Question

what is the final concentration of 925 ml of a solution made from 222 ml of 4.5 m hcl?

Explanation:

Step1: Recall dilution formula

The dilution formula is \( C_1V_1 = C_2V_2 \), where \( C_1 \) is the initial concentration, \( V_1 \) is the initial volume, \( C_2 \) is the final concentration, and \( V_2 \) is the final volume.

Step2: Identify known values

We know \( C_1 = 4.5 \, M \), \( V_1 = 222 \, mL \), and \( V_2 = 925 \, mL \). We need to find \( C_2 \).

Step3: Rearrange formula for \( C_2 \)

From \( C_1V_1 = C_2V_2 \), we can rearrange to \( C_2=\frac{C_1V_1}{V_2} \).

Step4: Substitute values into formula

Substitute \( C_1 = 4.5 \, M \), \( V_1 = 222 \, mL \), and \( V_2 = 925 \, mL \) into the formula:
\( C_2=\frac{4.5 \, M\times222 \, mL}{925 \, mL} \)
First, calculate the numerator: \( 4.5\times222 = 999 \)
Then, divide by the denominator: \( \frac{999}{925}\approx1.08 \, M \)

Answer:

\( 1.08 \)