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Answer :
To find the molarity of a solution made by dissolving 39.4 g of lithium chloride (LiCl) in 1.59 L of solution, follow these steps:
1. Calculate the Molar Mass of LiCl:
- The molar mass of lithium (Li) is given as 6.94 g/mol.
- The molar mass of chlorine (Cl) is given as 35.45 g/mol.
- Therefore, the molar mass of lithium chloride (LiCl) is:
[tex]\[
\text{Molar Mass of LiCl} = 6.94 \, \text{g/mol} + 35.45 \, \text{g/mol} = 42.39 \, \text{g/mol}
\][/tex]
2. Calculate the Moles of LiCl:
- Use the mass of LiCl and its molar mass to find the number of moles:
[tex]\[
\text{Moles of LiCl} = \frac{\text{Mass of LiCl}}{\text{Molar Mass of LiCl}} = \frac{39.4 \, \text{g}}{42.39 \, \text{g/mol}} \approx 0.929 \, \text{moles}
\][/tex]
3. Calculate the Molarity of the Solution:
- Molarity is defined as the number of moles of solute per liter of solution:
[tex]\[
\text{Molarity} = \frac{\text{Moles of LiCl}}{\text{Volume of solution in Liters}} = \frac{0.929 \, \text{moles}}{1.59 \, \text{L}} \approx 0.585 \, \text{M}
\][/tex]
Therefore, the molarity of the solution is approximately 0.585 M (moles per liter).
1. Calculate the Molar Mass of LiCl:
- The molar mass of lithium (Li) is given as 6.94 g/mol.
- The molar mass of chlorine (Cl) is given as 35.45 g/mol.
- Therefore, the molar mass of lithium chloride (LiCl) is:
[tex]\[
\text{Molar Mass of LiCl} = 6.94 \, \text{g/mol} + 35.45 \, \text{g/mol} = 42.39 \, \text{g/mol}
\][/tex]
2. Calculate the Moles of LiCl:
- Use the mass of LiCl and its molar mass to find the number of moles:
[tex]\[
\text{Moles of LiCl} = \frac{\text{Mass of LiCl}}{\text{Molar Mass of LiCl}} = \frac{39.4 \, \text{g}}{42.39 \, \text{g/mol}} \approx 0.929 \, \text{moles}
\][/tex]
3. Calculate the Molarity of the Solution:
- Molarity is defined as the number of moles of solute per liter of solution:
[tex]\[
\text{Molarity} = \frac{\text{Moles of LiCl}}{\text{Volume of solution in Liters}} = \frac{0.929 \, \text{moles}}{1.59 \, \text{L}} \approx 0.585 \, \text{M}
\][/tex]
Therefore, the molarity of the solution is approximately 0.585 M (moles per liter).
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