When 250 cm3 of a saturated solution of CuSO\(_4\) at 30°C was evaporated to dryness, 5.0 g of the salt was obtained.. What is the solubility of the salt at...

When 250 cm3 of a saturated solution of CuSO\(_4\) at 30°C was evaporated to dryness, 5.0 g of the salt was obtained..

What is the solubility of the salt at 30°C? [CuSO\(_4\) = 160]

Answer Details

The solubility of a salt is the maximum amount of the salt that can dissolve in a given amount of solvent at a particular temperature. In this question, we are given that 250 cm\(^3\) of a saturated solution of CuSO\(_4\) at 30°C was evaporated to dryness to obtain 5.0 g of the salt. Since the solution was saturated, all of the CuSO\(_4\) that could dissolve at 30°C was already dissolved in the solution. Therefore, the mass of CuSO\(_4\) obtained from evaporating the solution represents the maximum amount of CuSO\(_4\) that can dissolve in 250 cm\(^3\) of water at 30°C. To find the solubility of CuSO\(_4\) at 30°C, we can use the equation: Solubility = Maximum amount of solute / Volume of solvent The maximum amount of CuSO\(_4\) that can dissolve in 250 cm\(^3\) of water at 30°C is 5.0 g. The volume of solvent is 250 cm\(^3\), or 0.250 L. The molar mass of CuSO\(_4\) is 159.6 g/mol. Therefore, the solubility of CuSO\(_4\) at 30°C is: Solubility = Maximum amount of solute / Volume of solvent = 5.0 g / 0.250 L = 20 g/L To convert the solubility to g/100 mL, we can multiply by 10: Solubility = 20 g/L x 10 = 200 g/100 mL To find the solubility in g/100 mL, we also need to divide by the molar mass of CuSO\(_4\): Solubility = 200 g/100 mL / 159.6 g/mol = 1.253 mol/L Therefore, the solubility of CuSO\(_4\) at 30°C is 1.253 g/100 mL, which is approximately equal to 1.25 g/100 mL. The correct option is 0.125.