At what temperature is the concentration of a saturated solution of #KCl# (molar mass 74.5 g) approximately 3 molal?

1 Answer
Feb 10, 2016

Answer:

#0^@"C"#

Explanation:

In order to be able to answer this question, you need to have the solubility graph of potassium chloride, #"KCl"#, which looks like this

http://www.metafysica.nl/ontology/general_ontology_29m5d.html

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 molality is defined as moles of solute, which in your case is potassium chloride, divided by kilograms** of solvent, 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"#.