ATAR Notes: Forum

VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: Starlight on May 20, 2012, 04:51:52 pm

Title: How do you do this basic chemistry question? (to do with mol, mass etc.)
Post by: Starlight on May 20, 2012, 04:51:52 pm

I need to find the answer to the question: What mass of sodium chloride was in the original 100 mL solution (A)?


This is the information I have:

The addition of 100 mL of an aqueous sodium chloride solution (A) to 100 mL of an aqueous 0.100M solution of silver nitrate (B) produces a precipitate of mass 0.3886g. The concentration of sodium chloride in A is less than 0.100M


* Chemical reaction as described by NaCl + AgNO3 ---> AgCl + NaNo3

* Mass of chloride precipitated is 0.0962 g

Title: Re: How do you do this basic chemistry question? (to do with mol, mass etc.)
Post by: Hellrocks on May 20, 2012, 05:14:56 pm

Did you use the wrong molar mass in step 1?

Title: Re: How do you do this basic chemistry question? (to do with mol, mass etc.)
Post by: Hellrocks on May 20, 2012, 07:15:09 pm

But you used m(Cl-) rather than m(AgCl)

Title: Re: How do you do this basic chemistry question? (to do with mol, mass etc.)
Post by: Starlight on May 20, 2012, 08:12:25 pm

The answer is 0.159g btw.

Title: Re: How do you do this basic chemistry question? (to do with mol, mass etc.)
Post by: charmanderp on May 20, 2012, 08:21:45 pm

I give up.

In true idiot fashion I only read the first and last sentences of the question. The mass of the AgCl precipitate is 0.3886g rather than 0.0962g. So therefore the amount of AgCl, which is equal to the amount of Cl-, which is equal to the amount of NaCl, is 0.3886/143.32=0.002711415015mol.

So the mass of sodium chloride is 0.002711415015*(58.44)=0.1584550935g, which would get you to 0.159g if you rounded differently at various stages.

I'm so sorry :/ As Hellrocks pointed out, I should have used the molar mass of chlorine given that I took the mass of chloride precipitated to do my calculations. For some reason I didn't read the word 'chloride' there and just assumed it was the mass of AgCl precipitated.

Title: Re: How do you do this basic chemistry question? (to do with mol, mass etc.)
Post by: Starlight on May 20, 2012, 08:56:38 pm

thanks for trying :) , anyone else possibly know how to do this question?

Title: Re: How do you do this basic chemistry question? (to do with mol, mass etc.)
Post by: charmanderp on May 20, 2012, 09:06:48 pm

I'm not that bad! I did do the question correctly!

Title: Re: How do you do this basic chemistry question? (to do with mol, mass etc.)
Post by: charmanderp on May 20, 2012, 09:08:00 pm

Quote from: charmanderp on May 20, 2012, 08:21:45 pm

The mass of the AgCl precipitate is 0.3886g. So therefore the amount of AgCl, which is equal to the amount of Cl-, which is equal to the amount of NaCl (according to the balanced chemical equation), is 0.3886/143.32=0.002711415015mol.

So the mass of sodium chloride is 0.002711415015*(58.44)=0.1584550935g, which would get you to 0.159g if you rounded differently at various stages.

Title: Re: How do you do this basic chemistry question? (to do with mol, mass etc.)
Post by: Phy124 on May 20, 2012, 09:12:05 pm

Quote from: El2012 on May 20, 2012, 08:56:38 pm

thanks for trying :) , anyone else possibly know how to do this question?

charmanderp edited in the correct solution above.

edit: beaten

Actually, I'll take this opportunity to say something. charmanderp, you know there's an edit function, yeah? :P

Title: Re: How do you do this basic chemistry question? (to do with mol, mass etc.)
Post by: Starlight on May 20, 2012, 09:16:44 pm

Oh woops haha

Thanks, completely overlooked that!