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A set of four [tex]15.0 \, \text{mL}[/tex] solutions—each in a similar open dish—was left at room temperature, and the amount that remained was recorded after every hour. Which of the solutions do you expect to be the first to completely evaporate from the open dish?

1. [tex]1.5 \, \text{M} \, \text{NaOH}[/tex]
2. [tex]1.0 \, \text{M} \, \text{NaOH}[/tex]
3. [tex]0.5 \, \text{M} \, \text{Na}_2\text{SO}_4[/tex]
4. [tex]0.25 \, \text{M} \, \text{Na}_2\text{SO}_4[/tex]

Sagot :

GinnyAnswer
To determine which solution will evaporate first, we need to consider the concentration of the solutions. Here are the given concentrations:

1. [tex]\(1.5 \, m\, \text{NaOH}\)[/tex]
2. [tex]\(1.0 \, m\, \text{NaOH}\)[/tex]
3. [tex]\(0.5 \, m\, \text{Na}_2\text{SO}_4\)[/tex]
4. [tex]\(0.25 \, m\, \text{Na}_2\text{SO}_4\)[/tex]

The rate of evaporation is typically inversely related to the concentration of the solution: the solution with the highest concentration is expected to evaporate faster because it has fewer water molecules per unit volume as solutes are dominant.

Following this reasoning, we can list the solutions in descending order of their concentrations:

1. [tex]\(1.5 \, m\, \text{NaOH}\)[/tex]
2. [tex]\(1.0 \, m\, \text{NaOH}\)[/tex]
3. [tex]\(0.5 \, m\, \text{Na}_2\text{SO}_4\)[/tex]
4. [tex]\(0.25 \, m\, \text{Na}_2\text{SO}_4\)[/tex]

Based on their concentrations, the solution with [tex]\(1.5 \, m\, \text{NaOH}\)[/tex] is the most concentrated. Given that higher concentration solutions tend to evaporate first, we can conclude that the [tex]\(1.5 \, m\, \text{NaOH}\)[/tex] solution will be the first to completely evaporate from the open dish.

Therefore, the NaOH solution with a concentration of [tex]\(1.5 \, m\)[/tex] is expected to be the first to evaporate. This corresponds to the first solution listed, which is:

The [tex]\(1.5 \, m\, \text{NaOH}\)[/tex] solution (index 0 in the list) is expected to be the first to completely evaporate.
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