A set of solubility data is given below. What is the mass of the water (solvent) evaporated?

\begin{tabular}{|c|c|c|c|c|}
\hline
Sample & \begin{tabular}{l}
Temperature \\
[tex]$\left({ }^{\circ} C \right)$[/tex]
\end{tabular} & \begin{tabular}{l}
Boat Mass \\
(g)
\end{tabular} & \begin{tabular}{l}
Boat + \\
Solution (g)
\end{tabular} & \begin{tabular}{l}
Boat + \\
Dry (g)
\end{tabular} \\
\hline
2 & 30.1 & 0.730 & 0.929 & 0.816 \\
\hline
\end{tabular}

A. 0.816 g
B. 0.113 g
C. 0.0860 g
D. 0.761 g



Answer :

To find the mass of the water (solvent) evaporated, we can follow these steps using the provided data:

1. Identify the Mass of the Boat Before Adding the Solution:
The mass of the empty boat is given as [tex]\(0.730\)[/tex] grams.

2. Identify the Mass of the Boat with the Solution:
The mass of the boat after adding the solution is given as [tex]\(0.929\)[/tex] grams.

3. Identify the Mass of the Boat with the Dry Solution:
After drying the solution, the mass of the boat plus the dry residue (solute) is given as [tex]\(0.816\)[/tex] grams.

4. Calculate the Mass of the Dry Solution:
The mass of the dry solution (solute) itself can be found by subtracting the mass of the empty boat from the total mass after drying the solution.
[tex]\[ \text{Mass of the dry solution (solute)} = 0.816 \text{ g} - 0.730 \text{ g} = 0.086 \text{ g} \][/tex]

5. Calculate the Mass of the Solvent that was Initially Present:
This step isn't necessary for the problem at hand since we are only interested in the mass of the solvent that evaporated, not the initial mass.

6. Calculate the Mass of the Solvent that Evaporated:
The mass of the solvent (water) that evaporated can be found by subtracting the mass of the dry boat and dry solution from the mass of the boat with the initial solution.
[tex]\[ \text{Mass of solvent evaporated} = 0.929 \text{ g} - 0.816 \text{ g} = 0.113 \text{ g} \][/tex]

Therefore, the mass of the water (solvent) evaporated is [tex]\(0.113\)[/tex] grams.

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