Question

a student starts with a 6.0 m stock solution of nacl. 3.9 ml of this stock solution is removed and diluted to 50.0 ml, then 4.9 ml of this new solution is removed and diluted to 100.0 ml. what is the final concentration (m) of na (aq) after the two step serial dilution?

Answer 1

The final concentration (m) of Na⁺(aq) after the two-step serial dilution is 0.336 M.

Step-by-step explanation:

In the first step, 3.9 mL of the 6.0 M stock solution is removed and diluted to 50.0 mL. To calculate the new concentration (C₁) after the first dilution, we use the formula:

C₁V₁ = C₂V₂

Where C₁ is the initial concentration, V₁ is the initial volume, C₂ is the final concentration, and V₂ is the final volume.

6.0 M × 3.9 mL = C₂ × 50.0 mL

Solving for C₂ gives us the concentration after the first dilution.

In the second step, 4.9 mL of the solution from the first step is removed and diluted to 100.0 mL. Using the same dilution formula, we calculate the final concentration (C₃):

C₂V₂ = C₃V₃

`\[C₂ * 4.9 mL = C₃ * 100.0 mL\]`

Solving for C₃ gives us the final concentration after the second dilution.

To find the overall final concentration (C₄) after the two-step serial dilution, we can multiply the concentrations from the two steps:

`\[C₄ = C₁ * C₃\]`

Substitute the values of C₁ and C₃ into the equation to get the final concentration of Na⁺(aq).

Therefore, the final concentration (m) of Na⁺(aq) after the two-step serial dilution is 0.336 M.

Answer 2

The final concentration of NaCl
`(aq)` after the two-step serial dilution is
`\(1.17 * 10^(-2) \, \text{M}\)`.

Step-by-step explanation:

In the first step of the serial dilution process, 3.9 mL of a 6.0 M NaCl solution is withdrawn and diluted to 50.0 mL. The concentration of this intermediate solution is calculated using the formula
`\(C1V1 = C2V2\)`, where (C1) is the initial concentration, (V1) is the initial volume, (C2) is the final concentration, and (V2) is the final volume. This yields a concentration of
`\(0.468 \, \text{M}\)` for the intermediate solution.

Subsequently, in the second step, 4.9 mL of the intermediate solution is taken and further diluted to 100.0 mL. Applying the dilution formula again, the concentration of this second intermediate solution is found to be
`\(0.023 \, \text{M}\)`.

To determine the overall concentration after both dilutions, these two intermediate concentrations are multiplied
`(\(0.468 * 0.023\))`, resulting in a final concentration of
`\(1.17 * 10^(-2) \, \text{M}\)` for the NaCl in the original stock solution.