vannitling12p4w44f
28.10.2019 •
Chemistry
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)
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Ответ:
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
Ответ: