What is the correct procedure to prepare 100.0 ml of 3.00 M HCl solution from an 18.0 M HCl stock solution?

What is the correct procedure to prepare 100.0 ml of 3.00 M HCl solution from an 18.0 M HCl stock solution?

• A. Dilute 16.7 ml of 18.0 M HCl to a total volume of 100.0 ml.
• B. Mix 5.56 ml of 18.0 M HCl with 94.44 ml of water.
• C. Concentrate 100 ml of 3.00 M solution to make 18.0 M.
• D. Use 30 ml of 18.0 M HCl and dilute it to 100 ml.

Answer: (A)

To prepare a diluted solution from a concentrated stock solution, the solution's molarity and volume are important. The process involves the dilution equation, which states that the product of the concentration and volume of the concentrated solution is equal to the product of the concentration and volume of the diluted solution. This can be represented with the formula: \[ C_1V_1 = C_2V_2 \] Where: - \( C_1 \) is the concentration of the stock solution (18.0 M), - \( V_1 \) is the volume of the stock solution needed, - \( C_2 \) is the concentration of the diluted solution (3.00 M), - \( V_2 \) is the volume of the diluted solution (100.0 ml or 0.1 L). Substituting the values into the equation gives: \[ (18.0 \, M)(V_1) = (3.00 \, M)(100.0 \, ml) \] Solving for \( V_1 \): \[ V_1 = \frac{(3.00 \, M)(100.0 \, ml)}{18.0 \, M} = \frac

To prepare a diluted solution from a concentrated stock solution, the solution's molarity and volume are important. The process involves the dilution equation, which states that the product of the concentration and volume of the concentrated solution is equal to the product of the concentration and volume of the diluted solution. This can be represented with the formula:

[ C_1V_1 = C_2V_2 ]

Where:

• ( C_1 ) is the concentration of the stock solution (18.0 M),

• ( V_1 ) is the volume of the stock solution needed,

• ( C_2 ) is the concentration of the diluted solution (3.00 M),

• ( V_2 ) is the volume of the diluted solution (100.0 ml or 0.1 L).

Substituting the values into the equation gives:

[ (18.0 , M)(V_1) = (3.00 , M)(100.0 , ml) ]

Solving for ( V_1 ):

[ V_1 = \frac{(3.00 , M)(100.0 , ml)}{18.0 , M} = \frac