Question

Question 7

A chemistry student is given 2.00 L of a clear aqueous solution at 43.° C. He is told an unknown amount of a certain compound X is dissolved in the solution. The student allows the solution to cool to 25.° C. At that point, the student sees that a precipitate has formed. He pours off the remaining liquid solution, throws away the precipitate, and evaporates the water from the remaining liquid solution under vacuum. More precipitate forms. The student washes, dries and weighs the additional precipitate. It weighs 0.062 kg

1) Using only the information above, can you calculate the solubility of X in water at 25 degrees C?

2) If yes calculate it. Round answer to 2 significant digits

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Answer 1

Hi
Part 1
As given in the question when we are cooling the solution up to 25 o C ,we are getting some precipitate.
After this is removed and remaining water is evaporated some amount of precipitate is still obtained this means that some amount of the precipitate is still remaining in the solution that will be left in the container when we evaporate the remaining amount.

When solution is cooled at 25 o C the undissolved salt precipitates out and the remaining is the dissolved salt.

This dissolved salt will remain when the water is evaporated as the value of dissolved salt is given we can find the solubility.

Part 2

We have mass of precipitate remaining =0.063 g
We have 2 L solution, we can neglect the change is volume because of the precipitation.

So the solibility will be 0.063/2 = 0.0315 g/ L