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Please Help Solve This Chemistry Question by banky200: 7:27am On May 14 |
. If 50cm3 of a saturated solution of potassium chloride at 30oC yielded 18.62g of dry salt, calculate the solubility of the salt in mol/dm3 at 30oC |
Re: Please Help Solve This Chemistry Question by CodeTemplar: 8:05am On May 14 |
18.62 X mols in a gram. Incomplete question. A constant is missing. |
Re: Please Help Solve This Chemistry Question by Light78: 8:25am On May 14 |
5 mol/dm3 Potassium Chloride = KCL Solubility ( mol/dm3) = mass of solute (g) / molar mass ÷ volume of solvent.......equ 1 Remember Mole = Mass/Molar mass Molar mass of KCL = K is 39.10g/mol, and CL is 35.45g/mol. Adding them together, = 74.55 which is the molar mass of KCL. Mass of dry salt is 18.62g Mole of KCL = mass of dry salt/ molar mass of KCL = 18.62/74.55 = 0.249 mol We go back to equ 1 Solubility = Mole/ volume of solvent. Remember, the volume of solvent is given in Cm3, it should be converted to dm3; 50/1000 = 0.05 dm3 From equ 1 Solubility= mole of KCL/ volume of solvent in dm3 = 0.249/0.05 = 5 mol/dm3 Hence, the solubility of KCL in Mol/dm3 at 30°C is 5 mol/dm3
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