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Sagot :
Certainly! To determine the final volume when diluting a solution, we can apply the concept of conservation of moles. This is based on the principle that the number of moles of solute before dilution will be equal to the number of moles of solute after dilution. We use the dilution formula:
[tex]\[ C_1 V_1 = C_2 V_2 \][/tex]
where:
- [tex]\( C_1 \)[/tex] is the initial concentration (446 mM)
- [tex]\( V_1 \)[/tex] is the initial volume (43.7 mL)
- [tex]\( C_2 \)[/tex] is the final concentration (372 mM)
- [tex]\( V_2 \)[/tex] is the final volume we need to find
To find the final volume, we rearrange the formula to solve for [tex]\( V_2 \)[/tex]:
[tex]\[ V_2 = \frac{C_1 \cdot V_1}{C_2} \][/tex]
Substituting the values given:
- [tex]\( C_1 = 446 \, \text{mM} \)[/tex]
- [tex]\( V_1 = 43.7 \, \text{mL} \)[/tex]
- [tex]\( C_2 = 372 \, \text{mM} \)[/tex]
So,
[tex]\[ V_2 = \frac{446 \, \text{mM} \times 43.7 \, \text{mL}}{372 \, \text{mM}} \][/tex]
Simplifying this,
[tex]\[ V_2 = \frac{19492.2 \, \text{mL} \cdot \text{mM}}{372 \, \text{mM}} \][/tex]
[tex]\[ V_2 \approx 52.39301075268818 \, \text{mL} \][/tex]
To follow the instruction to round the answer to 3 significant digits:
[tex]\[ V_2 \approx 52.4 \, \text{mL} \][/tex]
Therefore, the final volume is approximately:
[tex]\[ \boxed{52.4} \, \text{mL} \][/tex]
[tex]\[ C_1 V_1 = C_2 V_2 \][/tex]
where:
- [tex]\( C_1 \)[/tex] is the initial concentration (446 mM)
- [tex]\( V_1 \)[/tex] is the initial volume (43.7 mL)
- [tex]\( C_2 \)[/tex] is the final concentration (372 mM)
- [tex]\( V_2 \)[/tex] is the final volume we need to find
To find the final volume, we rearrange the formula to solve for [tex]\( V_2 \)[/tex]:
[tex]\[ V_2 = \frac{C_1 \cdot V_1}{C_2} \][/tex]
Substituting the values given:
- [tex]\( C_1 = 446 \, \text{mM} \)[/tex]
- [tex]\( V_1 = 43.7 \, \text{mL} \)[/tex]
- [tex]\( C_2 = 372 \, \text{mM} \)[/tex]
So,
[tex]\[ V_2 = \frac{446 \, \text{mM} \times 43.7 \, \text{mL}}{372 \, \text{mM}} \][/tex]
Simplifying this,
[tex]\[ V_2 = \frac{19492.2 \, \text{mL} \cdot \text{mM}}{372 \, \text{mM}} \][/tex]
[tex]\[ V_2 \approx 52.39301075268818 \, \text{mL} \][/tex]
To follow the instruction to round the answer to 3 significant digits:
[tex]\[ V_2 \approx 52.4 \, \text{mL} \][/tex]
Therefore, the final volume is approximately:
[tex]\[ \boxed{52.4} \, \text{mL} \][/tex]
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