Question

you want to determine the concentration when 93 ml of a 2.03 m kf solution is diluted by adding 3920 ml h₂o. what is m₂ (molarity of the dilute solution) for this dilution? the unknown 3920 ml 2.03 m

Explanation:

Step1: Recall the dilution formula

The dilution formula is \( M_1V_1 = M_2V_2 \), where \( M_1 \) is the initial molarity, \( V_1 \) is the initial volume, \( M_2 \) is the final molarity (unknown here), and \( V_2 \) is the final volume.

Step2: Determine the values of \( M_1 \), \( V_1 \), and \( V_2 \)

• \( M_1 = 2.03 \, \text{M} \)
• \( V_1 = 93 \, \text{mL} \)
• \( V_2 = V_1 + \text{volume of water added} = 93 \, \text{mL} + 3920 \, \text{mL} = 4013 \, \text{mL} \)

Step3: Rearrange the dilution formula to solve for \( M_2 \)

From \( M_1V_1 = M_2V_2 \), we can solve for \( M_2 \) as \( M_2=\frac{M_1V_1}{V_2} \)

Step4: Substitute the values into the formula

Substitute \( M_1 = 2.03 \, \text{M} \), \( V_1 = 93 \, \text{mL} \), and \( V_2 = 4013 \, \text{mL} \) into the formula:
\( M_2=\frac{2.03 \, \text{M} \times 93 \, \text{mL}}{4013 \, \text{mL}} \)
First, calculate the numerator: \( 2.03\times93 = 2.03\times(90 + 3)=2.03\times90+2.03\times3 = 182.7+6.09 = 188.79 \)
Then, divide by the denominator: \( M_2=\frac{188.79}{4013}\approx0.0470 \, \text{M} \)

Answer:

\( \approx 0.0470 \, \text{M} \)