Fig. 1 shows a water-treatment worker testing how rapidly sodium thiosulfate reacts with dilute hydrochloric acid. A black cross is placed beneath a beaker....
Question 1 Report
Fig. 1 shows a water-treatment worker testing how rapidly sodium thiosulfate reacts with dilute hydrochloric acid. A black cross is placed beneath a beaker. Sulfur forms during the reaction and makes the liquid cloudy. The worker records the time at which the cross can no longer be seen through the solution. In all tests, the same observer views the cross from above.
| Water added / cm3 | Time for cross to disappear / s |
|---|---|
| 0 | 24 |
| 10 | 37 |
| 20 | 58 |
(a) Identify the dependent variable. [1]
(b) Calculate the rate for the test taking 24 s, using rate = 1/time. [2]
(c) Explain why adding water increases the time needed for the cross to disappear. [2]
Answer Details
(a) Time for the cross to disappear / time for solution to become cloudy [1] (b) 1 / 24 [1]; 0.042 s-1, accept 0.04 s-1 [1] (c) The sodium thiosulfate is less concentrated [1]; there are fewer collisions between reacting particles each second [1]
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