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What volume, in liters, of 1.5 M [tex]$CaCl_2$[/tex] solution can be made using 1200.0 g of [tex]$CaCl_2$[/tex]?

[tex]
\begin{array}{c}
CaCl_2: 110.98 \, \text{g/mol} \\
[?] \, \text{L}
\end{array}
\]


Sagot :

To determine the volume of a 1.5 M [tex]\( CaCl_2 \)[/tex] solution that can be made using 1200.0 grams of [tex]\( CaCl_2 \)[/tex], follow these steps:

1. Identify the molar mass of [tex]\( CaCl_2 \)[/tex]:
[tex]\[ \text{Molar mass of \( CaCl_2 \)} = 110.98 \, \text{g/mol} \][/tex]

2. Calculate the moles of [tex]\( CaCl_2 \)[/tex] needed:
First, divide the given mass of [tex]\( CaCl_2 \)[/tex] by its molar mass to find the number of moles.
[tex]\[ \text{Moles of \( CaCl_2 \)} = \frac{\text{Mass of \( CaCl_2 \)}}{\text{Molar mass of \( CaCl_2 \)}} \][/tex]
[tex]\[ \text{Moles of \( CaCl_2 \)} = \frac{1200.0 \, \text{g}}{110.98 \, \text{g/mol}} \][/tex]
From the previous analysis, it can be affirmed that:
[tex]\[ \text{Moles of \( CaCl_2 \)} \approx 10.812759055685708 \, \text{moles} \][/tex]

3. Use the molarity to find the volume of the solution:
Molarity (M) is defined as the number of moles of solute per liter of solution. Rearranging this definition, we find the volume:
[tex]\[ \text{Volume of solution (in liters)} = \frac{\text{Moles of solute}}{\text{Molarity}} \][/tex]
Given that the molarity is 1.5 M:
[tex]\[ \text{Volume of solution} = \frac{10.812759055685708 \, \text{moles}}{1.5 \, \text{M}} \][/tex]

4. Calculate the volume:
[tex]\[ \text{Volume of solution} = \frac{10.812759055685708}{1.5} \][/tex]
[tex]\[ \text{Volume of solution} \approx 7.208506037123805 \, \text{liters} \][/tex]

Therefore, using 1200.0 grams of [tex]\( CaCl_2 \)[/tex], you can prepare approximately 7.2085 liters of a 1.5 M [tex]\( CaCl_2 \)[/tex] solution.