Ammonia (NH3) dissolved in water is heated in a beaker. If initially there had been 60 grams dissolved in 100 grams of water, what would happen if the temperature reached 90℃ and why? A. The solution would become supersaturated and the ammonia would all remain dissolved . B. The solution would become saturated but the amount of ammonia dissolved would remain the same . C. about 50 more grams of ammonia would be able to dissolve as the solution has become unsaturated . D. about 50 grams of ammonia would come out of solution as the temperature caused the solubility to decrease .

C. About 50 more grams ammonia would be able to dissolve as the solution has become unsaturated

Explanation:

From the question, we have;

The mass of ammonia dissolved in water = 60 grams

The mass of the water solvent which dissolves the ammonia = 100 g

The temperature to which the ammonia water solution is raised = 90°C

Therefore, given that the solubility of a solute in a solvent increases with temperature, when the temperature is raised, we have, that the solution will more unsaturated and will be able to dissolve more ammonia, therefore, option 'C' is the most correct option that corresponds with the relationship of temperature and solubility

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%