Question

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?

Answer 1

You need to know moles, you have mass.

\[\text{moles} = \frac{\text{mass}}{\text{molar mass}}\]

Answer 2

@whpalmer , do i find the molar mass of Cl and divide it by 0.221 g? What about the 50 mL of water?

Answer 3

@whpalmer so 0.221g/35.35g/mol?

Answer 4

yes, that gives you the number of moles. what is the result?

Answer 5

I got 0.00631 moles

Answer 6

\[0.221/35.35 \approx 0.00625177\]not sure how you got \(0.00631\) from that

Answer 7

but you are in the right ballpark

Answer 8

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?

Answer 9

do i use the moles and divide it by the volume given?

Answer 10

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

Answer 11

So 0.00623/0.05L?

Answer 12

yep.

Answer 13

How about # 2? will i take the mass of the sample and divide it to the molar mass of Cl?

Answer 14

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?

Answer 15

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

Answer 16

Good, it was starting to look like work for me :-)

I have to go, good luck with the rest of it!

Answer 17

Thank you!