vheah
  • vheah
1.) A sample containing 0.221 g Cl- is dissolved in 50.0 mL water a) How many moles of Cl- ion are in the solution? b) What is the molarity of the Cl- ion in the solution? 2.) A solid chloride sample weighing 0.3147 g required 43.75 mL of 0.05273 M AgNO3 to reach the AgCrO4 end point. A) how many moles Cl- ion were present in the sample? B) how many grams Cl- ion were present? c) what was the mass percent cl- ion in the sample?
Chemistry
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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whpalmer4
  • whpalmer4
You need to know moles, you have mass. \[\text{moles} = \frac{\text{mass}}{\text{molar mass}}\]
vheah
  • vheah
@whpalmer , do i find the molar mass of Cl and divide it by 0.221 g? What about the 50 mL of water?
vheah
  • vheah
@whpalmer so 0.221g/35.35g/mol?

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whpalmer4
  • whpalmer4
yes, that gives you the number of moles. what is the result?
vheah
  • vheah
I got 0.00631 moles
whpalmer4
  • whpalmer4
\[0.221/35.35 \approx 0.00625177\]not sure how you got \(0.00631\) from that
whpalmer4
  • whpalmer4
but you are in the right ballpark
vheah
  • vheah
I only used 35, my bad. But I apologize, the molar mass for cl is 35.45 not 35.35. But how do i find the molarity of the Cl ion?
vheah
  • vheah
do i use the moles and divide it by the volume given?
whpalmer4
  • whpalmer4
well, 35.453 g/mol if you want to get picky :-) Okay, molarity is moles/liter, right? divide the number of moles by volume of solute
vheah
  • vheah
So 0.00623/0.05L?
whpalmer4
  • whpalmer4
yep.
vheah
  • vheah
How about # 2? will i take the mass of the sample and divide it to the molar mass of Cl?
whpalmer4
  • whpalmer4
2.) A solid chloride sample weighing 0.3147 g required 43.75 mL of 0.05273 M AgNO3 to reach the AgCrO4 end point. A) how many moles Cl- ion were present in the sample? B) how many grams Cl- ion were present? c) what was the mass percent cl- ion in the sample? For A, you need to find the number of moles of AgNO3, I think. But I am confused by AgCrO4 — should be perhaps be AgClO4?
vheah
  • vheah
I took this from a lab, i didn't read hard enough. I just realized that they gave me the equations! Thanks a lot for the help! xD
whpalmer4
  • whpalmer4
Good, it was starting to look like work for me :-) I have to go, good luck with the rest of it!
vheah
  • vheah
Thank you!

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