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295 mL of a 1.00 M ammonium chloride solution is diluted with [tex]$2,250 \, \text{mL}$[/tex] of water.

What is the new concentration in molarity?

[tex]\text{[?]} \, M \, \text{NH}_4\text{Cl}[/tex]

Hint: Notice the problem says "diluted with" rather than "diluted to."



Answer :

GinnyAnswer
To solve the problem of determining the new concentration of a 1.00 M ammonium chloride solution after dilution, follow these steps:

1. Identify the initial conditions:
- Initial volume (V_initial) of the ammonium chloride solution: 295 mL
- Initial molarity (M_initial) of the ammonium chloride solution: 1.00 M

2. Determine the volume of water added:
- Volume of water added (V_added): 2250 mL

3. Calculate the total final volume after dilution:
- The total final volume (V_final) is the sum of the initial volume of the solution and the volume of water added:
[tex]\[ V_{\text{final}} = V_{\text{initial}} + V_{\text{added}} \][/tex]
[tex]\[ V_{\text{final}} = 295\ \text{mL} + 2250\ \text{mL} \][/tex]
[tex]\[ V_{\text{final}} = 2545\ \text{mL} \][/tex]

4. Use the dilution formula to find the new molarity:
- The dilution formula is expressed as:
[tex]\[ M_{\text{initial}} \times V_{\text{initial}} = M_{\text{final}} \times V_{\text{final}} \][/tex]
- Rearranging to solve for the final molarity (M_final):
[tex]\[ M_{\text{final}} = \frac{M_{\text{initial}} \times V_{\text{initial}}}{V_{\text{final}}} \][/tex]

5. Substitute the known values into the equation:
[tex]\[ M_{\text{final}} = \frac{1.00\ \text{M} \times 295\ \text{mL}}{2545\ \text{mL}} \][/tex]

6. Calculate the final molarity:
[tex]\[ M_{\text{final}} \approx 0.11591355599214145\ \text{M} \][/tex]

Therefore, the new concentration of the ammonium chloride solution after dilution is approximately [tex]\( 0.116 \)[/tex] M.

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