To solve this problem, we will use the concept of dilution, which is based on the principle of conservation of moles. In a dilution, the number of moles of solute remains the same before and after the dilution process, whereas the volume changes. The relationship can be described using the dilution formula:
[tex]\[ M_1 \times V_1 = M_2 \times V_2 \][/tex]
Where:
- [tex]\( M_1 \)[/tex] is the molarity of the concentrated solution.
- [tex]\( V_1 \)[/tex] is the volume of the concentrated solution needed.
- [tex]\( M_2 \)[/tex] is the molarity of the dilute solution.
- [tex]\( V_2 \)[/tex] is the volume of the dilute solution.
Given:
- [tex]\( M_1 = 2.5 \)[/tex] M (molarity of the concentrated lithium nitrate solution)
- [tex]\( V_1 = ? \)[/tex] (volume of the concentrated lithium nitrate solution we need to find)
- [tex]\( M_2 = 1.0 \)[/tex] M (molarity of the dilute lithium nitrate solution)
- [tex]\( V_2 = 150 \)[/tex] mL (volume of the dilute lithium nitrate solution)
We need to determine [tex]\( V_1 \)[/tex].
### Step-by-Step Solution
1. Write the dilution formula:
[tex]\[ M_1 \times V_1 = M_2 \times V_2 \][/tex]
2. Substitute the given values into the formula:
[tex]\[ 2.5 \times V_1 = 1.0 \times 150 \][/tex]
3. Solve for [tex]\( V_1 \)[/tex]:
[tex]\[ V_1 = \frac{1.0 \times 150}{2.5} \][/tex]
4. Calculate the value:
[tex]\[ V_1 = \frac{150}{2.5} \][/tex]
[tex]\[ V_1 = 60 \text{ mL} \][/tex]
### Conclusion
The volume of the 2.5 M lithium nitrate solution needed to make 150 mL of a 1.0 M solution is 60 mL.
