You must print the exam that has your lab book number. If you choose the wrong n

Question

You must print the exam that has your lab book number. If you choose the wrong number your grade will be blank and graded as zero (0). Once you print your exam you may start answering it. if there is not room enough for your answer, attach additional pages. Show your work when calculation should be done. you must submit a hard copy of this exam in one week. LAB 42I student has weighed 0.91 g of an unknown salt and dissolves it in water at different temperature, recording the following data: Calculate the solubilities per 100 ml. water, for each saturation temperature (fill the last column) Prepare a solubility -temperature curve for the unknown salt on the graph paper on the next page (alternative you may use an excel graph). Estimate the solubility of unknown salt at 90 degreeC from the solubility - temperature curve you prepared in the former question

Explanation / Answer

I will hel you with the data of solubility. With that data, you can construct the graph in b) and then answer part C according to that.

All you have to do in part is divide the mass by the volume in there, that will give you the solubility in 1 mL of water, then with a rule of 3 you can determine the solubility in 100 mL, or simply use a conversion factor:

1. C = 0.91 g / 20 mL = 0.0455 g/mL
0.455 g -----> 1 mL water
X ----------> 100 mL water
0.455 * 100 / 1 = 45.5 g/100 mL water

2. C = 0.91 / 27.40964 mL water = 0.03319 g/mL * 100 mL water/ 1 mL = 3.3199 g/100 mL water

3. C = 0.91 / 39.73799 * 100 = 2.29 g/100 mL

4. C = 0.91/60.66667 * 100 = 1.4999 g/100 mL

5. C = 0.91/81.25 * 100 = 1.12 g/100 mL

6. C = 0.91/93.52518 * 100 = 0.9729 g/100mL

7. C = 0.91/144.4444 * 100 = 0.63 g/100mL

8. C = 0.91/202.2222 * 100 = 0.45 g/100mL

Hope this helps.

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