Asked by Heather
Using calculations, determine how you would make a 525 ml pg a 0.50 M solution of Cl- ions using CaCl2.
I am unsure how to do this question, I was going to use C1V1=C2V2 but with this I would be missing the first part of the equation. If that is the case, would I assume the solution was originally 1 M CaCl2 and find the volume necessary to make 525 ml of a 0.50 M solution?
C1V1=C2V2
(1 M)(V1)=(0.50 M)(525 ml)
V1 = (0.50 M)(525 ml)/(1 M)
V1 = 262.5 ml of 1 M CaCl2.
I am unsure how to do this question, I was going to use C1V1=C2V2 but with this I would be missing the first part of the equation. If that is the case, would I assume the solution was originally 1 M CaCl2 and find the volume necessary to make 525 ml of a 0.50 M solution?
C1V1=C2V2
(1 M)(V1)=(0.50 M)(525 ml)
V1 = (0.50 M)(525 ml)/(1 M)
V1 = 262.5 ml of 1 M CaCl2.
Answers
Answered by
Scott
525 mL of .5 M solution contains
(.525 * .5) moles of solute
since you get 2 Cl-ions from each CaCl₂
you will need...(.525 * .5 * .5) moles of CaCl₂, dissolved in 525 mL of solution
molar mass of CaCl₂ is 111 g
(.5 * .5 * .525 * 111) g of CaCl₂
(.525 * .5) moles of solute
since you get 2 Cl-ions from each CaCl₂
you will need...(.525 * .5 * .5) moles of CaCl₂, dissolved in 525 mL of solution
molar mass of CaCl₂ is 111 g
(.5 * .5 * .525 * 111) g of CaCl₂
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.