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Hydration and Crystallization of Salts

Chemistry ⇒ Acids, Bases, and Salts

Hydration and Crystallization of Salts starts at 9 and continues till grade 12. QuestionsToday has an evolving set of questions to continuously challenge students so that their knowledge grows in Hydration and Crystallization of Salts. How you perform is determined by your score and the time you take. When you play a quiz, your answers are evaluated in concept instead of actual words and definitions used.
See sample questions for grade 11
A sample of hydrated magnesium sulfate (MgSO4·xH2O) weighs 6.16 g. After heating, the residue weighs 3.00 g. Calculate the value of x (to the nearest whole number). (MgSO4 = 120 g/mol, H2O = 18 g/mol)
A sample of hydrated sodium carbonate (Na2CO3·xH2O) weighs 14.3 g. After heating, the residue weighs 5.3 g. Calculate the value of x. (Na2CO3 = 106 g/mol, H2O = 18 g/mol)
A student heats 5.0 g of hydrated copper(II) sulfate (CuSO4·5H2O) and obtains 3.2 g of anhydrous copper(II) sulfate. Calculate the mass of water lost during heating.
Describe the process of crystallization of a salt from its aqueous solution.
Describe the role of temperature in the crystallization of salts from solution.
Describe what happens to the structure of a hydrated salt when it is heated strongly.
Explain the difference between an anhydrous salt and a hydrated salt.
Explain why anhydrous copper(II) sulfate is used as a test for water.
Explain why hydrated salts often have different colors compared to their anhydrous forms.
Explain why some salts do not form hydrates.
State one use of hydrated salts in everyday life.
What is meant by the term 'hydrated salt'?
A 7.80 g sample of hydrated barium chloride (BaCl2·xH2O) is heated until all water is removed, leaving 6.00 g of anhydrous BaCl2. Calculate the value of x. (BaCl2 = 208 g/mol, H2O = 18 g/mol)
A hydrated salt has the formula FeSO4·xH2O. If 278 g of the hydrated salt yields 152 g of anhydrous FeSO4 after heating, calculate the value of x. (FeSO4 = 152 g/mol, H2O = 18 g/mol)
A student prepares a saturated solution of potassium alum (KAl(SO4)2·12H2O) at 60°C and allows it to cool slowly to room temperature. Large, well-formed crystals are obtained. Explain why slow cooling is important in the crystallization process.
Describe the difference between efflorescent, deliquescent, and hygroscopic substances, giving one example of each.
Explain, with reference to the structure of hydrated salts, why heating often leads to a change in color for transition metal salts.