To determine the volume of a 2.5 M lithium nitrate (LiNO₃) solution required to make 150 mL of a 1.0 M solution by dilution, we can use the dilution formula, which is given by:
[tex]\[ M_1 V_1 = M_2 V_2 \][/tex]
Where:
- [tex]\( M_1 \)[/tex] is the molarity of the concentrated solution.
- [tex]\( V_1 \)[/tex] is the volume of the concentrated solution that we need to find.
- [tex]\( M_2 \)[/tex] is the molarity of the dilute solution.
- [tex]\( V_2 \)[/tex] is the volume of the dilute solution.
Given:
- [tex]\( M_1 = 2.5 \, M \)[/tex] (molarity of the concentrated lithium nitrate solution)
- [tex]\( M_2 = 1.0 \, M \)[/tex] (molarity of the dilute lithium nitrate solution)
- [tex]\( V_2 = 150 \, \text{mL} \)[/tex] (volume of the dilute solution)
We need to find [tex]\( V_1 \)[/tex] (the volume of the concentrated solution).
Step-by-step solution:
1. Substitute the given values into the dilution formula:
[tex]\[ 2.5 \, M \times V_1 = 1.0 \, M \times 150 \, \text{mL} \][/tex]
2. Solve for [tex]\( V_1 \)[/tex]:
[tex]\[ V_1 = \frac{1.0 \, M \times 150 \, \text{mL}}{2.5 \, M} \][/tex]
3. Simplify the calculation:
[tex]\[ V_1 = \frac{150 \, \text{mL}}{2.5} \][/tex]
4. Perform the division:
[tex]\[ V_1 = 60 \, \text{mL} \][/tex]
Therefore, you would need 60 mL of a 2.5 M lithium nitrate solution to make 150 mL of a 1.0 M solution by dilution.
