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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 kg1) 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

Sagot :

Answer:

Follows are the solution to the given points:

Explanation:

In part 1:

As described and in the query, they become precipitated whenever the solutions are refrozen to [tex]25^{\circ} \ C[/tex].

Afterward, certain precipitate becomes replaced as well as the remaining water is evaporated, it implies that certain precipitate remained throughout the solution to just the container when the entire balance is evaporated.

The unrecoverable salt precipitates whenever the solvent is cooled at [tex]25^{\circ} \ C[/tex]and the remaining salt dissolves. It dissolved salt remains whenever the water is evaporated because as dissolved salt value is given that results can be achieved.

In part 2:

They have precipitation weight = [tex]0.063\ g[/tex]. They have a [tex]2 \ L[/tex] the solution, they may disregard the volume increases due to its precipitation. The intensity therefore is [tex]\frac{0.063}{2} = 0.0315 \ \frac{g}{L}[/tex]

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