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What is the molarity of a solution made by dissolving 39.4 g of lithium chloride in enough water to make 1.59 L of solution?

[tex]
\[
\begin{array}{c}
\text{Molar mass of } LiCl: 42.39 \, \text{g/mol} \\
\text{Molarity} \, \left[\frac{\text{mol}}{L}\right]
\end{array}
\]
[/tex]

Answer :

To determine the molarity of the solution, we follow these steps:

1. Calculate the number of moles of lithium chloride ([tex]$\text{LiCl}$[/tex]) using the formula:
[tex]$$
n = \frac{m}{M}
$$[/tex]
where
- [tex]$m = 39.4\,\text{g}$[/tex] is the mass of [tex]$\text{LiCl}$[/tex],
- [tex]$M = 42.39\,\text{g/mol}$[/tex] is the molar mass of [tex]$\text{LiCl}$[/tex].

Thus,
[tex]$$
n = \frac{39.4\,\text{g}}{42.39\,\text{g/mol}} \approx 0.92946\,\text{mol}.
$$[/tex]

2. Calculate the molarity ([tex]$C$[/tex]) using the definition of molarity:
[tex]$$
C = \frac{n}{V}
$$[/tex]
where [tex]$V = 1.59\,\text{L}$[/tex] is the volume of the solution.

Therefore,
[tex]$$
C = \frac{0.92946\,\text{mol}}{1.59\,\text{L}} \approx 0.58457\,\frac{\text{mol}}{\text{L}}.
$$[/tex]

Rounded appropriately, the molarity of the solution is approximately [tex]$0.5846\,\text{M}$[/tex].

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