Pearson Edexcel International GCSE (9–1) · CHEMISTRY · Chapter
States of Matter
102 practice questions
Topics covered
Practice questions
11–16 of 16 · Hard
Question 11
5 marks · Hard
Answer and see the solution →The solubility curves below show the solubility of salt X and salt Y in water at different temperatures.
Solubility of Salts X and Y in Water (a)[1 mark]State which salt is more soluble in water at 30°C.
(b)[1 mark]Calculate the difference in solubility between salt X and salt Y at 60°C.
(c)[2 marks]Describe how the solubility of each salt changes as the temperature increases from 10°C to 70°C.
(d)[1 mark]Explain why it is important to use the same amount of water when measuring the solubility of different salts.
Question 12
2 marks · Hard
Answer and see the solution →Describe one safety precaution and one step to ensure accuracy when carrying out a solubility experiment for a solid in water.
Question 13
2 marks · Hard
Answer and see the solution →A student obtains the following measurements while investigating the solubility of a solid in water: Mass of empty dish = 22.35 g Mass of dish + solution = 58.20 g Mass of dish + dry crystals = 39.10 g Calculate the solubility of the solid in 100 g of water at this temperature. Show your working.
Question 14
5 marks · Hard
Answer and see the solution →The table below shows the solubility of sodium sulfate at different temperatures.
Solubility of sodium sulfate at different temperatures Temperature /°C Solubility / g per 100 g of water 10 12.5 30 19.6 50 29.4 70 40.8 90 53.1 (a)[2 marks]On graph paper, plot a solubility curve for sodium sulfate using the data in the table. Label the axes appropriately.
(b)[1 mark]Estimate the solubility of sodium sulfate at 60°C using your curve.
(c)[2 marks]Describe the trend shown by the solubility curve as temperature increases.
Question 15
2 marks · Hard
Answer and see the solution →The solubility curve for potassium nitrate shows the following data: At 25°C, solubility = 38 g per 100 g of water At 60°C, solubility = 109 g per 100 g of water A solution is prepared by dissolving 70 g of potassium nitrate in 100 g of water at 60°C. Use the solubility curve to determine the temperature at which crystals of potassium nitrate will begin to form when this solution is cooled.
Question 16
8 marks · Hard
Answer and see the solution →A student is investigating the solubility of a salt, ammonium nitrate (NH₄NO₃), in water. They perform an experiment and obtain the following results:
(a)Experimental masses Item Mass / g Empty evaporating basin 26.47 Evaporating basin + solution 62.12 Evaporating basin + dry crystals 40.11 Using the results above, calculate the solubility of ammonium nitrate at 35°C in g per 100g of water.
(i)[1 mark]Calculate the mass of crystals and mass of water used.
(ii)[2 marks]Calculate the solubility (g per 100g of water).
(b)The student repeats this experiment at different temperatures and obtains the following solubility data:
Temperature/°C Solubility / g per 100g water 10 65 20 77 35 98 55 138 Explain, using your result from part (a), how the student could check that their measurement at 35°C fits the trend shown by the other data points. What pattern would you expect to see on a solubility curve for this salt?
(i)[1 mark]What does the solubility curve show about the relationship between temperature and solubility for ammonium nitrate?
(ii)[1 mark]State how a result that does NOT fit the trend would appear on a solubility curve.
(c)[1 mark]Give the correct term for a solution in which no more solute can dissolve at a particular temperature, and undissolved solid is present.
(d)[1 mark]State the equation used to calculate solubility in g per 100g of water from experimental mass data.
(e)[1 mark]Explain briefly why it is important to pour off only the solution and not any undissolved solid during the experiment.