To determine the volume of a 2.5 M lithium nitrate (LiNO_3) solution needed to make 150 mL of a 1.0 M solution by dilution, we can use the dilution equation:
[tex]\[ M_1 \cdot V_1 = M_2 \cdot V_2 \][/tex]
where:
- [tex]\( M_1 \)[/tex] is the initial molarity (2.5 M),
- [tex]\( V_1 \)[/tex] is the initial volume (which we need to find),
- [tex]\( M_2 \)[/tex] is the final molarity (1.0 M),
- [tex]\( V_2 \)[/tex] is the final volume (150 mL).
We need to solve for [tex]\( V_1 \)[/tex]:
[tex]\[ V_1 = \frac{M_2 \cdot V_2}{M_1} \][/tex]
Let's substitute the given values into the equation:
[tex]\[ V_1 = \frac{1.0 \, \text{M} \times 150 \, \text{mL}}{2.5 \, \text{M}} \][/tex]
[tex]\[ V_1 = \frac{150 \, \text{mL}}{2.5} \][/tex]
[tex]\[ V_1 = 60 \, \text{mL} \][/tex]
So, to make 150 mL of a 1.0 M lithium nitrate solution, you would need 60 mL of a 2.5 M lithium nitrate solution.
