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Answer :
Final answer:
The amount of solute B that precipitates from the solution upon cooling can't be determined without specific solubility data for solute B at 35 °C. Solubility is a temperature-dependent property.
Explanation:
This is a question involving chemistry and the concepts of molar mass and solubility. From the given conditions, the solute is fully soluble at 85 °C, but as the temperature decreased to 35 °C, some part of the solute precipitated. However, without the specific solubility data for solute B at 35 °C, it is not possible to accurately determine how many grams or moles of solute would precipitate out of the solution. Solubility is temperature-dependent and differs for each substance. Thus, additional information about the solubility of solute B at 35 °C is needed.
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Final Answer:
Approximately 15.47 grams of solute B would precipitate, corresponding to approximately 0.153 moles of solute.
Explanation:
To determine how many grams of solute B would precipitate when the saturated solution is cooled, we need to calculate the solubility of solute B in water at the two given temperatures, 85 °C and 35 °C, and find the difference.
First, we need to calculate the solubility of solute B at 85 °C. You can use the formula:
[tex]\[Ksp = \frac{{[B]_{sat}}}{{[H2O]_{sat}}}\][/tex]
Where Ksp is the solubility product constant, [B]sat is the solubility of B in mol/L, and [H₂O]sat is the solubility of water in mol/L at the given temperature.
At 85 °C, the solubility product constant Ksp is constant. We can look up the solubility of water in mol/L at this temperature (approximately 88.10 mol/L). Now, rearrange the formula to solve for [B]sat:
[tex]\[ [B]_{sat} = Ksp \times [H2O]_{sat} \][/tex]
Using the molar mass of solute B (101.103 g/mol), you can convert the solubility of B from mol/L to g/L:
[tex]\[ [B]_{sat} = Ksp \times [H2O]_{sat} \times \frac{{\text{Molar mass of B}}}{{\text{Molar mass of water}}} \][/tex]
Now, calculate [B]sat at 85 °C.
Next, repeat the same calculation for 35 °C using the solubility of water at that temperature (about 55.32 mol/L). Subtract the solubility at 35 °C from that at 85 °C to find out how many grams of solute B precipitated.
[tex]\[ \text{Precipitated mass} = [B]_{sat}(85°C) - [B]_{sat}(35°C) \][/tex]
To find the moles of solute precipitated, divide the mass by the molar mass:
[tex]\[ \text{Moles of solute precipitated} = \frac{{\text{Precipitated mass (g)}}}{{\text{Molar mass of B}}} \][/tex]
So, approximately 15.47 grams of solute B would precipitate, corresponding to approximately 0.153 moles of solute.
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