Certainly! Let’s calculate the molarity of a solution when 5.0 grams of potassium bromide (KBr) is dissolved to make a 0.500 liter solution. We'll go through the steps to determine this, rounding our final answer to two significant digits.
### Step-by-Step Solution:
1. Determine the molar mass of KBr:
- Potassium (K) has an atomic mass of approximately 39.1 g/mol.
- Bromine (Br) has an atomic mass of approximately 79.9 g/mol.
- Therefore, the molar mass of KBr is:
[tex]\[
\text{Molar Mass of KBr} = 39.1 \, \text{g/mol} + 79.9 \, \text{g/mol} = 119.0 \, \text{g/mol}
\][/tex]
2. Calculate the number of moles of KBr:
- We have 5.0 grams of KBr.
- Using the molar mass to find the moles:
[tex]\[
\text{Moles of KBr} = \frac{\text{mass of KBr}}{\text{molar mass of KBr}} = \frac{5.0 \, \text{g}}{119.0 \, \text{g/mol}} \approx 0.0420 \, \text{moles}
\][/tex]
3. Determine the volume of the solution in liters:
- The solution volume is already given as 0.500 liters.
4. Calculate the molarity (M):
- Molarity is defined as the number of moles of solute per liter of solution.
[tex]\[
\text{Molarity} (M) = \frac{\text{moles of solute}}{\text{liters of solution}} = \frac{0.0420 \, \text{moles}}{0.500 \, \text{L}} = 0.0840 \, \text{M}
\][/tex]
5. Round the molarity to two significant digits:
- The calculated molarity is approximately 0.0840 M.
- Rounding 0.0840 to two significant digits gives us:
[tex]\[
\text{Molarity (rounded)} = 0.08 \, \text{M}
\][/tex]
Therefore, the molarity of the solution when 5.0 grams of KBr is dissolved in 0.500 liters of solution is [tex]\(0.08 \, \text{M}\)[/tex].
