This question uses this chart to solve. part A) A student makes a potassium chlo

Question

This question uses this chart to solve.

part A) A student makes a potassium chloride solution with 6.40 x 103 g water at 50oC. She finds that she can still add an additional 144.0 grams of potassium chloride before no more will dissolve. What was the original molar concentration of the solution she made? Use the solubility chart above. Assume the density of the solution (at any concentration) is 1.11 g/mL.

part B) A student makes a saturated solution of potassium dichromate at 90.0oC with 1775 g water and then cools the solution down to 30.0oC. How many moles of potassium dichromate will precipitate out of solution when he cools his solution down?

Explanation / Answer

(A) According to the chart, at 50°C solubility of KCl = 42.5 g KCl / 100 g water

(42.5 g KCl / 100 g water) * 6400 g water = 2720 g KCl are allowed so dissolve.

Original mass = 2720 g - 144 g = 2576 g KCl --> we'll have to convert this to moles later.

    The original volume was = 8976 g / 1.11 g/ml = 8086 ml = 8.1 L

Original moles of KCl = mass / molar mass = 2576 g KCl / 74.55 g/mol = 34.55 moles

Original molar concentration = 34.55 moles / 8.1 L = 4.3 M

(B)

At this temperature, the amount of K2Cr2O7 in solution is:

Mass of K2Cr2O7 @ 90°C = (70 g K2Cr2O7/ 100 g water) * 1775 g water = 1242.5 g K2Cr2O7

At this temperature, the amount of K2Cr2O7 in solution is:

Mass of K2Cr2O7 @ 30°C = (16 g K2Cr2O7/ 100 g water) * 1775 g water = 198.8 g K2Cr2O7

Mass of precipitate = 1242.5 g - 198.8 g K2Cr2O7 = 1043.7 g

Moles of precipitate = mass / molar mass = 1043.7 g / 294.2 g/mol = 3.55 moles

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