The method and the results are on the solubility lab. Do the analysis yourself first, then check the answers below.
Quick answer: Solubility is the maximum mass of a solute that dissolves in 100 g of water at a given temperature. These answers go with the grade 7 science lab on solubility: the graph, the calculations, the uncertainties and the conclusion explained, for all 8 analysis questions.
These model answers use the sample class results from the tables. Your own values will differ, but the saturation point should fall close to the accepted value and the times should decrease as the temperature rises. Check your reasoning against the steps below.
15.0 g dissolved fully but 18.0 g left grains, so saturation happened between 15.0 g and 18.0 g. Between 15 g and 18 g.
Question 2 Convert to 100 mL
Multiply by 2: 15.0 × 2 = 30 g and 18.0 × 2 = 36 g. 30 g to 36 g per 100 mL.
Question 3 Estimate and compare
Middle = (30 + 36) ÷ 2 = 33 g. Difference = 36 – 33 = 3 g. 3 ÷ 36 × 100 = 8.3 %. The estimate is 33 g, about 8 % below the accepted value.
Question 4 Pattern in Part B
As the temperature increases from 10 °C to 60 °C, the time falls from 90 s to 24 s. Hotter water dissolves sugar faster.
Question 5 Dissolving rate
10 °C: 5.0 ÷ 90 = 0.056 g/s. 60 °C: 5.0 ÷ 24 = 0.208 g/s. 90 ÷ 24 = 3.75. 0.056 g/s and 0.21 g/s: about 3.8 times faster.
Question 6 Explain with particles
In hot water the water particles move faster, so they hit the sugar crystals harder and more often and carry the sugar particles away more quickly. Faster particles dissolve the sugar sooner.
Question 7 Speed or amount
Temperature affects the rate of dissolving, which is how fast, even when it does not affect the solubility, which is how much. Rate and solubility are different.
Question 8 Control variables
Mass of sugar, volume of water, stirring speed and crystal size. An error is that the water cools during the test or that the end point is judged by eye. Controls and error given.
