Question

dilution

1. how much concentrated 18m sulfuric acid is needed to prepare 250ml of a 6.0m solution?

1. how much concentrated 12m hydrochloric acid is needed to prepare 100ml of a 2.0m solution?

1. to what volume should 25 ml of 15m nitric acid be diluted to prepare a 3.0m solution?

1. to how much water should 50ml of 12m hydrochloric acid be added to produce a 4.0m solution?

1. to how much water should 100 ml of 18m sulfuric acid be added to prepare a 1.5m solution?

Explanation:

Question 1

Step1: Recall 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, and \( V_2 \) is the final volume. We need to find \( V_1 \).

Step2: Identify known values

\( M_1 = 18 \, M \), \( M_2 = 6.0 \, M \), \( V_2 = 250 \, mL \). Rearrange the formula to solve for \( V_1 \): \( V_1=\frac{M_2V_2}{M_1} \)

Step3: Substitute values

\( V_1=\frac{6.0 \, M\times250 \, mL}{18 \, M} \)
\( V_1=\frac{1500}{18} \, mL \approx 83.3 \, mL \)

Step1: Use dilution formula \( M_1V_1 = M_2V_2 \)

We need to find \( V_1 \).

Step2: Identify values

\( M_1 = 12 \, M \), \( M_2 = 2.0 \, M \), \( V_2 = 100 \, mL \). Formula: \( V_1=\frac{M_2V_2}{M_1} \)

Step3: Substitute

\( V_1=\frac{2.0 \, M\times100 \, mL}{12 \, M} \)
\( V_1=\frac{200}{12} \, mL \approx 16.7 \, mL \)

Step1: Apply dilution formula \( M_1V_1 = M_2V_2 \)

We need to find \( V_2 \).

Step2: Identify values

\( M_1 = 15 \, M \), \( V_1 = 25 \, mL \), \( M_2 = 3.0 \, M \). Rearrange: \( V_2=\frac{M_1V_1}{M_2} \)

Step3: Substitute

\( V_2=\frac{15 \, M\times25 \, mL}{3.0 \, M} \)
\( V_2 = 125 \, mL \)

Answer:

\( 83.3 \, mL \)

Question 2