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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]\[ \text{LiCl: 42.39 g/mol} \][/tex]

Answer :

We start by finding the number of moles of lithium chloride ([tex]$\text{LiCl}$[/tex]) using its mass and molar mass. The number of moles is given by

[tex]$$
\text{moles of LiCl} = \frac{\text{mass}}{\text{molar mass}} = \frac{39.4\ \text{g}}{42.39\ \frac{\text{g}}{\text{mol}}} \approx 0.9295\ \text{mol}.
$$[/tex]

Next, we calculate the molarity of the solution. Molarity is defined as the number of moles of solute divided by the volume of the solution in liters:

[tex]$$
\text{Molarity} = \frac{\text{moles of LiCl}}{\text{volume in liters}} = \frac{0.9295\ \text{mol}}{1.59\ \text{L}} \approx 0.5846\ \text{M}.
$$[/tex]

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

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