Chemistry Forum

Please answer my query concerning the solubility of crystal hydrates:

Add 80.0 mL of a 15.0% CuSO₄ solution with a density of 1.137 g/mL (T = 20 °C) to 30.0 g of solid CuSO₄*5H₂O and mix thoroughly. CuSO₄ solubility (T = 20 C) = 20.8 g / 100 g water
a.) Determine how many grams of CuSO₄*5H₂O remain undissolved (at 20 degrees Celsius).
b.) Determine the volume of 15.0% CuSO₄ solution with a density of 1.137 g/mL (T = 20 °C) that must be added to dissolve all of the added CuSO₄*5H₂O (30.0 g) at 20 °C.

Thank you in advance.

Separate the Masses of CuSO₄ and Water.
With the specific gravity of the 80 ml solution you get the mass of it with percentage the mass of CuSO₄. Calculate the mass of CuSO₄ from the 30 g CuSO₄×5 H₂O and also the mass of the 5 H₂O. So you will get two sum, one of the both CuSO₄ amounts and the water from the crystal and the solution. From the mass of water and the solubility of CuSO₄ you get the mass what will be dissolved. The difference will give the amount what will be not dissolved.
Problem CuSO₄ and CuSO₄ × H₂O is different. So 20,8 g CuSO₄/100 g H₂O = 32,54 g CuSO₄ x 5 H₂O/ 88,27 g H₂O

The solubility of crystal hydrates refers to the ability of these compounds to dissolve in a solvent and form a solution. Crystal hydrates are compounds that contain a specific number of water molecules in their crystal structure. The presence of these water molecules can affect the solubility of the compound, as they can form hydrogen bonds with the solvent molecules and influence the interactions between the solute and solvent. The solubility of crystal hydrates can be influenced by factors such as temperature, pressure, and the chemical nature of the solvent. Some crystal hydrates are highly soluble in water, while others are only sparingly soluble or insoluble.

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Thank you for your response. I will try my best next time.