygen in aerobic respiration, explaining the low dissolved oxygen concentration. [2]
(c) By point B, bacterial numbers have decreased as sewage is decomposed or diluted, so less oxygen is used in decomposition. Water plants also photosynthesise and release oxygen. [2]
(d) Screening, sedimentation, or filtration can remove solid waste before treated water enters a river. [2]
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