Question

2) What is the concentration of NaCl in a solution prepared by diluting 56.98 ml of 0.5894 M stock
solution to a new volume of 1.20 L? (45 pts)

Answer 1

Answer:

The NaCl concentration will be 0.03 M.

Explanation:

Given data:

Initial volume = V₁ = 56.98 mL (56.98/1000 = 0.05698 L)

Initial concentration = M₁= 0.5894 M

Final volume = V₂= 1.20 L

Final concentration = M₂= ?

Solution:

By diluting the solution volume of solution will increase while number of moles of solute remain the same.

Formula:

Initial concentration × Initial volume  = Final concentration × Final volume

M₁V₁ = M₂V₂

M₂ = M₁V₁ / V₂

M₂ = 0.5894 M × 0.05698 L / 1.20 L

M₂ = 0.0336 M /1.20  

M₂ = 0.03 M