To prepare a [tex]$2.00 \text{ L}$[/tex] solution of [tex]$0.100 \text{ M} \text{ NH}_4\text{NO}_3$[/tex] from a [tex]$1.75 \text{ M} \text{ NH}_4\text{NO}_3$[/tex] stock solution, you can use the dilution formula [tex]\( M_1 V_1 = M_2 V_2 \)[/tex], where:
- [tex]\( M_1 \)[/tex] is the molarity of the stock solution,
- [tex]\( V_1 \)[/tex] is the volume of the stock solution needed,
- [tex]\( M_2 \)[/tex] is the molarity of the final solution,
- [tex]\( V_2 \)[/tex] is the volume of the final solution.
Given:
- [tex]\( M_1 = 1.75 \text{ M} \)[/tex],
- [tex]\( M_2 = 0.100 \text{ M} \)[/tex],
- [tex]\( V_2 = 2.00 \text{ L} \)[/tex].
We need to find [tex]\( V_1 \)[/tex].
1. Using the formula [tex]\( M_1 V_1 = M_2 V_2 \)[/tex], we can rearrange to solve for [tex]\( V_1 \)[/tex]:
[tex]\[
V_1 = \frac{M_2 \cdot V_2}{M_1}
\][/tex]
2. Substitute the given values into the equation:
[tex]\[
V_1 = \frac{0.100 \text{ M} \times 2.00 \text{ L}}{1.75 \text{ M}}
\][/tex]
3. Calculate [tex]\( V_1 \)[/tex]:
[tex]\[
V_1 = \frac{0.200}{1.75} \approx 0.1142857142857143 \text{ L}
\][/tex]
4. Convert [tex]\( V_1 \)[/tex] from liters to milliliters (since [tex]\( 1 \text{ L} = 1000 \text{ mL} \)[/tex]):
[tex]\[
V_1 \approx 0.1142857142857143 \text{ L} \times 1000 \text{ mL/L} \approx 114.28571428571429 \text{ mL}
\][/tex]
To achieve the desired solution:
- Measure approximately [tex]\( 114 \text{ mL} \)[/tex] (specifically [tex]\( 114.28571428571429 \text{ mL} \)[/tex]) of the [tex]$1.75 \text{ M} \text{ NH}_4\text{NO}_3$[/tex] stock solution.
- Dilute this [tex]\( 114 \text{ mL} \)[/tex] of stock solution to a final volume of [tex]\( 2.00 \text{ L} \)[/tex].
Therefore, the correct procedure is:
Measure [tex]$114 \text{ mL}$[/tex] of the [tex]$1.75 \text{ M}$[/tex] solution, and dilute it to [tex]$2.00 \text{ L}$[/tex].
This ensures the final solution has the desired concentration of [tex]$0.100 \text{ M}$[/tex].
