Sure, let's walk through the solution step by step.
1. Convert the mass of LiBr from milligrams to grams:
- Given mass: 12.92 mg
- Conversion factor: 1 g = 1000 mg
So,
[tex]\[
\text{mass in grams} = \frac{12.92 \, \text{mg}}{1000} = 0.01292 \, \text{g}
\][/tex]
2. Convert the volume of the solution from milliliters to liters:
- Given volume: 194.1 mL
- Conversion factor: 1 L = 1000 mL
So,
[tex]\[
\text{volume in liters} = \frac{194.1 \, \text{mL}}{1000} = 0.1941 \, \text{L}
\][/tex]
3. Calculate the molar mass of LiBr:
- Li (Lithium) has a molar mass of approximately 6.94 g/mol
- Br (Bromine) has a molar mass of approximately 79.90 g/mol
So,
[tex]\[
\text{molar mass of LiBr} = 6.94 + 79.90 = 86.84 \, \text{g/mol}
\][/tex]
4. Calculate the number of moles of LiBr:
Using the formula:
[tex]\[
\text{moles of LiBr} = \frac{\text{mass of LiBr (g)}}{\text{molar mass of LiBr (g/mol)}}
\][/tex]
Substituting the values:
[tex]\[
\text{moles of LiBr} = \frac{0.01292 \, \text{g}}{86.84 \, \text{g/mol}} = 0.00014877936434822662 \, \text{mol}
\][/tex]
5. Calculate the molarity of the solution:
Molarity is defined as the number of moles of solute per liter of solution. Using the formula:
[tex]\[
\text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution (L)}}
\][/tex]
Substituting the values:
[tex]\[
\text{Molarity} = \frac{0.00014877936434822662 \, \text{mol}}{0.1941 \, \text{L}} = 0.0007665088322938002 \, \text{M}
\][/tex]
Therefore, the molarity of the solution is [tex]\(0.0007665 \, \text{M}\)[/tex] when expressed to four significant figures.
