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JAMB UTME - Chemistry - 1990

Question 1 Report

The solubility in mol dm-3 of 20g of CuSO4 dissolved in 100g of water at 180o is

[Cu = 64, S = 32, O =16]

Answer Details
The question is asking for the solubility of CuSO4 in water at a certain temperature. The given information is that 20g of CuSO4 is dissolved in 100g of water at 180°C. To solve this problem, we need to first calculate the molar mass of CuSO4, which is: Cu: 1 x 64 = 64 g/mol S: 1 x 32 = 32 g/mol O: 4 x 16 = 64 g/mol Total: 160 g/mol Next, we need to calculate the number of moles of CuSO4 that were dissolved in the water. We can do this using the formula: moles = mass/molar mass mass = 20g molar mass = 160 g/mol moles = 20/160 = 0.125 mol Finally, we can calculate the solubility of CuSO4 in water at this temperature using the formula: solubility = moles/volume We are given that the volume of the solution is 100g, but we need to convert this to dm3. Since the density of water is 1 g/cm3, we can assume that 100g of water has a volume of 100 cm3 or 0.1 dm3. solubility = 0.125/0.1 = 1.25 mol dm-3 Therefore, the solubility of CuSO4 in water at 180°C is 1.25 mol dm-3, which corresponds to.