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slega [8]
3 years ago
8

A chemistry student is given 250.0 mL of clear aqueous solution at 42 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 25C. 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 remaining liquid solution under vacuum. More precipitate forms. The student washes, dries, and weighs the additional precipitate. It weighs 8.75 g. Using only the information above, can you calculate the solubility of X at 25C? If yes, calculate it.
Chemistry
1 answer:
lara [203]3 years ago
7 0

Answer:

No

Explanation:

The solubility of a solid in water refers to the amount of that solid that dissolves in water.

It is not possible to calculate the solubility of the solid because the student threw away the first precipitate that formed. We already have the volume of water, but having lost some mass of precipitate, it has become impossible to accurately determine the solubility.

Hence the answer provided above.

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<h3>Further explanation</h3>

Given

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Required

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Solution

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\tt H_2\div O_2=\dfrac{mol~H_2}{coefficient}\div \dfrac{mol~O_2}{coefficient}\\\\=\dfrac{5}{2}\div \dfrac{3}{1}=2.5\div 3\rightarrow H_2~limiting~reactant(smaller~ratio)

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mol H₂O based on mol H₂, and from equation mol ratio H₂ : H₂O=2 : 2, so mol H₂O = 5 mol and the volume also 5 L

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