If 21 mL of 1.0 M NH4Cl is used what volume does it need to be diluted to in order to make a 0.12 M solution? (give your answer in mL to zero decimal places)

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### How to Calculate Dilution of a Solution

**Question:**
If 21 mL of 1.0 M NH₄Cl is used, what volume does it need to be diluted to in order to make a 0.12 M solution? (Give your answer in mL to zero decimal places)

**Solution:**

To solve this problem, we will use the concept of dilution, which can be calculated using the formula:
\[ C_1 \times V_1 = C_2 \times V_2 \]

Where:
- \( C_1 \) is the concentration of the initial solution (1.0 M)
- \( V_1 \) is the volume of the initial solution (21 mL)
- \( C_2 \) is the desired concentration of the final solution (0.12 M)
- \( V_2 \) is the volume of the final solution we need to find

Rearranging the equation to solve for \( V_2 \):
\[ V_2 = \frac{C_1 \times V_1}{C_2} \]

Now plug in the values:
\[ V_2 = \frac{1.0 \text{ M} \times 21 \text{ mL}}{0.12 \text{ M}} \]

\[ V_2 = \frac{21}{0.12} \]

\[ V_2 = 175 \text{ mL} \]

Therefore, you need to dilute the 21 mL of 1.0 M NH₄Cl to a final volume of **175 mL** to obtain a 0.12 M solution.
Transcribed Image Text:### How to Calculate Dilution of a Solution **Question:** If 21 mL of 1.0 M NH₄Cl is used, what volume does it need to be diluted to in order to make a 0.12 M solution? (Give your answer in mL to zero decimal places) **Solution:** To solve this problem, we will use the concept of dilution, which can be calculated using the formula: \[ C_1 \times V_1 = C_2 \times V_2 \] Where: - \( C_1 \) is the concentration of the initial solution (1.0 M) - \( V_1 \) is the volume of the initial solution (21 mL) - \( C_2 \) is the desired concentration of the final solution (0.12 M) - \( V_2 \) is the volume of the final solution we need to find Rearranging the equation to solve for \( V_2 \): \[ V_2 = \frac{C_1 \times V_1}{C_2} \] Now plug in the values: \[ V_2 = \frac{1.0 \text{ M} \times 21 \text{ mL}}{0.12 \text{ M}} \] \[ V_2 = \frac{21}{0.12} \] \[ V_2 = 175 \text{ mL} \] Therefore, you need to dilute the 21 mL of 1.0 M NH₄Cl to a final volume of **175 mL** to obtain a 0.12 M solution.
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