Question

What is the volume, in milliliters, of each of the following diluted solutions? a. A \(1.5 \mathrm{M} \mathrm{HCl}\) solution prepared from \(20.0 \mathrm{~mL}\) of a \(6.0 \mathrm{M}\) \(\mathrm{HCl}\) solution b. A \(2.0 \%(\mathrm{~m} / \mathrm{v}) \mathrm{LiCl}\) solution prepared from \(50.0 \mathrm{~mL}\) of a \(10.0 \%(\mathrm{~m} / \mathrm{v}) \mathrm{LiCl}\) solution c. A \(0.500 \mathrm{M} \mathrm{H}_{3} \mathrm{PO}_{4}\) solution prepared from \(50.0 \mathrm{~mL}\) of a \(6.00 \mathrm{M} \mathrm{H}_{3} \mathrm{PO}_{4}\) solution d. A \(5.0 \%(\mathrm{~m} / \mathrm{v})\) glucose solution prepared from \(75 \mathrm{~mL}\) of a \(12 \%(\mathrm{~m} / \mathrm{v})\) glucose solution

Short answer

a) 80 mL, b) 250 mL, c) 600 mL, d) 180 mL

Step-by-step solution

- Understanding the dilution formula

The dilution formula is given by: \[ C_1 V_1 = C_2 V_2 \] Here, - \( C_1 \) and \( V_1 \) are the concentration and volume of the initial solution, - \( C_2 \) and \( V_2 \) are the concentration and volume of the diluted solution.

- Solve part (a)

For part (a), we have: \( C_1 = 6.0 \, M \) \( V_1 = 20.0 \, mL \) \( C_2 = 1.5 \, M \) Using the dilution formula, \[ 6.0 \times 20.0 = 1.5 \times V_2 \] \[ V_2 = \frac{6.0 \times 20.0}{1.5} \] \[ V_2 = 80 \, mL \]

- Solve part (b)

For part (b), we have: \( C_1 = 10.0 \,\% (m/v) \) \( V_1 = 50.0 \, mL \) \( C_2 = 2.0 \,\% (m/v) \) Using the dilution formula, \[ 10.0 \times 50.0 = 2.0 \times V_2 \] \[ V_2 = \frac{10.0 \times 50.0}{2.0} \] \[ V_2 = 250 \, mL \]

- Solve part (c)

For part (c), we have: \( C_1 = 6.00 \, M \) \( V_1 = 50.0 \, mL \) \( C_2 = 0.500 \, M \) Using the dilution formula, \[ 6.00 \times 50.0 = 0.500 \times V_2 \] \[ V_2 = \frac{6.00 \times 50.0}{0.500} \] \[ V_2 = 600 \, mL \]

- Solve part (d)

For part (d), we have: \( C_1 = 12.0 \,\% (m/v) \) \( V_1 = 75.0 \, mL \) \( C_2 = 5.0 \,\% (m/v) \) Using the dilution formula, \[ 12.0 \times 75.0 = 5.0 \times V_2 \] \[ V_2 = \frac{12.0 \times 75.0}{5.0} \] \[ V_2 = 180 \, mL \]

Key concepts

molarity

Molarity, symbolized as \(\text{M}\), refers to the number of moles of solute per liter of solution. It is a vital concept in chemistry for expressing the concentration of a solution. To find molarity, use the formula: \[ \text{Molarity} (M) = \frac{\text{moles of solute}}{\text{liters of solution}} \] For example, in exercise part (a), we use molarity to find how much hydrochloric acid (\(\text{HCl}\)) is present in the solution. By understanding molarity, you can better grasp how the concentration changes when a solution is diluted.

percentage concentration

Percentage concentration indicates how much solute is present in a solution relative to the total solution volume or mass. It can be of various types, but for our purposes, mass/volume percentage (\(\text{m/v}\)) is often used. This is calculated as: \[ \text{Percentage concentration} (\text{m/v}) = \frac{\text{mass of solute}}{\text{volume of solution}} \times 100\] Consider part (b) of the exercise, where a 10\(\text{m/v}\) \(\text{LiCl}\) solution is diluted to 2\(\text{m/v}\). Knowing how to interpret and calculate percentage concentration can help you determine the amount of solute in diluted solutions and understand its significance in lab contexts.

solution volume

Solution volume refers to the total amount of space that the solution occupies. When diluting a solution, the volume changes but the amount of solute remains the same. This concept is crucial for understanding how concentrations change during dilution. Using the dilution formula \[ C_1 V_1 = C_2 V_2 \] we can solve for the final volume (\