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Report (any type)/Brief report
22/09/2021
##0^@”C”##
In order to be able to answer this question, you need to have the of potassium chloride, ##”KCl”##, which looks like this
Since the solubility of potassium chloride is given per ##”100 g”## of water, calculate how many moles of potassium chloride would make a ##”3.5-molal”## solution in that much water.
Keep in mind that is defined as moles of , which in your case is potassium chloride, divided by kilograms** of , which in your case is water.
##color(blue)(b = n_”solute”/m_”solvent”)##
This means that you have
##n_”solute” = b * m_”solvent”##
##n_”solute” = “3.5 mol” color(red)(cancel(color(black)(“kg”^(-1)))) * 100 * 10^(-3)color(red)(cancel(color(black)(“kg”))) = “0.35 moles KCl”##
Next, use potassium chloride’s molar mass to figure out how many grams would contain this many moles
##0.35color(red)(cancel(color(black)(“moles KCl”))) * “74.5 g”/(1color(red)(cancel(color(black)(“mole KCl”)))) = “26.1 g”##
Now take a look at the solubility graph and decide at which temperature dissolving ##”26.1 g”## of potassium chloride per ##”100 g”## of water will result in a saturated solution.
Practically speaking, you’re looking for the temperature at which the saturation line, drawn on the graph in ##color(blue)(“blue”)##, matches the value ##”26.1 g”##.
From the look of it, dissolving this much potassium chloride per ##”100 g”## of water will produce a ##”3.5-molal”## saturated solution at ##0^@”C”##.
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I want to thank your team (Jeff) for finding a new writer when you were unable to reach my existing writer to revise my paper. I appreciate your professionalism and efforts. Thanks
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