Q1: 1 kg of a saturated aqueous solution of a highly soluble component A at 60°C is cooled to 25°C. The solubility limits of A are (0.6 kgA/kg water) at 60°C and (0.2 kgA/kg water) at 25°C. The amount in kg of the crystals formed is

ANS:C - 0.2

To find the amount of crystals formed when the solution is cooled, we need to calculate the amount of component A that precipitates out of solution due to the decrease in temperature. The initial amount of component A in the solution at 60°C is the difference between the actual amount (1 kg) and the amount that would remain in solution if it were saturated at 60°C (0.6 kg). Initial amount of A = 1 kg - 0.6 kg = 0.4 kg At 25°C, the solubility limit of A is 0.2 kg/kg water. This means that the solution can only hold 0.2 kg of A per kg of water at this temperature. So, the amount of A that remains in solution at 25°C is: Amount of A at 25°C = 0.2 kg/kg water * 1 kg = 0.2 kg Therefore, the amount of A that precipitates out of solution upon cooling is the difference between the initial amount and the amount remaining in solution at 25°C: Amount of A precipitated = Initial amount of A - Amount of A at 25°C = 0.4 kg - 0.2 kg = 0.2 kg So, the amount of crystals formed is 0.2 kg.