- Identify molarity as \(2.00 \text{ mol/L}\) and volume as \(0.25 \text{ L}\).
- Use the formula \(\text{moles} = \text{molarity} \times \text{volume}\) to set up the calculation.
- Multiply the values: \(2.00 \text{ mol/L} \times 0.25 \text{ L} = 0.50 \text{ mol}\).
- The final answer is \(\boxed{0.50 \text{ mol}}\), corresponding to option a.
Explanation
- Analysis of the Problem
We are given a solution of barium nitrate \(\text{Ba(NO}_3)_2\) with a concentration (molarity) of \(2.00 \text{ M}\) and a volume of \(0.25 \text{ L}\). We need to determine the total number of moles of the solute, \(\text{Ba(NO}_3)_2\), present in this specific volume of the solution. The definition of molarity (\(M\)) is provided as the number of moles of solute divided by the volume of the solution in liters:
\[\text{Molarity} = \frac{\text{moles of solute}}{\text{liters of solution}}\]
- Setting Up the Calculation
To isolate the number of moles of the solute, we rearrange the given formula:
\[\text{moles of solute} = \text{molarity} \times \text{liters of solution}\]
Now, we substitute the given values:
- Molarity = \(2.00 \text{ M} = 2.00 \text{ mol/L}\)
- Volume = \(0.25 \text{ L}\)
Substituting these into our rearranged equation, we get:
\[\text{moles of solute} = 2.00 \text{ mol/L} \times 0.25 \text{ L}\]
- Final Calculation and Conclusion
Performing the multiplication:
\[\text{moles of solute} = 2.00 \times 0.25 = 0.50 \text{ mol}\]
Comparing this result with the given options:
A\(0.50 \text{ mol}\)
B\(2.25 \text{ mol}\)
C\(0.13 \text{ mol}\)
D\(8.0 \text{ mol}\)
The calculated result corresponds to option a. Therefore, there are \(0.50 \text{ moles}\) of \(\text{Ba(NO}_3)_2\) in the solution.
Examples
In a chemistry laboratory, you often need to prepare precise concentrations of solutions for experiments. For instance, if you are performing a titration and need exactly 0.50 moles of a reactant for a reaction to reach its equivalence point, you would calculate the required volume of a 2.00 M solution using this exact method. Understanding molarity is essential for scaling chemical recipes and ensuring that reactions proceed as intended without wasting expensive or hazardous chemicals.