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Answer :
To find the molarity of the lithium chloride solution, we first determine the number of moles of lithium chloride using its mass and molar mass, and then we calculate the molarity by dividing the obtained moles by the volume of the solution.
1. Calculate the number of moles ([tex]$n$[/tex]) using the formula:
[tex]$$ n = \frac{\text{mass}}{\text{molar mass}} $$[/tex]
Given a mass of [tex]$39.4\text{ g}$[/tex] and a molar mass of [tex]$42.39 \text{ g/mol}$[/tex],
[tex]$$ n = \frac{39.4 \text{ g}}{42.39 \text{ g/mol}} \approx 0.9295 \text{ mol} $$[/tex]
2. Calculate the molarity ([tex]$M$[/tex]) using the formula:
[tex]$$ M = \frac{n}{V} $$[/tex]
Here, [tex]$V = 1.59\text{ L}$[/tex], so:
[tex]$$ M = \frac{0.9295 \text{ mol}}{1.59 \text{ L}} \approx 0.5846\ \frac{\text{mol}}{\text{L}} $$[/tex]
Thus, the molarity of the solution is approximately
[tex]$$0.5846\ \frac{\text{mol}}{\text{L}}.$$[/tex]
1. Calculate the number of moles ([tex]$n$[/tex]) using the formula:
[tex]$$ n = \frac{\text{mass}}{\text{molar mass}} $$[/tex]
Given a mass of [tex]$39.4\text{ g}$[/tex] and a molar mass of [tex]$42.39 \text{ g/mol}$[/tex],
[tex]$$ n = \frac{39.4 \text{ g}}{42.39 \text{ g/mol}} \approx 0.9295 \text{ mol} $$[/tex]
2. Calculate the molarity ([tex]$M$[/tex]) using the formula:
[tex]$$ M = \frac{n}{V} $$[/tex]
Here, [tex]$V = 1.59\text{ L}$[/tex], so:
[tex]$$ M = \frac{0.9295 \text{ mol}}{1.59 \text{ L}} \approx 0.5846\ \frac{\text{mol}}{\text{L}} $$[/tex]
Thus, the molarity of the solution is approximately
[tex]$$0.5846\ \frac{\text{mol}}{\text{L}}.$$[/tex]
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